在Tomcat中7 [重复]从多形式的Servlet GET参数(Servlet get param

这个问题已经在这里有一个答案：

• 如何使用JSP / Servlet来上传文件到服务器？ 12个回答

将项目和需要的文件上传。 所以，我使用ENCTYPE = “的multipart / form-data的” 通知。

但我无法从请求解析参数。 我也试过getPart但它返回一个空字符串。

servlet代码 - >

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;

public class upload_wall extends HttpServlet {
    private static final long serialVersionUID = 1L;
    // location to store file uploaded
    private static final String UPLOAD_DIRECTORY = "image_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

    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        response.setContentType("text/html");
        PrintWriter out = response.getWriter();

        // 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_DIRECTORY;

        // creates the directory if it does not exist
        File uploadDir = new File(uploadPath);
        if (!uploadDir.exists()) {
            uploadDir.mkdir();
        }

        try {
            @SuppressWarnings("unchecked")
            List<FileItem> formItems = upload.parseRequest(request);
            String fileName1 = "";
            if (formItems != null && formItems.size() > 0) {
                for (FileItem item : formItems) {
                    // processes only fields that are not form fields
                    if (!item.isFormField()) {
                        String fileName = new File(item.getName()).getName();
                        fileName1 += fileName;
                        String filePath = uploadPath + File.separator + fileName;
                        File storeFile = new File(filePath);
                        // saves the file on disk
                        item.write(storeFile);
                    }
                }
            }
            String p_text = request.getParameter("p_data");
            String gallery_nm = request.getParameter("upload_wall_gallery");
            out.println(p_text);
            out.println(gallery_nm);
            //out.println("Upload has been done successfully!"+fileName1);
        } catch (Exception ex) {
            out.println(ex.getMessage());
        }

    }

}

我无法找到网或计算器的工作示例。

任何简单的解决方案？

Answer 1:

问题就在这里：

String p_text = request.getParameter("p_data");
String gallery_nm = request.getParameter("upload_wall_gallery");

从如何使用JSP / Servlet的文件上传到服务器？ 答案 ，你应该把你的参数FileItem这isFormField()方法返回true 。 从发布的答案相关的代码片段：

for (FileItem item : formItems) {
    if (item.isFormField()) {
        // Process regular form field (input type="text|radio|checkbox|etc", select, etc).
        String fieldname = item.getFieldName();
        String fieldvalue = item.getString();
        // ... (do your job here)
    } else {
        // Process form file field (input type="file").
        String fieldname = item.getFieldName();
        String filename = FilenameUtils.getName(item.getName());
        InputStream filecontent = item.getInputStream();
        // ... (do your job here)
    }
}

解决方案：移动你的其他请求参数处理给else打电话时，部分if (!item.isFormField())

for (FileItem item : formItems) {
    // processes only fields that are not form fields
    if (!item.isFormField()) {
        String fileName = new File(item.getName()).getName();
        fileName1+=fileName;
        String filePath = uploadPath + File.separator + fileName;
        File storeFile = new File(filePath);
        // saves the file on disk
        item.write(storeFile);
    } else {
        //here...
        String fieldname = item.getFieldName();
        String fieldvalue = item.getString();
        if (fieldname.equals("p_data")) {
            //logic goes here...
        } else if (fieldname.equals("upload_wall_gallery")) {
            //next logic goes here...
        }
    }
}

Answer 2:

由于它是一个multipart/form-data形成一个呼叫request.getParameter()总是会返回null。 在这种情况下，你必须做的是

 if (item.isFormField()) {
     System.out.println("Got a form field: " + item.getFieldName()+ " " +item.getString());
 } else {
     // it is a file process, the way you are doing it
 }

我希望每一件事应该可以正常工作。

文章来源: Servlet get parameter from multipart form in tomcat 7 [duplicate]