如何在PHP中回显cURL结果？

问题描述：

这里是一个网址→ https://ajax.googleapis.com/ajax/services/search/images?v=1.0&q=snoopy&rsz=1

这是我的PHP代码

<?php
    $url = "https://ajax.googleapis.com/ajax/services/search/images?"."v=1.0&q=snoopy";

    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    $body = curl_exec($ch);
    curl_close($ch);

    $json = json_decode($body);
?>

我想从结果中回显所有图像网址

（对于例如： http://img2.wikia.nocookie.net /__cb20110331075248/peanuts/images/6/62/Snoopy.gif ）

I wanna echo all of the image urls from its result
(For example: http://img2.wikia.nocookie.net/__cb20110331075248/peanuts/images/6/62/Snoopy.gif )

但是我不知道如何回显它们。

希望有人可以帮助我，谢谢！

But I have no idea how to echo them.
Hope someone could help me, thanks!

答

也许像这样：

<?php
    $url = "https://ajax.googleapis.com/ajax/services/search/images?" .
   "v=1.0&q=barack%20obama&userip=INSERT-USER-IP";

   // sendRequest
   // note how referer is set manually
   $ch = curl_init();
   curl_setopt($ch, CURLOPT_URL, $url);
   curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
   curl_setopt($ch, CURLOPT_REFERER, 'http://example.com');
   $body = curl_exec($ch);
   curl_close($ch);

   // now, process the JSON string
    $pics = json_decode($body);
    $ret='';
    if($pics){
       foreach($pics->responseData->results as $pic)
          $ret .= $pic->url.'<br>';
    }
    echo $ret;
?>

如何在PHP中回显cURL结果？

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