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Simple uploading and downloading of Web project files

2022-07-26 17:58:00 【Supra code heartbeat】

Simple uploading and downloading of files

One 、 Upload files

  1. need jar package （ You can download it once you search the Internet ）

Insert picture description here

  1. The front-end code 
<%@ page contentType="text/html;charset=UTF-8" language="java" %>
<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>Title</title>
    <!--  It specifies base label  -->
    <base href="<%=request.getContextPath()+"/"%>>">
    <style type="text/css">
        input[type="submit"] {
            outline: none;
            border-radius: 5px;
            cursor: pointer;
            background-color: #31B0D5;
            border: none;
            width: 70px;
            height: 35px;
            font-size: 20px;
        }

        img {
            border-radius: 50%;
        }

        form {
            position: relative;
            width: 200px;
            height: 200px;
        }

        input[type="file"] {
            position: absolute;
            left: 0;
            top: 0;
            height: 200px;
            opacity: 0;
            cursor: pointer;
        }
    </style>

    <script type="text/javascript">
        function prev(event) {
            // Get the area where the picture is displayed 
            var img = document.getElementById("prevView");
            // Get file object 
            var file = event.files[0];
            // Get file reader ： Js Of a class , It can be used directly 
            var reader = new FileReader();
            reader.readAsDataURL(file);
            reader.onload = function () {
                // to img Of src Set pictures url
                img.setAttribute("src", this.result);
            }
        }
    </script>

</head>
<body>
<!--  Form enctype Property to be set to multipart/form-data
    enctype="multipart/form-data"  Indicates that the submitted data is constructed in multiple parts , There are documents and text 
 -->

<form action="fileUploadServlet" method="post" enctype="multipart/form-data">
     chart : <img src="2.jpg" alt="" width="200" height="200" id="prevView">
    <input type="file" name="pic" id="" value="" οnchange="prev(this)"/>

     name : <input type="text" name="name"><br/>

    <input type="submit" value=" Upload "/>
</form>
</body>
</html>
  1. servlet Code 
package mzq.servlet;
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 utils.WebUtils;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.File;
import java.io.IOException;
import java.util.List;
import java.util.UUID;

public class FileUploadServlet extends HttpServlet {
    
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    
        //System.out.println("FileUploadServlet  Called ...");

        //1.  Determine whether it is a file form (enctype="multipart/form-data")
        if (ServletFileUpload.isMultipartContent(request)) {
    
            //System.out.println("OK");
            //2.  establish  DiskFileItemFactory  object ,  It is used to build a tool object for parsing uploaded data 
            DiskFileItemFactory diskFileItemFactory = new DiskFileItemFactory();
            //3.  Create a tool object to parse the uploaded data  
            ServletFileUpload servletFileUpload =
                    new ServletFileUpload(diskFileItemFactory);
            // Solve the problem that the received file name is Chinese garbled 
            servletFileUpload.setHeaderEncoding("utf-8");

            //4.  The key point , servletFileUpload  Object can send the data submitted by the form text /  file 
            //  Encapsulate it in  FileItem  In the file item 
            try {
    
                List<FileItem> list = servletFileUpload.parseRequest(request);
                for (FileItem fileItem : list) {
    
                    //System.out.println("fileItem=" + fileItem);
                    // Judge whether it's a file 
                    if (fileItem.isFormField()) {
    // If it is true It's text  input text
                        String name = fileItem.getString("utf-8");                       
                    } else {
    // It's a document 

                        // In one way 
                        // Get the name of the uploaded file 
                        String name = fileItem.getName();
                        System.out.println(" Uploaded file name =" + name);

                        // Upload this to   Server's  temp Save the files in the directory you specified 
                        //1. Specify a directory  ,  It's under the working directory of our website 
                        String filePath = "/upload/";
                        //2.  Get the full directory  [io/servlet Basics ]
                        //  This directory is related to your web Bound to the project running environment .  It's dynamic .
                        String fileRealPath =
                                request.getServletContext().getRealPath(filePath);

                        //3.  Create this uploaded Directory =>  Create directory ?=> Java Basics 
                        //  Teacher thinking ;  We also have a tool class , Can return  /2024/11/11  character string 
                        File fileRealPathDirectory = new File(fileRealPath + WebUtils.getYearMonthDay());
                        if (!fileRealPathDirectory.exists()) {
    // non-existent , create 
                            fileRealPathDirectory.mkdirs();// establish 
                        }

                        //4.  Copy the file to fileRealPathDirectory Catalog 
                        //  Build a complete path for uploading files  ： Catalog + file name 
                        //  Process the uploaded file name ,  Add a prefix before , Guarantee is the only thing ,  Pretty good 
                        name = UUID.randomUUID().toString() + "_" + name;
                        String fileFullPath = fileRealPathDirectory + "/" +name;
                        fileItem.write(new File(fileFullPath));

                        //5.  Prompt information 
                        response.setContentType("text/html;charset=utf-8");
                        response.getWriter().write(" Upload successful ~");
                    }
                }

            } catch (Exception e) {
    
                e.printStackTrace();
            }
        } else {
    
            System.out.println(" Not a file form ...");
        }
    }

    private void saveFile(){
    

    }

    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    
        doPost(request, response);
    }
}
  1. WebUtils
import java.time.LocalDateTime;

public class WebUtils {
    

    public static String getYearMonthDay() {
    
        // How to get the current date -> java Basics   date   Third generation class 
        LocalDateTime ldt = LocalDateTime.now();
        int year = ldt.getYear();
        int monthValue = ldt.getMonthValue();
        int dayOfMonth = ldt.getDayOfMonth();
        String yearMonthDay = year + "/" + monthValue + "/" + dayOfMonth + "/";
        return yearMonthDay;
    }
}

Be careful ：

Upload files , establish web/upload Folder , stay tomcat Startup time , Not in the out Under the table of contents establish Yes Ought to be upload Folder , as a result of tomcat The corresponding empty directory will not be in out Create the corresponding directory under , therefore , only Need to be in upload Under the table of contents , Just put a file , This is Idea + Tomcat The problem of , Actual development will not exist

Two 、 File download

  1. Front end simple code 
<%@ page contentType="text/html;charset=UTF-8" language="java" %>
<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title> File download </title>
    <base href="<%=request.getContextPath()+"/"%>>">
</head>
<body>
<h1> File download </h1>
<a href="fileDownLoadServlet?name=1.jpg"> Click download picture </a><br/><br/>
    <%--1.jpg  For the file name you want to download on the server    The location of the server here is download Next --%>
</body>
</html>
  1. FileDownloadServlet
package mzq.servlet;

import org.apache.commons.io.IOUtils;
import sun.misc.BASE64Encoder;

import javax.servlet.ServletContext;
import javax.servlet.ServletException;
import javax.servlet.ServletOutputStream;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;
import java.io.InputStream;
import java.net.URLEncoder;

public class FileDownloadServlet extends HttpServlet {
    
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
     
        //1.  First prepare the file to be downloaded [ Assume that these files are public resources ]
        //  important :  Promise to be our tomcat After starting , In the working directory out There are download Folder ,  There are files available for download !!
        //  If you don't see what you created download In the working directory out Next  rebuild project -> restart,  Just OK

        //2.  Get the name of the file to download 
        request.setCharacterEncoding("utf-8");
        String downLoadFileName = request.getParameter("name");
        //System.out.println("downLoadFileName= " + downLoadFileName);

        //3.  to http Respond to , Set the response header  Content-Type ,  It's a document MIME
        //  adopt servletContext  To get 
        ServletContext servletContext = request.getServletContext();
        String downLoadPath = "/download/"; // Download directory from  web Project root directory calculation  /download/1.jpg
        String downLoadFileFullPath = downLoadPath + downLoadFileName;
        String mimeType = servletContext.getMimeType(downLoadFileFullPath);
        System.out.println("mimeType= " + mimeType);
        response.setContentType(mimeType);

        //4.  to http Respond to , Set the response header  Content-Disposition
        //  There are many details to consider here , For example, different browsers write differently , Consider coding 
        // ff  yes   The file name in Chinese needs  base64,  and  ie/chrome  yes  URL code 
        //  Here we only need to know the principle 
        //(1) If it is Firefox  The Chinese code needs  base64
        //(2)Content-Disposition  Is the display form of the specified downloaded data  ,  If attachment  Then use the file download method 
        //(3) If it's something else ( Main stream ie/chrome)  Chinese code use URL code 
        if (request.getHeader("User-Agent").contains("Firefox")) {
    
            //  firefox  Base64 code 
            response.setHeader("Content-Disposition", "attachment; filename==?UTF-8?B?" +
                    new BASE64Encoder().encode(downLoadFileName.getBytes("UTF-8")) + "?=");
        } else {
    
            //  other ( Main stream ie/chrome) Use URL Coding operation 
            response.setHeader("Content-Disposition", "attachment; filename=" +
                    URLEncoder.encode(downLoadFileName, "UTF-8"));
        }

        //5.  Read the downloaded file data , Return to the client / browser 
        //(1)  Create a file to download , Associated input stream 
        InputStream resourceAsStream =
                servletContext.getResourceAsStream(downLoadFileFullPath);
        //(2)  Get the output stream of the returned data  [ Because most of the returned files are binary ( byte ), IO java Basics ]
        ServletOutputStream outputStream = response.getOutputStream();

        //(3)  Use tool class , The file associated with the input stream , Copy to the output stream , And return it to the client / browser 
        IOUtils.copy(resourceAsStream, outputStream);

    }

    protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    
        doPost(request, response);
    }
}
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本文为[Supra code heartbeat]所创，转载请带上原文链接，感谢
https://yzsam.com/2022/207/202207261703181220.html

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