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Java servlet如何在一个请求中处理多个上传的文件_Java_Tomcat_Servlets - Fatal编程技术网

Java servlet如何在一个请求中处理多个上传的文件

Java servlet如何在一个请求中处理多个上传的文件,java,tomcat,servlets,Java,Tomcat,Servlets,我知道在servlet中只上载一个文件时如何保存它。HTML <form action="storeArticle" method="post" enctype="multipart/form-data"> <input type="file" name="file"> ... </form> 例如,在中,用户可以通过单击仅一个图像图标来选择上载多个图像。但是,当一次上载多个文件时,servlet如何保存这些上载的文件?HTML <in

我知道在servlet中只上载一个文件时如何保存它。
HTML

<form action="storeArticle" method="post" enctype="multipart/form-data">
    <input type="file" name="file">
    ...
</form>
例如,在中,用户可以通过单击仅一个图像图标来选择上载多个图像。但是,当一次上载多个文件时,servlet如何保存这些上载的文件?
HTML

<input type="file" name="file[]" multiple >

HTML将显示3个元素,这不是我想要的。例如,在中,用户可以选择仅通过一个图像图标上载多个图像。
<input type="file" name="file[]" multiple >
<form action="storeArticle" method="post" enctype="multipart/form-data">
<input type="file" name="file">
<input type="file" name="file2">
<input type="file" name="file3">

</form>
Part part = request.getPart("file");
 File file = new File(filePath);
 try (InputStream inputStream= part.getInputStream()) { // save uploaded file
  Files.copy(inputStream, file.toPath());
 }


  Part part = request.getPart("file2");
  File file = new File(filePath);
 try (InputStream inputStream= part.getInputStream()) { // save uploaded file
  Files.copy(inputStream, file.toPath());
 }



  Part part = request.getPart("file3");
  File file = new File(filePath);
  try (InputStream inputStream= part.getInputStream()) { // save uploaded file
    Files.copy(inputStream, file.toPath());
   }