Question

我需要在链接中插入一个id变量＆＃34; https://api.quickpay.net/payments/9727866/link＆＃34; （而不是＆＃34; 9727866＆＃34;）但我似乎无法使语法正确。

无论我做什么，它似乎都不接受我的变量..谁知道怎么做？

   $params = array(
       "amount"            => 100,
       "order_id"       => 999999
   );                                                                    
   $data_string = http_build_query($params, '&'); 

   $headers = array(
          'Accept-Version: v10',
          'Accept: application/json', 
          'Authorization: Basic ' . base64_encode(":HIDDEN_API_KEY")
      );                                                                                  

   $ch = curl_init('https://api.quickpay.net/payments/9727866/link');                                                                      
   curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "PUT");                                                                     
   curl_setopt($ch, CURLOPT_POSTFIELDS, $data_string);                                                                  
   curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);      
   curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);     
   curl_setopt($ch, CURLOPT_HEADER, true);
   curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);

   $result = curl_exec($ch);

   if(!curl_exec($ch)){
       die('Error: "' . curl_error($ch) . '" - Code: ' . curl_errno($ch));
   }
   curl_close($ch);

   print_r($result);

Answer 1

这可能是因为你使用的是单引号。请尝试以下任何一种方法：

 $ch = curl_init("https://api.quickpay.net/payments/$variable/link");

或

$ch = curl_init("https://api.quickpay.net/payments/{$variable}/link");

或

$ch = curl_init("https://api.quickpay.net/payments/" . $variable . "/link");

或

$str = "https://api.quickpay.net/payments/" . $variable . "/link";
$ch = curl_init($str);

或

sprintf("https://api.quickpay.net/payments/%s/link", $variable);

无论你的船漂浮着什么。

将curible插入curl_init（）

1 个答案: