如何形成以下命令行PHP卷曲的要求吗？ [英] How to form PHP curl request for the following command line?

问题描述

有人可以帮助我形成在PHP中，这commndline要求？

$卷曲-u [用户名：密码] https://subs.pinpayments.com/api/v4/meresheep/subscribers/7388.xml

我曾尝试以下，并只得到'假'回来......

  //查询用户引脚的细节支付...
 $ curl_url =htt​​ps://subs.pinpayments.com/api/v4/xyz​​-site/subscribers/32.xml; //打开连接
 $ CH = curl_init（）; //设置URL，POST数瓦尔，POST数据
 curl_setopt（$ CH，CURLOPT_URL，$网址）;
 curl_setopt（$ CH，CURLOPT_RETURNTRANSFER，1）;
 curl_setopt（$ CH，CURLOPT_SSL_VERIFYPEER，FALSE）;
 curl_setopt（$ CH，CURLOPT_HTTPHEADER，阵列（'授权：-u [用户名：密码]'））; //实行岗位
 $结果= curl_exec（$ CH）;
 curl_close（$ CH）;
 

解决方案

您需要更改 CURLOPT_HTTPHEADER 行：

  curl_setopt（$的数据，CURLOPT_HTTPHEADER，阵列（
    授权：基本。 base64_en code（'用户名：密码'）
））;
 

Can someone help me to form this commndline request in PHP ?

$ curl -u [USER:PASSWORD] https://subs.pinpayments.com/api/v4/meresheep/subscribers/7388.xml

I have tried the following and getting only 'false' returned...

 // Query the user to pin payments for the details...
 $curl_url = "https://subs.pinpayments.com/api/v4/xyz-site/subscribers/32.xml";

 //open connection
 $ch = curl_init();

 //set the url, number of POST vars, POST data
 curl_setopt($ch, CURLOPT_URL, $url);
 curl_setopt($ch, CURLOPT_RETURNTRANSFER,1);
 curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
 curl_setopt($ch, CURLOPT_HTTPHEADER, array('Authorization: -u [USER:PASSWORD]'));

 //execute post
 $result = curl_exec($ch);
 curl_close($ch);

解决方案

You need to change the CURLOPT_HTTPHEADER line:

curl_setopt($data, CURLOPT_HTTPHEADER, array(
    'Authorization: Basic ' . base64_encode('username:password')
));

这篇关于如何形成以下命令行PHP卷曲的要求吗？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！