点击Api并使用http发布和curl获取响应

Question

大家好，我是一名初级php开发人员，我正在将Java代码转换为php ..在Java api上命中并正确获得响应，现在我试图在php中使用curl http post命中这是我在我的软件公司plz中的任务帮我

我要向您展示我的Java代码正确运行，然后我的php代码无法正常工作并且不解析该API的参数，因此请您指导我

这是我的Java代码，它工作正常，我想从php做同样的工作

import java.io.*;

import java.util.jar.JarException;
import org.apache.http.HttpResponse;
import org.apache.http.client.HttpClient;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.*;

class MyCode{

            public static void main(String[] args) throws JarException, JSONException
            {

                        testCustomerApiIsExposed();

            }


            public static void testCustomerApiIsExposed() throws JarException, JSONException {

               try {

                   @SuppressWarnings("deprecation")

                   HttpClient c = new DefaultHttpClient();

                   HttpPost p = new
                   HttpPost("http://link");



                    String payload = "{id:\"" + 1 + "\","  + "method:\"" + "customerApi.getApiToken" + "\", params:[\"teabonezenminddemo1partner@gmail.com\", \"demo1234!\", \"\", \"\", \"\",     \"\", \"\", false, \"\", \"\"" + "]}";

                   String mimeType="";

                   /*There is something here. What constructor are we really calling here? */

                   // p.setEntity(new StringEntity( payload,ContentType.create("application/json")));

                                    p.setEntity(new StringEntity(payload));



                   HttpResponse r = c.execute(p);



                   BufferedReader reader = new BufferedReader(new InputStreamReader(r.getEntity().getContent(), "UTF-8"));

                   StringBuilder builder = new StringBuilder();

                   for (String line = null; (line = reader.readLine()) != null;) {

                       builder.append(line).append("\n");

                   }

                   JSONTokener tokener = new JSONTokener("[" + builder.toString() + "]");

                   JSONArray finalResult = new JSONArray(tokener);

                   JSONObject o = finalResult.getJSONObject(0);


                   //Getting names of the JSON object here
                   System.out.println(o.names());

                   String apiToken = (String) o.get("result");


                   System.out.println(apiToken);

               }

               catch(IOException e) {

                   System.out.println(e);

               }

            }
}

现在我在php上对此进行编码，但是没有得到响应检查它并指导我我使用curl http post和getApiToken方法帮助我解决这个问题，我非常紧张。

这是我的PHP代码

<?php

$data = array(params);

$ch = curl_init('http://link');
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt($ch, CURLOPT_POSTFIELDS, http_build_query($data));
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);

$result = curl_exec($ch);
echo $result;
?>

Answer 1

您正在从Java代码发布JSON。 因此也可以在PHP上使用json（确保格式正确）：

$payload = "{params}";

卷曲选项将是

// as you are posting JSON, so tell server that you are sending json
curl_setopt($ch, CURLOPT_HTTPHEADER, array("Content-Type: application/json"));

// Let server know that you are doing HTTP POST request
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $payload);

使用这些，我得到了示例响应：

{"id":"1","result":"results"}