的HttpClient

Question

我正在使用Chrome插件，可以成功使我的服务工作：

enter image description here

我现在正试图在Java调用中使用同样的东西。我遇到了将Raw部分放入Java服务的问题。有什么想法吗？

  httpPost = new HttpPost(URL);
  nvps = new ArrayList<NameValuePair>();
  nvps.add(new BasicNameValuePair("Authorization", "Bearer " + token));
  nvps.add(new BasicNameValuePair("Content-Type", "application/json"));

  //I know this part is incorrect, but I don't know what to do with it
  nvps.add(new BasicNameValuePair("{\"pCode\": \"\", \"rType\": \"Sales Case\", \"subject\": \"test3\", \"description\": \"test4\", \"lookupInfo\": \"test5\", \"aaNum\": \"\"}", ""));

  try
  {
    httpPost.setEntity(new UrlEncodedFormEntity(nvps, "UTF-8"));
  } catch (UnsupportedEncodingException e)
  {
    e.printStackTrace();
  }

  HttpResponse response;
  try {
    response = client.execute(httpPost);
    BufferedReader in = new BufferedReader(new InputStreamReader(response.getEntity().getContent(), "UTF-8"));
    String builder = "";

    String line = in.readLine();
    System.out.println("line = " + line);
  } catch (ClientProtocolException e)
  {
    e.printStackTrace();
  } catch (IOException e)
  {
    e.printStackTrace();
  }
}

Answer 1

的HttpClient

我不知道您使用什么来进行http调用，但为了简单和理智，请使用apache中的httpclient。

实施例

    HttpPost post = new HttpPost(URL);

    post.setHeader("Content-type", "application/json");
    post.setHeader("Authorization", "Bearer " + token));
    post.setHeader("Cache-Control", "no-cache"));

    FileBody fileBody = new FileBody(file);
    StringBody pCode = new StringBody("SOME TYPE OF VALUE", ContentType.MULTIPART_FORM_DATA);
    StringBody rType = new StringBody("SOME TYPE OF VALUE", ContentType.MULTIPART_FORM_DATA);
    //
    MultipartEntityBuilder builder = MultipartEntityBuilder.create();
    builder.setMode(HttpMultipartMode.BROWSER_COMPATIBLE);
    builder.addPart("pCode", pCode);
    builder.addPart("rType", rType);
    HttpEntity entity = builder.build();
    try
    {
        httpPost.setEntity(new UrlEncodedFormEntity(nvps, "UTF-8"));
    }catch (UnsupportedEncodingException e)
    {
       e.printStackTrace();
    }

    HttpResponse response;
    try {
       response = client.execute(httpPost);
       BufferedReader in = new BufferedReader(new InputStreamReader(response.getEntity().getContent(), "UTF-8"));
       String builder = "";

       String line = in.readLine();
       System.out.println("line = " + line);
    } catch (ClientProtocolException e)
    {
      e.printStackTrace();
    } catch (IOException e)
    {
    e.printStackTrace();
  }

}

此代码尚未经过测试。

更新

JSON只是一个奇特的字符串（在java的角度来看）所以只需添加以下内容：

builder.addPart("someName", new StringBody(json, ContentType.TEXT_PLAIN));

更新2

上面将JSON添加为POST变量，而不是正文.. Look AT This Post

Answer 2

您正在混合标题和内容：

这些是标题：

nvps.add(new BasicNameValuePair("Authorization", "Bearer " + token));
nvps.add(new BasicNameValuePair("Content-Type", "application/json"));
nvps.add(new BasicNameValuePair("Cache-Control", "no-cache"));

这就是内容：

nvps.add(new BasicNameValuePair("{\"pCode\": \"\", \"rType\": \"Sales Case\", \"subject\": \"test3\", \"description\": \"test4\", \"lookupInfo\": \"test5\", \"aaNum\": \"\"}", ""));

在此设置内容：

httpPost.setEntity(new UrlEncodedFormEntity(nvps, "UTF-8"));

您不需要执行多部分，只需设置字符串内容而不用对其进行编码，并使用httpPost.addHeader()设置标题。

Answer 3

您必须将标题信息和数据保存在适当的位置。不要像现在这样混淆它们。

了解Http请求的结构：

一个起始线 start-line = Request-Line | Status-Line

例如。 GET /hello.htm HTTP/1.1（这是客户发送的请求行） HTTP/1.1 200 OK（这是服务器发送的状态行）

零个或多个标题字段后跟CRLF（标题信息在此处）其中message-header = field-name ":" [ field-value ]

例如。

   User-Agent: curl/7.16.3 libcurl/7.16.3 OpenSSL/0.9.7l zlib/1.2.3  
   Host: www.example.com
   Accept-Language: en, mi    
   Date: Mon, 27 Jul 2009 12:28:53 GMT
   Server: Apache
   Last-Modified: Wed, 22 Jul 2009 19:15:56 GMT
   ETag: "34aa387-d-1568eb00"
   Accept-Ranges: bytes
   Content-Length: 51
   Vary: Accept-Encoding
   Content-Type: text/plain

表示标题字段结尾的空行（即CRLF前面没有任何内容的行）
可选择邮件正文（内容信息在此处）

您可以尝试以下操作：

public void postData() {

    // Create a new HttpClient and Post Header
    HttpClient httpclient = new DefaultHttpClient();
    // pass the url as parameter and create HttpPost object.
    HttpPost post = new HttpPost(url);

    // Add header information for your request - no need to create 
    // BasicNameValuePair() and Arraylist.
    post.setHeader("Authorization", "Bearer " + token);
    post.setHeader("Content-Type", "application/json");
    post.setHeader("Cache-Control", "no-cache");    

    try {
    // put your content as follows:
    List<NameValuePair> form_data = new ArrayList<NameValuePair>(); 
    form_data.add(new BasicNameValuePair("pCode", ""));
    form_data.add(new BasicNameValuePair("rType", "Sales Case"));
    form_data.add(new BasicNameValuePair("subject", "test3"));
    form_data.add(new BasicNameValuePair("description", "test4"));
    form_data.add(new BasicNameValuePair("lookupInfo", "test5"));
    form_data.add(new BasicNameValuePair("aaNum", ""));

    // pass the content as follows:
    post.setEntity(new UrlEncodedFormEntity(form_data,
                        HTTP.UTF_8));

    // Execute HTTP Post Request
    HttpResponse response = httpclient.execute(post);

    // TODO: Process your response as you would like.

    } catch (IOException e) {
        // TODO Auto-generated catch block
    }
}

RESTful服务电话

3 个答案:

的HttpClient

实施例

更新

更新2