Java 如何使用JSP/Servlet将文件上载到服务器?
如何使用JSP/Servlet将文件上载到服务器?我试过这个: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
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
}
}
}
}