Question

大家好我是一名初级php开发人员我正在努力将java代码转换为php ..对Java api命中并正确获取响应现在我正试图在php中使用curl http post这是我在我的任务软件公司PLZ帮帮我

我要告诉你我的java代码是否正常工作然后我的PHP代码无效并且没有解析params到那个api所以请指导我

这是我的Java代码 这是正常的，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上编码，但是没有得到回复检查它并指导我使用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

您正在从您的Java代码发布JSON。所以在PHP中也使用json（确保格式正常）：

$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，我得到了样本回复：

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

点击Api并使用http post和curl获取响应

1 个答案: