【问题标题】:cURL in java for instagram API用于 instagram API 的 java 中的 cURL
【发布时间】:2015-10-27 23:48:34
【问题描述】:

我如何使用 java 发布这个 cURL?

 curl -F 'client_id=CLIENT_ID' \
    -F 'client_secret=CLIENT_SECRET' \
    -F 'grant_type=authorization_code' \
    -F 'redirect_uri=AUTHORIZATION_REDIRECT_URI' \
    -F 'code=CODE' \
    https://api.instagram.com/oauth/access_token

【问题讨论】:

    标签: java api curl instagram


    【解决方案1】:

    这是您在 Java 中发布的方式(来自 here

        public static String httpPost(String urlStr, String[] paramName,
    String[] paramVal) throws Exception {
      URL url = new URL(urlStr);
      HttpURLConnection conn =
          (HttpURLConnection) url.openConnection();
      conn.setRequestMethod("POST");
      conn.setDoOutput(true);
      conn.setDoInput(true);
      conn.setUseCaches(false);
      conn.setAllowUserInteraction(false);
      conn.setRequestProperty("Content-Type",
          "application/x-www-form-urlencoded");
    
      // Create the form content
      OutputStream out = conn.getOutputStream();
      Writer writer = new OutputStreamWriter(out, "UTF-8");
      for (int i = 0; i < paramName.length; i++) {
        writer.write(paramName[i]);
        writer.write("=");
        writer.write(URLEncoder.encode(paramVal[i], "UTF-8"));
        writer.write("&");
      }
      writer.close();
      out.close();
    
      if (conn.getResponseCode() != 200) {
        throw new IOException(conn.getResponseMessage());
      }
    
      // Buffer the result into a string
      BufferedReader rd = new BufferedReader(
          new InputStreamReader(conn.getInputStream()));
      StringBuilder sb = new StringBuilder();
      String line;
      while ((line = rd.readLine()) != null) {
        sb.append(line);
      }
      rd.close();
    
      conn.disconnect();
      return sb.toString();
    }
    

    在 Java 中使用 Instagram API 的更好方法是使用其他人编写的名为 JInstagram 的 Java 库。

    【讨论】:

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