当前位置: 首页 > 面试题库 >

将上传的文件保存在特定位置

秦锐
2023-03-14
问题内容

我有以下代码可以处理服务器上的文件上传。但是如何将文件保存到服务器上的特定位置

import gwtupload.server.UploadAction;
import gwtupload.server.exceptions.UploadActionException;

import org.apache.commons.fileupload.FileItem;

import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.Hashtable;
import java.util.List;

import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;




/**
 * This is an example of how to use UploadAction class.
 *  
 * This servlet saves all received files in a temporary folder, 
 * and deletes them when the user sends a remove request.
 * 
 * @author Manolo Carrasco Moñino
 *
 */
public class SampleUploadServlet extends UploadAction {

  private static final long serialVersionUID = 1L;

  Hashtable<String, String> receivedContentTypes = new Hashtable<String, String>();
  /**
   * Maintain a list with received files and their content types. 
   */
  Hashtable<String, File> receivedFiles = new Hashtable<String, File>();

  /**
   * Override executeAction to save the received files in a custom place
   * and delete this items from session.  
   */
  @Override
  public String executeAction(HttpServletRequest request, List<FileItem> sessionFiles) throws UploadActionException {
    String response = "";
    int cont = 0;
    for (FileItem item : sessionFiles) {
      if (false == item.isFormField()) {
        cont++;
        try {
          /// Create a new file based on the remote file name in the client
          // String saveName = item.getName().replaceAll("[\\\\/><\\|\\s\"'{}()\\[\\]]+", "_");
          // File file =new File("/tmp/" + saveName);

          /// Create a temporary file placed in /tmp (only works in unix)
          // File file = File.createTempFile("upload-", ".bin", new File("/tmp"));

          /// Create a temporary file placed in the default system temp folder
          File file = File.createTempFile("upload-", ".bin");
          item.write(file);

          /// Save a list with the received files
          receivedFiles.put(item.getFieldName(), file);
          receivedContentTypes.put(item.getFieldName(), item.getContentType());

          /// Compose a xml message with the full file information
          response += "<file-" + cont + "-field>" + item.getFieldName() + "</file-" + cont + "-field>\n";
          response += "<file-" + cont + "-name>" + item.getName() + "</file-" + cont + "-name>\n";
          response += "<file-" + cont + "-size>" + item.getSize() + "</file-" + cont + "-size>\n";
          response += "<file-" + cont + "-type>" + item.getContentType() + "</file-" + cont + "type>\n";
        } 
        catch (Exception e) 
        {
        }
      }
    }

    /// Remove files from session because we have a copy of them
    removeSessionFileItems(request);

    /// Send information of the received files to the client.
    return "<response>\n" + response + "</response>\n";
  }

  /**
   * Get the content of an uploaded file.
   */
  @Override
  public void getUploadedFile(HttpServletRequest request, HttpServletResponse response) throws IOException {
    String fieldName = request.getParameter(PARAM_SHOW);
    File f = receivedFiles.get(fieldName);
    if (f != null) {
      response.setContentType(receivedContentTypes.get(fieldName));
      FileInputStream is = new FileInputStream(f);
      copyFromInputStreamToOutputStream(is, response.getOutputStream());
    } else {
      renderXmlResponse(request, response, ERROR_ITEM_NOT_FOUND);
   }
  }

  /**
   * Remove a file when the user sends a delete request.
   */
  @Override
  public void removeItem(HttpServletRequest request, String fieldName)  throws UploadActionException {
    File file = receivedFiles.get(fieldName);
    receivedFiles.remove(fieldName);
    receivedContentTypes.remove(fieldName);
    if (file != null) {
      file.delete();
    }
  }

}

问题答案:

您应该使用File#createTempFile()which代替一个目录。

File file = File.createTempFile("upload-", ".bin", new File("/path/to/your/uploads"));
item.write(file);

或者,如果你真的想临时文件移动到另一个位置 之后
,使用File#renameTo()

File destination = new File("/path/to/your/uploads", file.getName());
file.renameTo(destination);


 类似资料:
  • 我的项目是一个web应用程序。 目前,它有一种生成文本文件并将其发送到客户端浏览器下载的方法。 我需要修改它,以强制浏览器将文件保存在客户端计算机上的自定义(预定义)位置。

  • 问题 上传文件,并将其保存到预先设定的某个目录下。 方法 import web urls = ('/upload', 'Upload') class Upload: def GET(self): web.header("Content-Type","text/html; charset=utf-8") return """<html><head></he

  • 我想上传文件并保存到特定的目录。而且我对文件的概念是陌生的。当我从我的页面上传文件时,它们保存在另一个目录(C:\ Users \ ROOTCP ~ 1 \ AppData \ Local \ Temp \ multipartbody 989135345617811478 astemporary file)而不是指定的目录。我不能设置它。请帮我找到解决办法。提前感谢所有的帮助。 上传的文件为什么保

  • 我想将一个文件保存在与应用程序jar文件所在位置相同的文件夹中。我正试图通过以下方式实现这一点: 如果我从Intellij运行项目,返回的路径是,如果我从命令行运行它,返回的路径是。这是应该发生的吗?另外,如何才能将jar路径仅返回到?

  • 问题内容: 我一直试图将纯文本文件保存到Android上Google云端硬盘中的特定文件夹中。 到目前为止,使用Google云端硬盘上的文档和快速入门指南,我已经可以做一些正确的事情,首先,我可以创建一个纯文本文件: 我已经能够使用以下命令在Google云端硬盘的基本目录中创建一个新文件夹: 我还能够通过以下方式列出用户的Google云端硬盘帐户中的所有文件夹: 但是,我对如何将文本文件保存到Go