Java HTTPPost请求与Apache HttpClient

我需要一个Java程序来生成以下请求。我正在使用Apache HttpClient库，但仍然无法产生像这样的请求：Java HTTPPost请求与Apache HttpClient

这是我的python程序生成的，我写了一个等价的java程序。但其投掷403.

2012-09-10 15：12：05G INFO：G2OAuth验证数据=“3，0.0.0.0，0.0.0.0，1347289925，3223833979，crlakamai” 2012-09-10 15： 12：05G INFO：G2OAuth sign string =“3，0.0.0.0，0.0.0.0，1347289925，3223833979，akamai/182228 \ nx-akamai-acs-action：version = 1 & action = dir & format = xml \ n”

send: 'POST /182228 HTTP/1.1\r\nHost: crl.api.akamailab.com\r\nAccept-Encoding: identity\r\nX-Akamai-ACS-Auth-Data: 3, 0.0.0.0, 0.0.0.0, 1347289925, 3223833979, crlsymc\r\nX-Akamai-ACS-Auth-Sign: eFnWtJBIyj4rxV3V0axF3w==\r\nX-Akamai-ACS-Action: version=1&action=dir&format=xml\r\n\r\n' 

reply: 'HTTP/1.1 200 OK\r\n' 
header: Server: Apache 
header: Content-Type: text/html 
header: Date: Mon, 10 Sep 2012 15:12:09 GMT 
header: Content-Length: 31 
header: Connection: keep-alive

的回应是这样的：

<?xml version="1.0" encoding="ISO-8859-1"?> 
<stat directory="/182232"> 
        <file type="file" name="log4j.properties" mtime="1346780907" size="301" md5="c92268157f1732a05c2027d151fc539a"/> 
</stat>

这是我的Java代码：

final HttpHost targetHost = new HttpHost("a.host.com", 80, "http"); 
    final DefaultHttpClient httpClient = new DefaultHttpClient(); 
    final Credentials credentials = new UsernamePasswordCredentials("user","pass"); 
    httpClient.getCredentialsProvider().setCredentials(new AuthScope(targetHost.getHostName(), targetHost.getPort()), credentials); 


    final HttpPost httpPostRequest = new HttpPost("akamai/182232"); 

    //Add your Data 
    final List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(6); 
    nameValuePairs.add(new BasicNameValuePair("Host: ", "a.host.com"); 
    nameValuePairs.add(new BasicNameValuePair("Accept-Encoding: ", "identity")); 
    nameValuePairs.add(new BasicNameValuePair("Content-Length: ", "6")); 

    httpPostRequest.setEntity(new UrlEncodedFormEntity(nameValuePairs)); 

    final HttpResponse response = httpClient.execute(targetHost, httpPostRequest); 

    if (response.getStatusLine().getStatusCode() != 200) { 
     throw new RuntimeException("Failed : HTTP error code : " + response.getStatusLine().getStatusCode()); 
    }

我的回复如下所示。

2012-09-10 11:31:22,600 DEBUG [wire] >> "POST /182228/a.crl HTTP/1.1[\r][\n]" 
2012-09-10 11:31:22,601 DEBUG [wire] >> "Content-Length: 394[\r][\n]" 
2012-09-10 11:31:22,601 DEBUG [wire] >> "Content-Type: application/x-www-form-urlencoded; charset=ISO-8859-1[\r][\n]" 
2012-09-10 11:31:22,601 DEBUG [wire] >> "Host: crl.api.symclab.com:80[\r][\n]" 
2012-09-10 11:31:22,601 DEBUG [wire] >> "Connection: Keep-Alive[\r][\n]" 
2012-09-10 11:31:22,601 DEBUG [wire] >> "User-Agent: Apache-HttpClient/4.1.3 (java 1.5)[\r][\n]" 
2012-09-10 11:31:22,602 DEBUG [wire] >> "[\r][\n]"

我想Accept-Encoding另一个标题作为职位的一部分，我如何添加他们？它一直是我相信的发布请求的一部分，而不是http头。

来源

2012-09-10 ankit249

的Accept-Encoding是HTTP头的一部分，除了你误发送的POST参数的参数：

下面是如何使用HTTP客户端发送：

httpPostRequest.setHeader("Content-Length", "6"); 
httpPostRequest.setHeader("Accept-Encoding", "identity"); 
httpPostRequest.setHeader("Host", "a.host.com");

来源

2012-09-10 21:03:12 SiN

如果您要更改的HTTP头的Accept-Encoding，你常这样

httpPostRequest.setHeader("Accept-Encoding", "identity");

你的代码

final List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(6); 
nameValuePairs.add(new BasicNameValuePair("Host: ", "a.host.com"); 
nameValuePairs.add(new BasicNameValuePair("Accept-Encoding: ", "identity")); 
nameValuePairs.add(new BasicNameValuePair("Content-Length: ", "6")); 

httpPostRequest.setEntity(new UrlEncodedFormEntity(nameValuePairs));

用于不要发送值对在HTTP请求体，在头

来源

2013-08-12 13:45:36

Java HTTPPost请求与Apache HttpClient

回答

相关问题