json_encodeとjson_decodeのカスタムphp apiがjson文字列を返さない

Question

私は最近、異なるサーバー間で複数のWebサイトから自分のデータベースにデータを挿入するためのカスタムAPIを書いています...しかし、それはいくつかの問題があり、私もリモートでAPIと通信するためのクラスを作成しました...そして、これがどこであるかもしれない...ここで

は私のAPIコードの抜粋のサンプル..です

if (function_exists($_GET['method'])) { 
     // function exists, so lets run it. 
     $_GET['method'](); 
    } else { 
     // function does not exist so lets throw an error. 
     $data['respCode'] = "100"; 
     $data['respMsg'] = "The method you have called does not exist. Please reformat your call."; 
     echo json_encode($data); 
    } 

// methods 

function newProspect() { 
    // lets first check for the lead in the database.. 

    $sql = mysql_query("SELECT * FROM leads WHERE email = '".$_POST['email']."' LIMIT 1"); 
    if (mysql_num_rows($sql) >= 1) { 
     // duplicate found, so lets just send the data.. 
     $data = mysql_fetch_assoc($sql); 
     $data['respCode'] = "200"; 
     $data['respMsg'] = "Duplicate Lead. Information echoed."; 
     echo json_encode($data); 
    } else { 
     // no duplicate found, so lets insert the data.. 
     $sql = "INSERT INTO leads SET "; 
     foreach ($_POST as $key => $value) { 
      $sql .= $key." = '".$value."',"; 
     } 
     $sql .= "register_date = '".$rdate."'"; 
     $sql = mysql_query($sql); 
     if (!$sql) { 
      // could not insert the info into the database so lets throw an error. 
      $data['respCode'] = "102"; 
      $data['respMsg'] = "Could not intert into database."; 
      $data['errorMsg'] = mysql_error(); 

      echo json_encode($data); return json_encode($data); 
     } else { 
      // lead was inserted into the database just fine, so lets pass back the data. 
      $id = mysql_insert_id($sql); 
      $sql = mysql_query("SELECT * FROM leads WHERE `idleads` =".$id.""); 
      $data = mysql_fetch_assoc($id); 
      $data['respCode'] = '200'; 
      $data['respMsg'] = "Lead Entered into Database Successfully."; 

      echo json_encode($data); 
     } 
    } 
} 

可能性があり問題は..

<?php 

class theApi { 
public $apIdomain = 'http://www.domain.com/api/index.php'; // ie: https://www.mydomain.com/admin/ 

public $APData = array(); 
public $postUrl = ''; 

public function __construct() { 

} 

function method ($meth = 'newProspect') { 
    $this->postUrl = $this->apIdomain."?method=".$meth; 
    return $this; 
} 

function postData($pdata) { 
    foreach ($pdata as $key => $val) { 
     if ($key != 'submit') { 
      $this->APData[$key] = $val; 
     } 
    } 
    return $this; 
} 

function process() { 
    $this->APData['ipaddress'] = ip2long($_SERVER['REMOTE_ADDR']); 

    $ch = curl_init(); 
    curl_setopt($ch, CURLOPT_URL, $this->postUrl); 
    curl_setopt($ch, CURLOPT_HEADER, 0); 
    curl_setopt($ch, CURLOPT_POST, 1); 
    curl_setopt($ch, CURLOPT_POSTFIELDS, $this->APData); 
    curl_setopt($ch, CURLOPT_RETURNTRANSFER,0); 
    curl_setopt($ch, CURLOPT_TIMEOUT,30); 

    $rawresponse = curl_exec($ch); 

    curl_close($ch); 

    return $rawresponse; 

} 
} 

$ap = new theApi(); 

その後、私はちょうどそれをテストするために私の提出のアプリを持っている...私はhttp://www.differentdomain.com/api-test/index.php?method=newProspectに行くことによって、コードを実行すると

<?php 
$_method = $_GET['method']; 
require_once("api.php"); 
switch ($_method) { 

    case 'newProspect': 
     $data['name'] = "TestB"; 
    $data['phone'] = "555-555-5555"; 
    $data['email'] = "testB@testerdomain1.com"; 

    $d = $ap->method('newProspect')->postData($data)->process(); 
break; 

case 'updateProspect': 

break; 

case 'saveProject': 
break; 

case 'finishApplication': 
break; 
} 

は、私は、ブラウザの出力で取得：{"idleads":"1886","classid":"1","ipaddress":"-949980134","register_date":"0000-00-00 00:00:00","name":"TestB","first_name":null,"last_name":null,"phone":"555-555-5555","phone2":null,"email":"testB@testerdomain1.com","age":null,"zip":null,"traffic_source":"1","affiliateid":null,"sversion":null,"purpose":null,"amount":null,"description":null,"respCode":"200","respMsg":"Duplicate Lead. Information echoed."}は

だから、API自体は動作しています...しかし、私はどこでもjson_decodeを実行することはできませんし、CURLがAPIからすべてのデータを取得していないかのように見えます。これに関する助けは大いに評価されています。ここにそれを働かせるために。

ありがとう。変更に

新しい出力はカールする：

object(stdClass)#2 (20) { ["idleads"]=> string(4) "1886" ["classid"]=> string(1) "1" ["ipaddress"]=> string(10) "-949980134" ["register_date"]=> string(19) "0000-00-00 00:00:00" ["name"]=> string(5) "TestB" ["first_name"]=> NULL ["last_name"]=> NULL ["phone"]=> string(12) "555-555-5555" ["phone2"]=> NULL ["email"]=> string(33) "justinblacktest@testerdomain1.com" ["age"]=> NULL ["zip"]=> NULL ["traffic_source"]=> string(1) "1" ["affiliateid"]=> NULL ["sversion"]=> NULL ["purpose"]=> NULL ["amount"]=> NULL ["description"]=> NULL ["respCode"]=> string(3) "200" ["respMsg"]=> string(36) "Duplicate Lead. Information echoed." }

Answer 1

設定CURL_RETURNTRANSFER1に、むしろそれを画面にエコーよりも、HTTP要求出力を返すように。

$ch = curl_init(); 
curl_setopt($ch, CURLOPT_URL, $this->postUrl); 
curl_setopt($ch, CURLOPT_HEADER, 0); 
curl_setopt($ch, CURLOPT_POST, 1); 
curl_setopt($ch, CURLOPT_POSTFIELDS, $this->APData); 
// Set to 1 
curl_setopt($ch, CURLOPT_RETURNTRANSFER,1); 
curl_setopt($ch, CURLOPT_TIMEOUT,30); 

// now $rawresponse actually contains the JSON retrieved 
// from the API call. 
$rawresponse = curl_exec($ch); 

これは、私は下の@私のOPでのvar_dumpを実行したとき..私は新しい出力を入れてちょっと働いin the curl_exec() documentation.