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Java 和 GWT 的新手。这是我的问题。我收到此错误消息:“OutputStream 已提交,无法再写入。” 当我试图通过 REST API 将 xml 发布到远程服务器时。

它发生在下面代码的这一行:

out.write("Content-Type: application/x-www-form-urlencoded\r\n");

这适用于终端:

curl -d "OPERATION_NAME=ADD_REQUEST&TECHNICIAN_KEY=DxxxxxxxxxxxxxxxxxxxB6&INPUT_DATA=<?xml version=%221.0%22 encoding=%22utf-8%22?><Operation><Details><requester>Me</requester><subject>Test</subject><description>Testing curl input again</description></Details></Operation>" http://app.company.com/sdpapi/request/ 

我无法将上面的 curl 命令转换为下面的代码。我从教程中获得了以下代码,但我不确定如何正确传递 URL 和参数。任何建议或其他故障排除方法将不胜感激。我在谷歌上没有找到很多东西。

package com.gwt.HelpDeskTest.server;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.net.URL;
import java.net.URLConnection;
import com.google.gwt.user.server.rpc.RemoteServiceServlet;
import com.gwt.HelpDeskTest.client.HelpDeskTestService;
import com.gwt.HelpDeskTest.shared.HelpDeskTestException;

@SuppressWarnings("serial")
public class HelpDeskTestImpl extends RemoteServiceServlet implements
    HelpDeskTestService {

@Override
public String getFromRemoteServer(String serviceUrl)
        throws HelpDeskTestException {


    String result = "";

    try {
        final URL url = new URL(serviceUrl);

        final BufferedReader in= new BufferedReader(new InputStreamReader(url.openStream()));

        String inputLine;
        while ((inputLine= in.readLine()) != null) {

            result+= inputLine;
        }

        in.close();
        return result;

    } catch (final Exception e) {
        throw new HelpDeskTestException();

    }

}

@Override
public String postToRemoteServer(String serviceUrl)
        throws HelpDeskTestException {

    try {




        final String serverHost= "http://app.company.com/";
        final String serverPath= "http://app.company.com/sdpapi/request/";


        final String serverParameters= 
                "OPERATION_NAME=ADD_REQUEST&TECHNICIAN_KEY=Dxxxxxxxxxx6&INPUT_DATA=%3C?xml%20version=%25221.0%2522%20encoding=%2522utf-8%2522?%3E%3COperation%3E%3CDetails%3E%3Crequester%3EMe%3C/requester%3E%3Csubject%3ETest%3C/subject%3E%3Cdescription%3ETesting%20GWT%20input%20again%3C/description%3E%3C/Details%3E%3C/Operation%3E"; //put parameters here for testing.

        final  URL url = new URL(serverHost);

        final URLConnection connection= url.openConnection();
        connection.setDoOutput(true);
        connection.setConnectTimeout(10000); //added this to see if I can address the timeout issue.
        connection.setReadTimeout(10000);

        final OutputStreamWriter out= new OutputStreamWriter(connection.getOutputStream());

        final BufferedReader in= new BufferedReader(new InputStreamReader(
                connection.getInputStream()));

        out.write("POST " + serverPath + "\r\n");
        out.write("Host: " + serverHost + "\r\n");
        out.write("Accept-Encoding: identity\r\n");
        out.write("Connection: close\r\n");
        out.write("Content-Type: application/x-www-form-urlencoded\r\n"); //This    is where the error is occuring
        out.write("Content-Length: " + serverParameters.length() + "\r\n\r\n" +
            serverParameters + "\r\n");


        String result = "";
        String inputLine;

        while ((inputLine=in.readLine()) != null) {
            result+= inputLine;
        }

        in.close();
        out.close();

        return result;

    }  catch (final Exception e) {
        System.out.println(e.getMessage());
        throw new HelpDeskTestException();

    }


}

}

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1 回答 1

1

我认为使用 HttpURLConnection 而不是普通的 URLConnection 会更容易。它具有设置标题和其他属性的便捷方法。有关如何使用它进行 POST 的一个很好的示例,请参阅此问题的已接受答案:

Java - 通过 POST 方法轻松发送 HTTP 参数

于 2012-10-09T15:25:51.517 回答