POST 请求发送 JSON 数据 Java HttpUrlConnection答案

【问题标题】：POST request send JSON data Java HttpUrlConnectionPOST 请求发送 JSON 数据 Java HttpUrlConnection
【发布时间】：2014-01-28 11:23:08
【问题描述】：

我开发了一个 Java 代码，它使用 URL 和 HttpUrlConnection 将以下 cURL 转换为 Java 代码。卷曲是：

curl -i 'http://url.com' -X POST -H "Content-Type: application/json" -H "Accept: application/json" -d '{"auth": { "passwordCredentials": {"username": "adm", "password": "pwd"},"tenantName":"adm"}}'

我已经编写了这段代码，但它总是给出 HTTP 代码 400 错误请求。我找不到丢失的东西。

String url="http://url.com";
URL object=new URL(url);

HttpURLConnection con = (HttpURLConnection) object.openConnection();
con.setDoOutput(true);
con.setDoInput(true);
con.setRequestProperty("Content-Type", "application/json");
con.setRequestProperty("Accept", "application/json");
con.setRequestMethod("POST");

JSONObject cred   = new JSONObject();
JSONObject auth   = new JSONObject();
JSONObject parent = new JSONObject();

cred.put("username","adm");
cred.put("password", "pwd");

auth.put("tenantName", "adm");
auth.put("passwordCredentials", cred.toString());

parent.put("auth", auth.toString());

OutputStreamWriter wr = new OutputStreamWriter(con.getOutputStream());
wr.write(parent.toString());
wr.flush();

//display what returns the POST request

StringBuilder sb = new StringBuilder();  
int HttpResult = con.getResponseCode(); 
if (HttpResult == HttpURLConnection.HTTP_OK) {
    BufferedReader br = new BufferedReader(
            new InputStreamReader(con.getInputStream(), "utf-8"));
    String line = null;  
    while ((line = br.readLine()) != null) {  
        sb.append(line + "\n");  
    }
    br.close();
    System.out.println("" + sb.toString());  
} else {
    System.out.println(con.getResponseMessage());  
}

【问题讨论】：

Java 冗长的漂亮插图。

标签： java json post curl httpurlconnection

【解决方案1】：

您的 JSON 不正确。而不是

JSONObject cred = new JSONObject();
JSONObject auth=new JSONObject();
JSONObject parent=new JSONObject();
cred.put("username","adm");
cred.put("password", "pwd");
auth.put("tenantName", "adm");
auth.put("passwordCredentials", cred.toString()); // <-- toString()
parent.put("auth", auth.toString());              // <-- toString()

OutputStreamWriter wr= new OutputStreamWriter(con.getOutputStream());
wr.write(parent.toString());

写

JSONObject cred = new JSONObject();
JSONObject auth=new JSONObject();
JSONObject parent=new JSONObject();
cred.put("username","adm");
cred.put("password", "pwd");
auth.put("tenantName", "adm");
auth.put("passwordCredentials", cred);
parent.put("auth", auth);

OutputStreamWriter wr= new OutputStreamWriter(con.getOutputStream());
wr.write(parent.toString());

因此，JSONObject.toString() 应该只为外部对象调用一次。

另一件事（很可能不是你的问题，但我想提一下）：

为确保不遇到编码问题，你应该指定编码，如果不是UTF-8:

con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
con.setRequestProperty("Accept", "application/json");

// ...

OutputStream os = con.getOutputStream();
os.write(parent.toString().getBytes("UTF-8"));
os.close();

【讨论】：

在我的情况下，设置请求属性的内容类型至关重要：con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
没有什么对我有用。我正在发送输入，但在 API 端我收到的是空白。

【解决方案2】：

private JSONObject uploadToServer() throws IOException, JSONException {
            String query = "https://example.com";
            String json = "{\"key\":1}";

            URL url = new URL(query);
            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            conn.setConnectTimeout(5000);
            conn.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
            conn.setDoOutput(true);
            conn.setDoInput(true);
            conn.setRequestMethod("POST");

            OutputStream os = conn.getOutputStream();
            os.write(json.getBytes("UTF-8"));
            os.close();

            // read the response
            InputStream in = new BufferedInputStream(conn.getInputStream());
            String result = org.apache.commons.io.IOUtils.toString(in, "UTF-8");
            JSONObject jsonObject = new JSONObject(result);


            in.close();
            conn.disconnect();

            return jsonObject;
    }

【讨论】：

【解决方案3】：

您可以使用此代码使用 http 和 json 进行连接和请求

try {
         
        URL url = new URL("https://www.googleapis.com/youtube/v3/playlistItems?part=snippet"
                + "&key="+key
                + "&access_token=" + access_token);
        HttpURLConnection conn = (HttpURLConnection) url.openConnection();
        conn.setDoOutput(true);
        conn.setRequestMethod("POST");
        conn.setRequestProperty("Content-Type", "application/json");
 
        String input = "{ \"snippet\": {\"playlistId\": \"WL\",\"resourceId\": {\"videoId\": \""+videoId+"\",\"kind\": \"youtube#video\"},\"position\": 0}}";
 
        OutputStream os = conn.getOutputStream();
        os.write(input.getBytes());
        os.flush();
 
        if (conn.getResponseCode() != HttpURLConnection.HTTP_CREATED) {
            throw new RuntimeException("Failed : HTTP error code : "
                + conn.getResponseCode());
        }
 
        BufferedReader br = new BufferedReader(new InputStreamReader(
                (conn.getInputStream())));
 
        String output;
        System.out.println("Output from Server .... \n");
        while ((output = br.readLine()) != null) {
            System.out.println(output);
        }
 
        conn.disconnect();
 
      } catch (MalformedURLException e) {
 
        e.printStackTrace();
 
      } catch (IOException e) {
 
        e.printStackTrace();
 
     }

【讨论】：

【解决方案4】：

正确答案很好，但是

OutputStreamWriter wr= new OutputStreamWriter(con.getOutputStream());
wr.write(parent.toString());

不适合我，而不是它，使用：

byte[] outputBytes = rootJsonObject.getBytes("UTF-8");
OutputStream os = con.getOutputStream();
os.write(outputBytes);

【讨论】：

它对你不起作用，因为你忘记关闭 OutputStreamWriter

【解决方案5】：

我有一个类似的问题，我得到了 400，只有 PUT 的错误请求，而 POST 请求完全没问题。

下面的代码在 POST 上运行良好，但对 PUT 发出了 BAD Request：

conn.setRequestProperty("Content-Type", "application/json");
os.writeBytes(json);

进行以下更改后，POST 和 PUT 都可以正常工作

conn.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
os.write(json.getBytes("UTF-8"));

【讨论】：

【解决方案6】：

这是完整的代码和解决方案

PostJSONWithHttpURLConnection.java 类

import android.util.Log;
import org.json.JSONException;
import org.json.JSONObject;
import java.io.BufferedInputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.net.HttpURLConnection;
import java.net.URL;

public class PostJSONWithHttpURLConnection {
    String charset = "UTF-8";
    HttpURLConnection con;
    URL urlObj;
    JSONObject jObj = null;
    StringBuilder result;

    public JSONObject makeHttpRequest(String url,
                                      String paramsJSON) {
        try {
            urlObj = new URL(url);
            con = (HttpURLConnection) urlObj.openConnection();
            con.setRequestMethod("POST");
            con.setRequestProperty("Content-Type", "application/json; utf-8");
            con.setRequestProperty("Accept", "application/json");

            con.setDoOutput(true);
            con.setReadTimeout(60000);
            con.setConnectTimeout(60000);

            try (OutputStream os = con.getOutputStream()) {
                byte[] input = paramsJSON.getBytes(charset);
                os.write(input, 0, input.length);
            }

            int code = con.getResponseCode();
            Log.d("HTTP CODE", String.valueOf(code));
        } catch (IOException e) {
            e.printStackTrace();
        }

        try {
            //Receive the response from the server
            InputStream in = new BufferedInputStream(con.getInputStream());
            BufferedReader reader = new BufferedReader(new InputStreamReader(in));
            result = new StringBuilder();
            String line;
            while ((line = reader.readLine()) != null) {
                result.append(line);
            }

            Log.d("JSON Parser", "result: " + result.toString());

        } catch (IOException e) {
            e.printStackTrace();
        }

        con.disconnect();

        // try parse the string to a JSON object
        try {
            jObj = new JSONObject(result.toString());
        } catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data " + e.toString());
        }

        return jObj;
    }
}

doInBackground(String... strArr) 上代码的使用

ObjectMapper objectMapper = new ObjectMapper();
objectMapper.configure(SerializationFeature.FAIL_ON_EMPTY_BEANS, false);
String writeValueAsString = objectMapper.writeValueAsString(jSONObject);   
PostJSONWithHttpURLConnection jsonPOST = new PostJSONWithHttpURLConnection();
JSONObject json = jsonPOST.makeHttpRequest(SERVER_ADDRESS + "YOUR_API", writeValueAsString);

jSONObject 是一个 Json Like：

{
"Password":"PASSWORD",
"FullName":"Full Name",
"Username":"USER_NAME"
}

【讨论】：