Java 如何使用JSP/Servlet将文件上载到服务器?

Java 如何使用JSP/Servlet将文件上载到服务器?,java,jsp,jakarta-ee,servlets,file-upload,Java,Jsp,Jakarta Ee,Servlets,File Upload,如何使用JSP/Servlet将文件上载到服务器?我试过这个: FileItemFactory factory = new DiskFileItemFactory(); ServletFileUpload upload = new ServletFileUpload(factory); List items = upload.parseRequest(request); // This line is where it died. 但是,我只获取文件名,而不获取文件内容。将enctype=mu

如何使用JSP/Servlet将文件上载到服务器?我试过这个:

FileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload upload = new ServletFileUpload(factory);
List items = upload.parseRequest(request); // This line is where it died.
但是,我只获取文件名,而不获取文件内容。将enctype=multipart/form数据添加到时,request.getParameter返回null

在研究过程中,我偶然发现。我试过这个:

FileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload upload = new ServletFileUpload(factory);
List items = upload.parseRequest(request); // This line is where it died.
不幸的是,servlet在没有明确消息和原因的情况下引发了异常。以下是stacktrace:

SEVERE: Servlet.service() for servlet UploadServlet threw exception
javax.servlet.ServletException: Servlet execution threw an exception
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:313)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
    at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233)
    at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
    at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
    at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
    at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
    at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:298)
    at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:852)
    at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:588)
    at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:489)
    at java.lang.Thread.run(Thread.java:637)
介绍 要浏览并选择要上载的文件,您需要表单中的HTML字段。如中所述,您必须使用POST方法,并且表单的enctype属性必须设置为multipart/form data

提交此类表单后,在未设置enctype的情况下,二进制多部分表单数据在请求正文中可用

在Servlet3.0之前,ServletAPI本机不支持多部分/表单数据。它只支持应用程序的默认表单类型/x-www-form-urlencoded。使用多部分表单数据时,request.getParameter和consors都将返回null。这就是著名人物出现的地方

不要手动解析它! 从理论上讲,您可以自己分析请求主体。然而,这是一项精确而乏味的工作,需要精确的知识。你不应该试图自己做这件事,也不应该复制粘贴一些在互联网上其他地方找到的本地无库代码。许多在线资源在这方面都失败了,比如roseindia.net。另见。您应该使用一个真正的库,它被使用并隐式测试!多年来,数以百万计的用户。这样一个库已经证明了它的健壮性

当您已经使用Servlet 3.0或更新版本时,请使用本机API 如果您至少在使用Servlet 3.0 Tomcat 7、Jetty 9、JBoss AS 6、GlassFish 3等,那么您可以使用提供的标准API来收集单个多部分表单数据项,大多数Servlet 3.0实现实际上都在为此目的使用Apache Commons FileUpload!。此外,getParameter还可以通过通常的方式使用标准格式字段

首先用注释您的servlet,以便它识别和支持多部分/表单数据请求,从而使getPart工作:

然后,按如下方式实现其doPost:

@RequestMapping(value = "/upload", method = RequestMethod.POST)
public void handleUpload(@RequestParam("file") MultipartFile file) throws IOException {
    if (!file.isEmpty()) {
            byte[] bytes = file.getBytes(); // alternatively, file.getInputStream();
            // application logic
    }
}
受保护的无效doPostHttpServletRequest请求、HttpServletResponse响应引发ServletException、IOException{ String description=request.getParameterdescription;//检索 Part filePart=request.getPartfile;//检索 字符串文件名=path.getfilePart.getSubmittedFileName.getFileName.toString;//MSIE修复。 InputStream fileContent=filePart.getInputStream; //…在这里做你的工作 } 注意PathgetFileName。这是一个关于获取文件名的MSIE修复程序。此浏览器不正确地沿名称而不是仅沿文件名发送完整文件路径

private static String getSubmittedFileName(Part part) {
    for (String cd : part.getHeader("content-disposition").split(";")) {
        if (cd.trim().startsWith("filename")) {
            String fileName = cd.substring(cd.indexOf('=') + 1).trim().replace("\"", "");
            return fileName.substring(fileName.lastIndexOf('/') + 1).substring(fileName.lastIndexOf('\\') + 1); // MSIE fix.
        }
    }
    return null;
}
如果您有一个用于多文件上载的方法,请按如下方式收集它们。遗憾的是,没有request.getPartsfile这样的方法:

请注意MSIE关于获取文件名的修复。此浏览器不正确地沿名称而不是仅沿文件名发送完整文件路径

private static String getSubmittedFileName(Part part) {
    for (String cd : part.getHeader("content-disposition").split(";")) {
        if (cd.trim().startsWith("filename")) {
            String fileName = cd.substring(cd.indexOf('=') + 1).trim().replace("\"", "");
            return fileName.substring(fileName.lastIndexOf('/') + 1).substring(fileName.lastIndexOf('\\') + 1); // MSIE fix.
        }
    }
    return null;
}
当您还没有使用Servlet3.0时,请使用ApacheCommonsFileUpload 如果您还没有使用Servlet 3.0,那么是时候升级了吗?通常的做法是使用解析多部分表单数据请求。它有一个很好的方法,仔细地处理这两个问题。还有O'Reilly MultipartRequest,但它有一些小错误,多年来不再积极维护。我不建议使用它。Apache Commons FileUpload仍在积极维护,目前非常成熟

为了使用Apache Commons FileUpload,您的webapp的/WEB-INF/lib中至少需要有以下文件:

您最初的尝试很可能失败,因为您忘记了commons IO

下面是使用Apache Commons FileUpload时UploadServlet的doPost的启动示例:

受保护的无效doPostHttpServletRequest请求、HttpServletResponse响应引发ServletException、IOException{ 试一试{ List items=new ServletFileUploadnew DiskFileItemFactory.parseRequestrequest; 对于FileItem项目:项目{ if item.isFormField{ //处理常规表单字段输入类型=文本|收音机|复选框|等,选择等。 字符串fieldName=item.getFieldName; 字符串字段值=item.getString; //…在这里做你的工作 }否则{ //流程表单文件字段输入类型=文件。 字符串fieldName=item.getFieldName; 字符串文件名=FilenameUtils.getNameitem.getName; InputStream fileContent=item.getInputStream; //…在这里做你的工作 } } }捕获文件上载异常{ 抛出新的ServletException无法解析多部分请求,e; } // ... } 这是我的 重要的是,不要事先在同一请求上调用getParameter、getParameterMap、getParameterValues、getInputStream、getReader等。否则,servlet容器将读取并解析请求主体,因此Apache Commons FileUpload将获得一个空的请求主体。另见a.o

请注意文件名utilsgetName。这是一个关于获取文件名的MSIE修复程序。此浏览器不正确地沿名称而不是仅沿文件名发送完整文件路径

private static String getSubmittedFileName(Part part) {
    for (String cd : part.getHeader("content-disposition").split(";")) {
        if (cd.trim().startsWith("filename")) {
            String fileName = cd.substring(cd.indexOf('=') + 1).trim().replace("\"", "");
            return fileName.substring(fileName.lastIndexOf('/') + 1).substring(fileName.lastIndexOf('\\') + 1); // MSIE fix.
        }
    }
    return null;
}
或者,您也可以将所有内容包装在一个过滤器中,该过滤器自动解析所有内容,并将内容放回请求的parametermap中,这样您就可以继续使用request.getParameter,并通过request.getAttribute检索上载的文件

getParameter仍然返回null的GlassFish3错误的解决方法 请注意,早于3.1.2的Glassfish版本中的getParameter仍然返回null。如果您的目标是这样一个容器,并且无法升级它,那么您需要借助以下实用程序方法从getPart中提取值:

private static String getValue(Part part) throws IOException {
    BufferedReader reader = new BufferedReader(new InputStreamReader(part.getInputStream(), "UTF-8"));
    StringBuilder value = new StringBuilder();
    char[] buffer = new char[1024];
    for (int length = 0; (length = reader.read(buffer)) > 0;) {
        value.append(buffer, 0, length);
    }
    return value.toString();
}
保存上传的文件不要使用getRealPath或part.write! 有关将获取的InputStream正确保存到fileContent变量的详细信息,请参阅以下答案,如上面的代码片段所示:

服务上传文件 有关将保存的文件从磁盘或数据库正确送达客户端的详细信息,请参阅以下答案:

将形式抽象化 下面回答如何使用Ajax和jQuery上传。请注意,收集表单数据的servlet代码不需要为此更改!只有您的响应方式可能会改变,但这是非常简单的,即不转发到JSP,只需打印一些JSON或XML,甚至是纯文本,这取决于负责Ajax调用的脚本所期望的内容

希望这些都能有所帮助:

您需要在lib目录中包含common-io.1.4.jar文件,或者如果您在任何编辑器(如NetBeans)中工作,那么您需要转到项目属性,只需添加jar文件即可


要获取common.io.jar文件,只需通过谷歌搜索它,或者直接访问Apache网站,在那里可以免费下载该文件。但请记住一件事:如果您是Windows用户,请下载二进制ZIP文件。

我对每个Html表单使用公共Servlet,无论它是否有附件。 这个Servlet返回一个树状图,其中键是jsp名称参数,值是用户输入,并将所有附件保存在固定目录中,然后您可以重命名您选择的目录。我想这对你有帮助

public class ServletCommonfunctions extends HttpServlet implements
        Connections {

    private static final long serialVersionUID = 1L;

    public ServletCommonfunctions() {}

    protected void doPost(HttpServletRequest request,
            HttpServletResponse response) throws ServletException,
            IOException {}

    public SortedMap<String, String> savefilesindirectory(
            HttpServletRequest request, HttpServletResponse response)
            throws IOException {
        // Map<String, String> key_values = Collections.synchronizedMap( new
        // TreeMap<String, String>());
        SortedMap<String, String> key_values = new TreeMap<String, String>();
        String dist = null, fact = null;
        PrintWriter out = response.getWriter();
        File file;
        String filePath = "E:\\FSPATH1\\2KL06CS048\\";
        System.out.println("Directory Created   ????????????"
            + new File(filePath).mkdir());
        int maxFileSize = 5000 * 1024;
        int maxMemSize = 5000 * 1024;
        // Verify the content type
        String contentType = request.getContentType();
        if ((contentType.indexOf("multipart/form-data") >= 0)) {
            DiskFileItemFactory factory = new DiskFileItemFactory();
            // maximum size that will be stored in memory
            factory.setSizeThreshold(maxMemSize);
            // Location to save data that is larger than maxMemSize.
            factory.setRepository(new File(filePath));
            // Create a new file upload handler
            ServletFileUpload upload = new ServletFileUpload(factory);
            // maximum file size to be uploaded.
            upload.setSizeMax(maxFileSize);
            try {
                // Parse the request to get file items.
                @SuppressWarnings("unchecked")
                List<FileItem> fileItems = upload.parseRequest(request);
                // Process the uploaded file items
                Iterator<FileItem> i = fileItems.iterator();
                while (i.hasNext()) {
                    FileItem fi = (FileItem) i.next();
                    if (!fi.isFormField()) {
                        // Get the uploaded file parameters
                        String fileName = fi.getName();
                        // Write the file
                        if (fileName.lastIndexOf("\\") >= 0) {
                            file = new File(filePath
                                + fileName.substring(fileName
                                        .lastIndexOf("\\")));
                        } else {
                            file = new File(filePath
                                + fileName.substring(fileName
                                        .lastIndexOf("\\") + 1));
                        }
                        fi.write(file);
                    } else {
                        key_values.put(fi.getFieldName(), fi.getString());
                    }
                }
            } catch (Exception ex) {
                System.out.println(ex);
            }
        }
        return key_values;
    }
}

如果您使用Geronimo及其嵌入式Tomcat,则会出现此问题的另一个根源。在这种情况下,在多次测试commons io和commons fileupload之后,问题来自处理commons xxx JAR的父类加载器。这是必须防止的。事故始终发生在:

fileItems = uploader.parseRequest(request);
请注意,当前版本的commons fileupload将fileItems的列表类型更改为Specially List,而以前的版本为generic List

我在Eclipse项目中添加了commons fileupload和commons io的源代码,以跟踪实际错误,并最终获得了一些见解。首先,抛出的异常属于Throwable类型,而不是声明的FileIOException,甚至这些异常都不会被捕获。其次,错误消息是模糊的,因为它声明找不到类,因为axis2找不到commons io。Axis2根本不在我的项目中使用,而是作为标准安装的一部分作为Geronimo存储库子目录中的文件夹存在

最后,我找到了一个地方,提出了一个有效的解决方案,成功地解决了我的问题。您必须在部署计划中对父加载程序隐藏JAR。这被放入geronimo-web.xml中,我的完整文件如下所示

Pasted from <http://osdir.com/ml/user-geronimo-apache/2011-03/msg00026.html> 



<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<web:web-app xmlns:app="http://geronimo.apache.org/xml/ns/j2ee/application-2.0" xmlns:client="http://geronimo.apache.org/xml/ns/j2ee/application-client-2.0" xmlns:conn="http://geronimo.apache.org/xml/ns/j2ee/connector-1.2" xmlns:dep="http://geronimo.apache.org/xml/ns/deployment-1.2" xmlns:ejb="http://openejb.apache.org/xml/ns/openejb-jar-2.2" xmlns:log="http://geronimo.apache.org/xml/ns/loginconfig-2.0" xmlns:name="http://geronimo.apache.org/xml/ns/naming-1.2" xmlns:pers="http://java.sun.com/xml/ns/persistence" xmlns:pkgen="http://openejb.apache.org/xml/ns/pkgen-2.1" xmlns:sec="http://geronimo.apache.org/xml/ns/security-2.0" xmlns:web="http://geronimo.apache.org/xml/ns/j2ee/web-2.0.1">
    <dep:environment>
        <dep:moduleId>
            <dep:groupId>DataStar</dep:groupId>
            <dep:artifactId>DataStar</dep:artifactId>
            <dep:version>1.0</dep:version>
            <dep:type>car</dep:type>
        </dep:moduleId>

<!--Don't load commons-io or fileupload from parent classloaders-->
        <dep:hidden-classes>
            <dep:filter>org.apache.commons.io</dep:filter>
            <dep:filter>org.apache.commons.fileupload</dep:filter>
        </dep:hidden-classes>
        <dep:inverse-classloading/>        


    </dep:environment>
    <web:context-root>/DataStar</web:context-root>
</web:web-app>
<form action="Controller" method="post" enctype="multipart/form-data">
     <label class="file-upload"> Click here to upload an Image </label>
     <input type="file" name="file" id="file" required> 
</form>

为文件发送多个文件我们必须使用enctype=multipart/form data 要发送多个文件,请在输入标记中使用multiple=multiple

<form action="upload" method="post" enctype="multipart/form-data">
 <input type="file" name="fileattachments"  multiple="multiple"/>
 <input type="submit" />
</form>
Tomcat 6 o 7中没有组件或外部库

在web.xml文件中启用上载:


根据servlet要求编辑代码,如最大文件大小、最大请求大小和可以设置的其他选项…

您可以使用jsp/servlet上载文件

<form action="UploadFileServlet" method="post">
  <input type="text" name="description" />
  <input type="file" name="file" />
  <input type="submit" />
</form>
您必须从该对象获取文件项和字段,然后才能将其存储到服务器中,如下所示:

String loc="./webapps/prjct name/server folder/"+contentid+extension;
File uploadFile=new File(loc);
item.write(uploadFile);

下面是一个使用apache commons fileupload的示例:

// apache commons-fileupload to handle file upload
DiskFileItemFactory factory = new DiskFileItemFactory();
factory.setRepository(new File(DataSources.TORRENTS_DIR()));
ServletFileUpload fileUpload = new ServletFileUpload(factory);

List<FileItem> items = fileUpload.parseRequest(req.raw());
FileItem item = items.stream()
  .filter(e ->
  "the_upload_name".equals(e.getFieldName()))
  .findFirst().get();
String fileName = item.getName();

item.write(new File(dir, fileName));
log.info(fileName);

如果您碰巧使用了Spring MVC,以下是如何: 我把这个留在这里,以防有人觉得有用

使用enctype属性设置为与BalusC答案相同的多部分/表格数据的表格

<form action="upload" method="post" enctype="multipart/form-data">
    <input type="file" name="file" />
    <input type="submit" value="Upload"/>
</form>
您可以使用MultipartFile的getOriginalFilename和getSize获取文件名和大小

我已经用SpringVersion4.1.1.RELEASE对此进行了测试。

HTML页面

<html>
<head>
<title>File Uploading Form</title>
</head>
<body>
<h3>File Upload:</h3>
Select a file to upload: <br />
<form action="UploadServlet" method="post"
                        enctype="multipart/form-data">
<input type="file" name="file" size="50" />
<br />
<input type="submit" value="Upload File" />
</form>
</body>
</html> 
SERVLET文件

// Import required java libraries
import java.io.*;
import java.util.*;

import javax.servlet.ServletConfig;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.apache.commons.fileupload.FileItem;
import org.apache.commons.fileupload.FileUploadException;
import org.apache.commons.fileupload.disk.DiskFileItemFactory;
import org.apache.commons.fileupload.servlet.ServletFileUpload;
import org.apache.commons.io.output.*;

public class UploadServlet extends HttpServlet {

   private boolean isMultipart;
   private String filePath;
   private int maxFileSize = 50 * 1024;
   private int maxMemSize = 4 * 1024;
   private File file ;

   public void init( ){
      // Get the file location where it would be stored.
      filePath = 
             getServletContext().getInitParameter("file-upload"); 
   }
   public void doPost(HttpServletRequest request, 
               HttpServletResponse response)
              throws ServletException, java.io.IOException {
      // Check that we have a file upload request
      isMultipart = ServletFileUpload.isMultipartContent(request);
      response.setContentType("text/html");
      java.io.PrintWriter out = response.getWriter( );
      if( !isMultipart ){
         out.println("<html>");
         out.println("<head>");
         out.println("<title>Servlet upload</title>");  
         out.println("</head>");
         out.println("<body>");
         out.println("<p>No file uploaded</p>"); 
         out.println("</body>");
         out.println("</html>");
         return;
      }
      DiskFileItemFactory factory = new DiskFileItemFactory();
      // maximum size that will be stored in memory
      factory.setSizeThreshold(maxMemSize);
      // Location to save data that is larger than maxMemSize.
      factory.setRepository(new File("c:\\temp"));

      // Create a new file upload handler
      ServletFileUpload upload = new ServletFileUpload(factory);
      // maximum file size to be uploaded.
      upload.setSizeMax( maxFileSize );

      try{ 
      // Parse the request to get file items.
      List fileItems = upload.parseRequest(request);

      // Process the uploaded file items
      Iterator i = fileItems.iterator();

      out.println("<html>");
      out.println("<head>");
      out.println("<title>Servlet upload</title>");  
      out.println("</head>");
      out.println("<body>");
      while ( i.hasNext () ) 
      {
         FileItem fi = (FileItem)i.next();
         if ( !fi.isFormField () )  
         {
            // Get the uploaded file parameters
            String fieldName = fi.getFieldName();
            String fileName = fi.getName();
            String contentType = fi.getContentType();
            boolean isInMemory = fi.isInMemory();
            long sizeInBytes = fi.getSize();
            // Write the file
            if( fileName.lastIndexOf("\\") >= 0 ){
               file = new File( filePath + 
               fileName.substring( fileName.lastIndexOf("\\"))) ;
            }else{
               file = new File( filePath + 
               fileName.substring(fileName.lastIndexOf("\\")+1)) ;
            }
            fi.write( file ) ;
            out.println("Uploaded Filename: " + fileName + "<br>");
         }
      }
      out.println("</body>");
      out.println("</html>");
   }catch(Exception ex) {
       System.out.println(ex);
   }
   }
   public void doGet(HttpServletRequest request, 
                       HttpServletResponse response)
        throws ServletException, java.io.IOException {

        throw new ServletException("GET method used with " +
                getClass( ).getName( )+": POST method required.");
   } 
}
web.xml

上面编译 servlet上传servlet并在web.xml文件中创建所需条目,如下所示

<servlet>
   <servlet-name>UploadServlet</servlet-name>
   <servlet-class>UploadServlet</servlet-class>
</servlet>

<servlet-mapping>
   <servlet-name>UploadServlet</servlet-name>
   <url-pattern>/UploadServlet</url-pattern>
</servlet-mapping>
对于Spring MVC 我已经试了好几个小时了 并设法有了一个更简单的版本,用于以表格形式输入数据和图像

<form action="/handleform" method="post" enctype="multipart/form-data">
  <input type="text" name="name" />
  <input type="text" name="age" />
  <input type="file" name="file" />
  <input type="submit" />
</form>

希望它能有所帮助:

最简单的文件和输入控件方法,有10亿个库:

  <%
  if (request.getContentType()==null) return;
  // for input type=text controls
  String v_Text = 
  (new BufferedReader(new InputStreamReader(request.getPart("Text1").getInputStream()))).readLine();    

  // for input type=file controls
  InputStream inStr = request.getPart("File1").getInputStream();
  char charArray[] = new char[inStr.available()];
  new InputStreamReader(inStr).read(charArray);
  String contents = new String(charArray);
  %>

首先必须将表单的enctype属性设置为multipart/formdata

如下所示

Pasted from <http://osdir.com/ml/user-geronimo-apache/2011-03/msg00026.html> 



<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<web:web-app xmlns:app="http://geronimo.apache.org/xml/ns/j2ee/application-2.0" xmlns:client="http://geronimo.apache.org/xml/ns/j2ee/application-client-2.0" xmlns:conn="http://geronimo.apache.org/xml/ns/j2ee/connector-1.2" xmlns:dep="http://geronimo.apache.org/xml/ns/deployment-1.2" xmlns:ejb="http://openejb.apache.org/xml/ns/openejb-jar-2.2" xmlns:log="http://geronimo.apache.org/xml/ns/loginconfig-2.0" xmlns:name="http://geronimo.apache.org/xml/ns/naming-1.2" xmlns:pers="http://java.sun.com/xml/ns/persistence" xmlns:pkgen="http://openejb.apache.org/xml/ns/pkgen-2.1" xmlns:sec="http://geronimo.apache.org/xml/ns/security-2.0" xmlns:web="http://geronimo.apache.org/xml/ns/j2ee/web-2.0.1">
    <dep:environment>
        <dep:moduleId>
            <dep:groupId>DataStar</dep:groupId>
            <dep:artifactId>DataStar</dep:artifactId>
            <dep:version>1.0</dep:version>
            <dep:type>car</dep:type>
        </dep:moduleId>

<!--Don't load commons-io or fileupload from parent classloaders-->
        <dep:hidden-classes>
            <dep:filter>org.apache.commons.io</dep:filter>
            <dep:filter>org.apache.commons.fileupload</dep:filter>
        </dep:hidden-classes>
        <dep:inverse-classloading/>        


    </dep:environment>
    <web:context-root>/DataStar</web:context-root>
</web:web-app>
<form action="Controller" method="post" enctype="multipart/form-data">
     <label class="file-upload"> Click here to upload an Image </label>
     <input type="file" name="file" id="file" required> 
</form>

@buhake sindi hey如果我使用live server,或者通过将文件上载到服务器live server中的任何目录来运行我的项目,那么文件路径应该是什么。如果您编写代码在servlet中创建目录,那么将在live SrverNot find.jar but.zip中创建目录。你是说.zip吗?我们如何处理getPartfileattachments,以获得一个部件数组?我不认为多个文件的getPart可以工作?啊,很抱歉,我看到了request.getPartsfile,并将x_x与Servlet 3.0混淆,如果违反了MultipartConfig条件,例如:maxFileSize,调用request.getParameter返回null。这是故意的吗?如果在调用getPart并检查IllegalStateException之前获取一些常规文本参数,该怎么办?这会导致在我有机会检查IllegalStateException之前抛出NullPointerException。@BalusC我创建了一篇与此相关的文章,你知道我如何从文件API webKitDirectory检索额外信息吗。这里有更多详细信息是的,如果有人试图在tomcat 7中使用3.0部分中的代码,他们可能会遇到字符串fileName=path.getfilePart.getSubmittedFileName.getFileName.toString;中的问题MSIE fix.part类似于me@aaa:当您出于不清楚的原因使用读卡器和/或写卡器将字节转换为字符时,可能会发生这种情况。不要那样做。在读取和写入上传的文件时,在所有位置使用InputStream/OutputStream,而无需将字节转换为字符。PDF文件不是基于字符的文本文件。这是一个二进制文件。如果我没有弄错的话,这需要你在服务器的应用程序配置中配置一个bean……也许这篇文章会有所帮助:@Adam:他们从我的答案中复制了一些内容,并在上面添加了一个广告侦探,试图用它赚钱。是的,很棒的文章…不,实际上没有被复制。我写了那篇文章的第一稿以及补充代码。核心参考文档可以在这里找到:并链接到文章中引用。示例部分转载自核心参考文件,即参考点文件,即参考点文件,但并非全部。请注意,参考文件并未详细说明。谢谢您能分享选择的图片表单DBMySQL并在jsp/html上显示吗?这个解决方案是不同的。另一个解决方案是使用一个库来处理没有第三方jar文件的文件。这已经包含在当前接受的答案中。你读过了吗?自2009年12月以来,本机API已经存在。顺便说一句,你关闭溪流的方式也是一种遗产。自2011年7月推出Java7以来,您可以使用trywithresources语句,而不是在finally中处理null检查。
// Import required java libraries
import java.io.*;
import java.util.*;

import javax.servlet.ServletConfig;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.apache.commons.fileupload.FileItem;
import org.apache.commons.fileupload.FileUploadException;
import org.apache.commons.fileupload.disk.DiskFileItemFactory;
import org.apache.commons.fileupload.servlet.ServletFileUpload;
import org.apache.commons.io.output.*;

public class UploadServlet extends HttpServlet {

   private boolean isMultipart;
   private String filePath;
   private int maxFileSize = 50 * 1024;
   private int maxMemSize = 4 * 1024;
   private File file ;

   public void init( ){
      // Get the file location where it would be stored.
      filePath = 
             getServletContext().getInitParameter("file-upload"); 
   }
   public void doPost(HttpServletRequest request, 
               HttpServletResponse response)
              throws ServletException, java.io.IOException {
      // Check that we have a file upload request
      isMultipart = ServletFileUpload.isMultipartContent(request);
      response.setContentType("text/html");
      java.io.PrintWriter out = response.getWriter( );
      if( !isMultipart ){
         out.println("<html>");
         out.println("<head>");
         out.println("<title>Servlet upload</title>");  
         out.println("</head>");
         out.println("<body>");
         out.println("<p>No file uploaded</p>"); 
         out.println("</body>");
         out.println("</html>");
         return;
      }
      DiskFileItemFactory factory = new DiskFileItemFactory();
      // maximum size that will be stored in memory
      factory.setSizeThreshold(maxMemSize);
      // Location to save data that is larger than maxMemSize.
      factory.setRepository(new File("c:\\temp"));

      // Create a new file upload handler
      ServletFileUpload upload = new ServletFileUpload(factory);
      // maximum file size to be uploaded.
      upload.setSizeMax( maxFileSize );

      try{ 
      // Parse the request to get file items.
      List fileItems = upload.parseRequest(request);

      // Process the uploaded file items
      Iterator i = fileItems.iterator();

      out.println("<html>");
      out.println("<head>");
      out.println("<title>Servlet upload</title>");  
      out.println("</head>");
      out.println("<body>");
      while ( i.hasNext () ) 
      {
         FileItem fi = (FileItem)i.next();
         if ( !fi.isFormField () )  
         {
            // Get the uploaded file parameters
            String fieldName = fi.getFieldName();
            String fileName = fi.getName();
            String contentType = fi.getContentType();
            boolean isInMemory = fi.isInMemory();
            long sizeInBytes = fi.getSize();
            // Write the file
            if( fileName.lastIndexOf("\\") >= 0 ){
               file = new File( filePath + 
               fileName.substring( fileName.lastIndexOf("\\"))) ;
            }else{
               file = new File( filePath + 
               fileName.substring(fileName.lastIndexOf("\\")+1)) ;
            }
            fi.write( file ) ;
            out.println("Uploaded Filename: " + fileName + "<br>");
         }
      }
      out.println("</body>");
      out.println("</html>");
   }catch(Exception ex) {
       System.out.println(ex);
   }
   }
   public void doGet(HttpServletRequest request, 
                       HttpServletResponse response)
        throws ServletException, java.io.IOException {

        throw new ServletException("GET method used with " +
                getClass( ).getName( )+": POST method required.");
   } 
}
<servlet>
   <servlet-name>UploadServlet</servlet-name>
   <servlet-class>UploadServlet</servlet-class>
</servlet>

<servlet-mapping>
   <servlet-name>UploadServlet</servlet-name>
   <url-pattern>/UploadServlet</url-pattern>
</servlet-mapping>
<form action="/handleform" method="post" enctype="multipart/form-data">
  <input type="text" name="name" />
  <input type="text" name="age" />
  <input type="file" name="file" />
  <input type="submit" />
</form>
@Controller
public class FormController {
    @RequestMapping(value="/handleform",method= RequestMethod.POST)
    ModelAndView register(@RequestParam String name, @RequestParam int age, @RequestParam MultipartFile file)
            throws ServletException, IOException {

        System.out.println(name);
        System.out.println(age);
        if(!file.isEmpty()){
            byte[] bytes = file.getBytes();
            String filename = file.getOriginalFilename();
            BufferedOutputStream stream =new BufferedOutputStream(new FileOutputStream(new File("D:/" + filename)));
            stream.write(bytes);
            stream.flush();
            stream.close();
        }
        return new ModelAndView("index");
    }
}
  <%
  if (request.getContentType()==null) return;
  // for input type=text controls
  String v_Text = 
  (new BufferedReader(new InputStreamReader(request.getPart("Text1").getInputStream()))).readLine();    

  // for input type=file controls
  InputStream inStr = request.getPart("File1").getInputStream();
  char charArray[] = new char[inStr.available()];
  new InputStreamReader(inStr).read(charArray);
  String contents = new String(charArray);
  %>
<form action="Controller" method="post" enctype="multipart/form-data">
     <label class="file-upload"> Click here to upload an Image </label>
     <input type="file" name="file" id="file" required> 
</form>
@MultipartConfig

public class Controller extends HttpServlet {

    protected void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        insertImage(request, response);
    }

    private void addProduct(HttpServletRequest request, HttpServletResponse response) {
        Part filePart = request.getPart("file");
        String imageName = Paths.get(filePart.getSubmittedFileName()).getFileName().toString();

        String imageSavePath = "specify image path to save image"; //path to save image
        FileOutputStream outputStream = null;
        InputStream fileContent = null;

        try {
            outputStream = new FileOutputStream(new File(imageSavePath + File.separator + imageName));
            //creating a new file with file path and the file name
            fileContent = filePart.getInputStream();
            //getting the input stream
            int readBytes = 0;
            byte[] readArray = new byte[1024];
            //initializing a byte array with size 1024

            while ((readBytes = fileContent.read(readArray)) != -1) {
                outputStream.write(readArray, 0, readBytes);
            }//this loop will write the contents of the byte array unitl the end to the output stream
        } catch (Exception ex) {
            System.out.println("Error Writing File: " + ex);
        } finally {
            if (outputStream != null) {
                outputStream.close();
                //closing the output stream
            }
            if (fileContent != null) {
                fileContent.close();
                //clocsing the input stream
            }
        }
    }
}