Android HTTPURLConnection-400错误请求
我检查一下。但我仍然不明白逻辑错误可能是什么。我仍然得到这个错误。我试图转储模拟器流量。但我还是不知道问题出在哪里 从流量转储中，这是Android作为请求发送到服务器的内容。您也可以看到响应：Android HTTPURLConnection-400错误请求,android,http,bad-request,Android,Http,Bad Request,我检查一下。但我仍然不明白逻辑错误可能是什么。我仍然得到这个错误。我试图转储模拟器流量。但我还是不知道问题出在哪里 从流量转储中，这是Android作为请求发送到服务器的内容。您也可以看到响应： GET /Authenticate/ HTTP/1.1 Authorization: Basic asdfasdfasdfas Accept-Charset: UTF-8 Host: www.domain.com User-Agent: Dalvik/1.4.0 (Linux; U; And
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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PÙ[ ˜ároP"8«‹ HTTP/1.1 400 Bad Request
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
[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/
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();
}
这是我自己想出来的。这是设置标题顺序的问题 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);
我也面临这个问题，但我通过更改代码来修复它，现在我使用以下代码行
BufferedReader reader;
StringBuffer buffer;
String res = null;
try {
URL url = new URL(request_url);
HttpURLConnection con = (HttpURLConnection) url.openConnection();
con.setReadTimeout(40000);
con.setConnectTimeout(40000);
con.setRequestMethod("GET");
con.setRequestProperty("Accept","application/json");
con.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
con.setDoInput(true);
con.setDoOutput(true);
int status = con.getResponseCode();
InputStream inputStream;
if (status == HttpURLConnection.HTTP_OK) {
inputStream = con.getInputStream();
} else {
inputStream = con.getErrorStream();
}
reader = new BufferedReader(new InputStreamReader(inputStream));
buffer = new StringBuffer();
String line = "";
while ((line = reader.readLine()) != null) {
buffer.append(line);
}
res = buffer.toString();
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (ProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return res;
你的URL的价值是什么？我明白你想说什么。但是我的URL没有任何查询字符串。这只是简单的基本身份验证标头，格式正确。然后您可以使用此查询而无需任何参数。只需提及您的URL
静态最终字符串_URL=“”据我所知，setDoOutput覆盖了他试图用“POST”生成的“GET”。。。尽管如此，我不知道它是否真的有效。。。
BufferedReader reader;
StringBuffer buffer;
String res = null;
try {
URL url = new URL(request_url);
HttpURLConnection con = (HttpURLConnection) url.openConnection();
con.setReadTimeout(40000);
con.setConnectTimeout(40000);
con.setRequestMethod("GET");
con.setRequestProperty("Accept","application/json");
con.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
con.setDoInput(true);
con.setDoOutput(true);
int status = con.getResponseCode();
InputStream inputStream;
if (status == HttpURLConnection.HTTP_OK) {
inputStream = con.getInputStream();
} else {
inputStream = con.getErrorStream();
}
reader = new BufferedReader(new InputStreamReader(inputStream));
buffer = new StringBuffer();
String line = "";
while ((line = reader.readLine()) != null) {
buffer.append(line);
}
res = buffer.toString();
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (ProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return res;