Servlet file upload


Servlet can be used with the HTML form tag to allow users to upload files to the server. The uploaded file can be a text file or an image file or any document.

Create a file upload form

The following HTML code creates a file upload form. The following points need to be noted:

  • The form method attribute should be set to the POST method, and the GET method cannot be used.

  • Formenctype The property should be set to multipart/form-data.

  • form# The ##action attribute should be set to the Servlet file that handles file uploads on the backend server. The following example uses UploadServlet Servlet to upload files.

  • To upload a single file, you should use a single <input .../> tag with the attribute type="file". To allow multiple file uploads, include multiple input tags with different name attribute values. Input tags have different values ​​for the name attribute. The browser associates a browse button with each input tag. 

    •  <html>
<head>
<title>文件上传表单</title>
</head>
<body>
<h3>文件上传：</h3>
请选择要上传的文件：<br />
<form action="UploadServlet" method="post"
                        enctype="multipart/form-data">
<input type="file" name="file" size="50" />
<br />
<input type="submit" value="上传文件" />
</form>
</body>
</html>
This will display the following results, allowing the user to select a file from the local computer. When the user clicks "Upload File", the form will be submitted along with the file selected from the local computer: 

 <b>文件上传：</b> 
请选择要上传的文件：<br /> 
<input type="file" name="file" size="50" /> 
<br /> 
<input type="button" value="上传文件" /> 
<br /> 
注：这只是虚拟的表单，不会正常工作。

Writing background Servlet

The following is the Servlet

UploadServlet, which will accept the uploaded file and store it in the directory <Tomcat-installation-directory>/webapps/ in data. This directory name can also be added using external configuration, such as the context-param element in web.xml, as shown below: 

 <web-app>
....
<context-param> 
    <description>Location to store uploaded file</description> 
    <param-name>file-upload</param-name> 
    <param-value>
         c:\apache-tomcat-5.5.29\webapps\data\
     </param-value> 
</context-param>
....
</web-app>

The following is the source code of UploadServlet, which can process multiple files at once Upload of files. Before proceeding, please confirm the following:

  • The example below depends on FileUpload, so be sure to have the latest version of

    commons-fileupload on your classpath. x.x.jar file. Can be downloaded from http://commons.apache.org/fileupload/.

  • FileUpload relies on Commons IO, so be sure to have the latest version of the

    commons-io-x.x.jar file in your classpath. Can be downloaded from http://commons.apache.org/io/.

  • When testing the following example, the size of the file you upload cannot be larger than

    maxFileSize, otherwise the file will not be uploaded.

  • Please ensure that the directories c:\temp and c:\apache-tomcat-5.5.29\webapps\data have been created in advance. 

    • // 导入必需的 java 库
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( ){
      // 获取文件将被存储的位置
      filePath = 
             getServletContext().getInitParameter("file-upload"); 
   }
   public void doPost(HttpServletRequest request, 
               HttpServletResponse response)
              throws ServletException, java.io.IOException {
      // 检查我们有一个文件上传请求
      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();
      // 文件大小的最大值将被存储在内存中
      factory.setSizeThreshold(maxMemSize);
      // Location to save data that is larger than maxMemSize.
      factory.setRepository(new File("c:\temp"));

      // 创建一个新的文件上传处理程序
      ServletFileUpload upload = new ServletFileUpload(factory);
      // 允许上传的文件大小的最大值
      upload.setSizeMax( maxFileSize );

      try{ 
      // 解析请求，获取文件项
      List fileItems = upload.parseRequest(request);
	
      // 处理上传的文件项
      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 () )	
         {
            // 获取上传文件的参数
            String fieldName = fi.getFieldName();
            String fileName = fi.getName();
            String contentType = fi.getContentType();
            boolean isInMemory = fi.isInMemory();
            long sizeInBytes = fi.getSize();
            // 写入文件
            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.");
   } 
}
Compile and run the Servlet

Compile the above Servlet UploadServlet and create the required entries in the web.xml file as follows: 

 <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>

Now try to upload a file using the HTML form you created above. When you visit: http://localhost:8080/UploadFile.htm in your browser, it will display the following result, which will help you to upload any file from your local computer. 

 <b>文件上传：</b> 
请选择要上传的文件：<br /> 
<input type="file" name="file" size="50" /> 
<br /> 
<input type="button" value="上传文件" />

If your Servelt script works properly, your files will be uploaded to the c:\apache-tomcat-5.5.29\webapps\data\ directory.

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