Android上的HTTP POST请求 [英] HTTP POST request on Android

问题描述

在过去的两小时，我一直试图做一个POST请求这个网页。的http：/ /www.halebop.se/butik/byt_behall_nummer/ 并试图向其发送的 numberToPort 。但是我得到了一堆饼干和一个302移动暂时回来了。

所有我想要做的就是发送一个带有编号的POST请求，并获得最后一页回来。在iOS上，我做到这一点使用ASIHTT prequest它处理重定向和饼干。

的iOS code：

 的NSString * halebopURLString = @http://www.halebop.se/kontantkort/byt_behall_nummer/#
ASIFormDataRequest *请求= [ASIFormDataRequest requestWithURL：[NSURL URLWithString：halebopURLString]];
[要求setPostValue：halebopNumber forKey：@numberToPort];
[要求setPostValue：@继续forKey：@行动];
[要求setPostValue：@提交forKey：@提交];
[请求startSynchronous]
 

我怎么做这在Android？

作为替代方案，一个PHP的解决方案是可接受的。

编辑：试过，它没有给输出及无异常。我有上网权限。预期结果：发送POST，获得302和饼干回来，发送cookie从302到URL并获得HTML背面（经过与萤火虫），但是我什么也得不到。

  {尝试
        InputStream的myInputStream = NULL;
        URL网址;
        URL =新的URL（http://www.halebop.se/kontantkort/byt_behall_nummer/#）;
        HttpURLConnection的康恩=（HttpURLConnection类）url.openConnection（）;
        conn.setDoOutput（真）;
        conn.setInstanceFollowRedirects（真）;
        conn.setRequestMethod（POST）;
        OutputStreamWriter WR =新OutputStreamWriter（conn.getOutputStream（））;
        wr.write（numberToPort =+ N +＆放大器;行动=继续和放大器;提交=提交）;
        wr.flush（）;
        myInputStream = conn.getInputStream（）;
        wr.close（）;        RD的BufferedReader =新的BufferedReader（新的InputStreamReader（myInputStream），4096）;
        串线;
        StringBuilder的sbResult =新的StringBuilder（）;
        而（（行= rd.readLine（））！= NULL）{
            sbResult.append（线）;
            Log.d（TAG，线+线）;
        }
        rd.close（）;
        串contentOfMyInputStream = sbResult.toString（）;
        Log.d（TAG，输出+ contentOfMyInputStream）;
    }赶上（例外五）{
        Log.d（TAG，e.getMessage（））;
    }
 

解决方案

下面是如何设置岗位参数：

  HttpPost httpost =新HttpPost（LOGIN_URL）;
    清单＆LT;＆的NameValuePair GT; nvps =新的ArrayList＆LT;＆的NameValuePair GT;（）;
    nvps.add（新BasicNameValuePair（测试1，测试1））;
    nvps.add（新BasicNameValuePair（test2的，测试2））;            httpost.setEntity（新UrlEn codedFormEntity（nvps，HTTP.UTF_8））;    响应= GETRESPONSE（httpost）;
 

这里是详细解释关于code。

我还解释如何从你的回答中检索的HTTP cookies，并将它们设置成要求的这里

For the last two hours i've been trying to make a POST request to this page http://www.halebop.se/butik/byt_behall_nummer/ and tried to send numberToPort. However i get a bunch of cookies and a 302 moved temporarily back.

All i want to do is send the POST request with the number and get the final page back. On iOS, i do this using ASIHTTPRequest which handles the redirect and cookies.

iOS code:

NSString *halebopURLString = @"http://www.halebop.se/kontantkort/byt_behall_nummer/#";
ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:[NSURL URLWithString:halebopURLString]];
[request setPostValue:halebopNumber forKey:@"numberToPort"];
[request setPostValue:@"continue" forKey:@"action"];
[request setPostValue:@"submit" forKey:@"submit"];
[request startSynchronous];

How do i do this on Android?

As an alternative, a PHP solution is acceptable.

Edit: Tried this, it gives no output and no exceptions. I have the internet permission. Expected result: Send POST, get 302 and cookies back, send cookies to URL from 302 and get HTML back (Checked with FireBug) however i get nothing.

try {
        InputStream myInputStream =null;
        URL url;
        url = new URL("http://www.halebop.se/kontantkort/byt_behall_nummer/#");
        HttpURLConnection conn = (HttpURLConnection) url.openConnection();
        conn.setDoOutput(true);
        conn.setInstanceFollowRedirects(true);
        conn.setRequestMethod("POST");
        OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
        wr.write("numberToPort="+n+"&action=continue&submit=submit");
        wr.flush();
        myInputStream = conn.getInputStream();
        wr.close();

        BufferedReader rd = new BufferedReader(new InputStreamReader(myInputStream), 4096);
        String line;
        StringBuilder sbResult =  new StringBuilder();
        while ((line = rd.readLine()) != null) {
            sbResult.append(line);
            Log.d(TAG, "Line "+line);
        }
        rd.close();
        String contentOfMyInputStream = sbResult.toString();
        Log.d(TAG, "Output "+contentOfMyInputStream);
    } catch (Exception e) {
        Log.d(TAG,e.getMessage());
    }

解决方案

Here is how you can set the post parameters:

            HttpPost httpost = new HttpPost(LOGIN_URL);
    List <NameValuePair> nvps = new ArrayList <NameValuePair>();
    nvps.add(new BasicNameValuePair("test1","test1" ));
    nvps.add(new BasicNameValuePair("test2", "test2" ));

            httpost.setEntity(new UrlEncodedFormEntity(nvps, HTTP.UTF_8));

    response = getResponse(httpost);

Here is the detailed explanation about the code.

I also explained How to retrieve HTTP cookies from your response and set them into request here

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