如何在J2ee中上载该图像后在Html文件中显示图像 [英] How Do I Show A Image In A Html File After Uploading That Image In J2ee
问题描述
我想在上传这些图片后显示图片文件..我有一个html文件和Servlet文件，我正在上传一些图片，但我想通过一个html页面显示这些图片，用户可以在那里查看它们。它是可能的?? PLZ指导我该怎么办？我正在发送html页面和servlet页面，PLZ帮助...
HTML文件
< html>
< body>
< form action =imagemethod =postenctype =multipart / form-data>
选择文件：< input type =filename =fname/>
< input type =submitvalue =upload />
< / form>
< / body>
< / html>
和Servlet文件
import java.io.File;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.List;
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.disk.DiskFileItemFactory;
import org.apache.commons.fileupload.servlet.ServletFileUpload;
/ **
*用于处理客户端文件上传的Java servlet。
*
* @author www.codejava.net
* /
公共类图像扩展HttpServlet {
私有静态最终长版serialVersionUID = 1L;
/ /上传存储文件的位置
private static final String UPLOAD_DIR =upload;
//上传设置
private static final int MEMORY_THRESHOLD = 1024 * 1024 * 3 ; // 3MB
private static final int MAX_FILE_SIZE = 1024 * 1024 * 40; // 40MB
private static final int MAX_REQUEST_SIZE = 1024 * 1024 * 50; // 50MB
/ **
*收到文件上传提交后，解析请求阅读
*上传数据并将文件保存在磁盘上。
* /
protected void doPost（HttpServletRequest请求，
HttpServletResponse响应）抛出ServletException，IOException {
//检查请求是否实际包含上传文件
if（！ServletFileUpload.isMultipartContent（request））{
//如果不是，我们在此停止
PrintWriter writer = response.getWriter（）;
writer.println（错误：表单必须具有enctype = multipart / form-data。）;
writer.flush（）;
返回;
}
/ /配置上传设置
DiskFileItemFactory factory = new DiskFileItemFactory（）;
//设置内存阈值 - 超出该阈值les存储在磁盘中
factory.setSizeThreshold（MEMORY_THRESHOLD）;
//设置存储文件的临时位置
factory.setRepository（new File （System.getProperty（java.io.tmpdir）））;
ServletFileUpload upload = new ServletFileUpload（factory）;
//设置上传文件的最大大小
upload.setFileSizeMax（MAX_FILE_SIZE）;
//设置最大值请求的大小（包括文件+表单数据）
upload.setSizeMax（MAX_REQUEST_SIZE）;
//构造商店上传的目录路径文件
//此路径相对于应用程序的目录
字符串uploadPath = getServletContext（）。getRealPath（）
+ File.separator + UPLOAD_DIR;
//如果目录不存在则创建目录
文件uploadDir =新文件（uploadPath）;
if（！uploadDir.exists（））{
uploadDir.mkdir（）;
}
尝试{
//解析请求的内容以提取文件数据
@SuppressWarnings （未选中）
列表< fileitem> formItems = upload.parseRequest（request）;
if（formItems！= null&& formItems.size（）> 0）{
//遍历表单的字段
for（FileItem item：formItems）{
//仅处理非表单字段的字段
if（！item.isFormField（））{
String fileName = new File（item.getName（））。getName（）;
String filePath = uploadPath + File .separator + fileName;
文件storeFile =新文件（filePath）;
item.write（ storeFile）;
request.setAttribute（message，
上传已成功完成！）;
}
}
}
} catch（exception ex）{
request.setAttribute（message，
有一个错误：+ ex.getMessage（））;
}
getServletContext（）。getRequestDispatcher（/ message.jsp）。forward（request，response）;
}
}
真的说我从互联网上获得了这个servlet代码..如果这段代码错了，请帮助... plz plz。 ..
I want to show image files after uploading these images..i have a html file and Servlet file where i'm uploading some images,but i want to show those images through a html page where user can view them..can it be possible??plz guide me what should i do??i'm sending html page and servlet page,plz help...
HTML file
<html>
<body>
<form action="image" method="post" enctype="multipart/form-data">
Select File:<input type="file" name="fname"/>
<input type="submit" value="upload"/>
</form>
</body>
</html>
and Servlet File
import java.io.File;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.List;
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.disk.DiskFileItemFactory;
import org.apache.commons.fileupload.servlet.ServletFileUpload;
/**
* A Java servlet that handles file upload from client.
*
* @author www.codejava.net
*/
public class image extends HttpServlet {
private static final long serialVersionUID = 1L;
// location to store file uploaded
private static final String UPLOAD_DIR = "upload";
// upload settings
private static final int MEMORY_THRESHOLD = 1024 * 1024 * 3; // 3MB
private static final int MAX_FILE_SIZE = 1024 * 1024 * 40; // 40MB
private static final int MAX_REQUEST_SIZE = 1024 * 1024 * 50; // 50MB
/**
* Upon receiving file upload submission, parses the request to read
* upload data and saves the file on disk.
*/
protected void doPost(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
// checks if the request actually contains upload file
if (!ServletFileUpload.isMultipartContent(request)) {
// if not, we stop here
PrintWriter writer = response.getWriter();
writer.println("Error: Form must has enctype=multipart/form-data.");
writer.flush();
return;
}
// configures upload settings
DiskFileItemFactory factory = new DiskFileItemFactory();
// sets memory threshold - beyond which files are stored in disk
factory.setSizeThreshold(MEMORY_THRESHOLD);
// sets temporary location to store files
factory.setRepository(new File(System.getProperty("java.io.tmpdir")));
ServletFileUpload upload = new ServletFileUpload(factory);
// sets maximum size of upload file
upload.setFileSizeMax(MAX_FILE_SIZE);
// sets maximum size of request (include file + form data)
upload.setSizeMax(MAX_REQUEST_SIZE);
// constructs the directory path to store upload file
// this path is relative to application's directory
String uploadPath = getServletContext().getRealPath("")
+ File.separator + UPLOAD_DIR;
// creates the directory if it does not exist
File uploadDir = new File(uploadPath);
if (!uploadDir.exists()) {
uploadDir.mkdir();
}
try {
// parses the request's content to extract file data
@SuppressWarnings("unchecked")
List<fileitem> formItems = upload.parseRequest(request);
if (formItems != null && formItems.size() > 0) {
// iterates over form's fields
for (FileItem item : formItems) {
// processes only fields that are not form fields
if (!item.isFormField()) {
String fileName = new File(item.getName()).getName();
String filePath = uploadPath + File.separator + fileName;
File storeFile = new File(filePath);
item.write(storeFile);
request.setAttribute("message",
"Upload has been done successfully!");
}
}
}
} catch (Exception ex) {
request.setAttribute("message",
"There was an error: " + ex.getMessage());
}
getServletContext().getRequestDispatcher("/message.jsp").forward(request, response);
}
}
Truly speaking i got this servlet code from internet..if this code wrong plz help...plz plz ...
推荐答案
将任务视为单独的。
任务一：
使用您的servlet或您希望的任何方式上传文件。您应该控制文件名和位置以供以后使用。
任务二：在页面呈现后从服务器加载图像文件。
使用AJAX上传文件，您可以保留当前页面。
使用表格，你需要加载一个新页面，即使它是同一页面。
在任何一种情况下，你拥有图像文件的名称和位置。
AJAX：将图像文件上传到你想要的位置。
形式：当你调用时你就是新的页面，将图像文件的名称构建到标题中，并使用javaScript加载图像。
Consider the tasks as separate.
Task one:
upload the file with your servlet, or by any means you wish. You should control the file name and location for later use.
Task Two: load an image file from your server after the page has already rendered.
Use AJAX to upload the file and you can stay on your current page.
Use a form and you'll need to load a new page, even if it's the same page.
In either case, you have the image file's name and location.
AJAX: upload the image file by to the location where you want it to go.
forms: when you invoke you're new page, build the image files' name into the header and use javaScript to load the image.
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