HTTPURLConnection - 400错误请求

我检查了这个Post。但我不明白什么是逻辑错误。我仍然收到这个错误。我试图转储模拟器流量。但是我仍然不明白问题的症结所在。HTTPURLConnection - 400错误请求

从流量转储中，这是Android向服务器发送的请求。你也可以看到反应：

GET /Authenticate/ HTTP/1.1 
Authorization: Basic asdfasdfasdfas 

Accept-Charset: UTF-8 
Host: www.domain.com 
User-Agent: Dalvik/1.4.0 (Linux; U; Android 2.3.3; sdk Build/GRI34) 
Connection: Keep-Alive 
Accept-Encoding: gzip 

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Date: Thu, 13 Sep 2012 04:47:42 GMT 
Server: Apache/2.2.15 (CentOS) 
Content-Length: 310 
Connection: close 
Content-Type: text/html; charset=iso-8859-1 

<!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.0//EN"> 
<html><head> 
<title>400 Bad Request</title> 
</head><body> 
<h1>Bad Request</h1> 
<p>Your browser sent a request that this server could not understand.<br /> 
</p> 
<hr> 
<address>Apache/2.2.15 (CentOS) Server at www.domain.com Port 80</address> 
</body></html> 
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    text/html; charset=iso-8859-1Bad Request

我不知道那些额外的字符是什么意思。但我试图从中找出问题。

这是基本的代码：

String credentials = username + ":" + password; 
byte[] toencode = null; 
try { 
    toencode = credentials.getBytes("UTF-8"); 
} catch (UnsupportedEncodingException e1) { 
    e1.printStackTrace(); 
} 
HttpURLConnection conn = null; 
try { 
    //Utilities.isNetworkAvailable(context); 
    URL url = new URL(params[0]); 
    Log.d(params[0],"UR"); 
    conn = (HttpURLConnection) url.openConnection(); 
    conn.setRequestProperty("Authorization", "Basic " + Base64.encodeToString(toencode, Base64.DEFAULT)); 
    conn.setRequestProperty("Accept-Charset", "UTF-8"); 
    conn.setRequestProperty("Host", "www.domain.com"); 
    conn.setRequestMethod("GET"); 
    conn.setDoOutput(true); 

    String data = conn.getInputStream().toString(); 
    return data; 
}

任何想法？

更新

我检查网络服务器日志，看是否请求被击中服务器，如果有一个与请求的任何问题。这是我从错误日志中看到的：

[Thu Sep 13 10:05:24 2012] [error] [client 91.222.195.132] client sent HTTP/1.1 request without hostname (see RFC2616 section 14.23): /Authenticate/ 
[Thu Sep 13 23:11:57 2012] [error] [client 91.222.195.132] client sent HTTP/1.1 request without hostname (see RFC2616 section 14.23): /Authenticate/ 
[Thu Sep 13 23:12:03 2012] [error] [client 91.222.195.132] client sent HTTP/1.1 request without hostname (see RFC2616 section 14.23): /Authenticate/

但是，我正在设置请求的标头属性。

任何想法？

来源

2012-09-13 Kevin Rave

你的URL的价值是什么？ – Lucifer

其http://www.domain.com/Authenticate/ –

我想通了这个自己。它与设置标题的顺序有关。

编辑：我使用的顺序。

URL url = new URL(strUrl); 
conn = (HttpURLConnection) url.openConnection(); 
conn.setRequestMethod("GET"); 
conn.setRequestProperty("Host", "myhost.com"); 
conn.setRequestProperty("Authorization", "Basic " + Base64.encodeToString(toencode, Base64.DEFAULT)); 
conn.setRequestProperty("User-Agent", "Mozilla/5.0 (Macintosh; U; PPC; en-US; rv:1.3.1)"); 
conn.setRequestProperty("Accept-Charset", "UTF-8"); 

conn.setConnectTimeout (5000) ; 
conn.setDoOutput(true); 
conn.setDoInput(true);

来源

2012-09-21 21:58:09

正确的顺序是？ – wangqi060934

更新了我的答案！ –

据我所知，setDoOutput覆盖了他正在试图用“POST”做出的“GET”......这样说，我不知道它是否工作肯定... – jcaruso

试试这个

static final String _url = "http://www.google.com"; 
static final String charset = "UTF-8"; 

// to build the query string that will send the message 
private static String buildRequestString(String param1, 
     String param2, String param3, String param4, String param5) 
     throws UnsupportedEncodingException { 
    String[] params = new String[5]; //customize this as per your need 
    params[0] = param1; 
    params[1] = param2; 
    params[2] = param3; 
    params[3] = param4; 
    params[4] = param5; 

    String query = String.format(
      "uid=%s&pwd=%s&msg=%s&phone=%s&provider=%s", 
      URLEncoder.encode(params[0], charset), 
      URLEncoder.encode(params[1], charset), 
      URLEncoder.encode(params[2], charset), 
      URLEncoder.encode(params[3], charset), 
      URLEncoder.encode(params[4], charset)); 
    return query; 

} 

public static void doSomething(String param1, String param2, 
     String param3, String param4, String param5) throws Exception { 
    // To establish the connection and perform the post request 
    URLConnection connection = new URL(_url 
      + "?" 
      + buildRequestString(param1, param2, param3, param4, 
        param5)).openConnection(); 
    connection.setRequestProperty("Accept-Charset", charset); 
    // This automatically fires the request and we can use it to determine 
    // the response status 
    InputStream response = connection.getInputStream(); 
    BufferedReader br = new BufferedReader(new InputStreamReader(response)); 
    // This stores the response code. 
    // Any call to br.readLine() after this is null. 
    responsecode = br.readLine(); 
    // And this logs the already stored response code 
    Log.d("ServerResponse", responsecode); 
    responseInt = Integer.valueOf(responsecode).intValue(); 
}

来源

2012-09-13 05:10:20

我明白你想说什么。但我的网址没有任何查询字符串。它只是直接http://www.domain.com/Authenticate/与正在形成正确的基本身份验证标题。 –

然后你可以使用这个查询没有任何参数。只要提及你的网址'static final String _url =“<你的URL HERE>”; ' –

没用。它已经是一个字符串。 –

HTTPURLConnection - 400错误请求

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