如何从CURL响应PHP API的JSON字符串中删除引用，不需要的数据

Question

Guys How To Remove Unwanted Data Like Double Double Quotes And String(126) From Curl Response I Need Only Url and i am getting This Curl Respnose:

string(126)"http://www.mydomainxyz.co/router/offers/74938e12xxxxxf519d57bd63252d/pre_entry?sdk=false&uid=805xxxx03&ai_id=15xx719261"

I need Only Plain Url Please Help I Have Allready Tried Many Things But No Good Results Code Below:

    <?php

  $url_send ="https://www.mydomainxyz.co/supply_api/surveys/offer?";


 function sendPostData($url, $post){
  $ch = curl_init($url);
  curl_setopt($ch,CURLOPT_USERAGENT,'Content-type: application/json');
  curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
  curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1 );
  curl_setopt($ch, CURLOPT_POST, 1);
  curl_setopt($ch, CURLOPT_POSTFIELDS, "api_token=eca8c17xxxxxxxxx486ec8a7271&user_identifier=80xxxxxx9203");

  $jsonStr = curl_exec($ch);
  curl_close($ch);

  $json = json_decode($jsonStr, true);
  var_dump($json['offer_url']);
}

echo " " . sendPostData($url_send, $str_data);

?>

Answer 1

In your function sendPostData() your using

var_dump($json['offer_url']);

Which is where the problem comes. You really just need to return the value and this will be printed out in your echo at the end. So change the var_dump() to...

return $json['offer_url'];