如何在php中编写curl以访问php中的某些数据？

Question

我有一个卷曲代码，如下所示：

curl --request GET \
--user 60CF3Ce97nRS1Z1Wp5m9kMmzHHEh8Rkuj31QCtVxjPWGYA9FymyqsK0Enm1P6mHJf0THbR:API-P4ss \
https://api.sandbox.ewaypayments.com/Transaction/44DD7aVwPYUPemGRf7pcWxyX2FJS-0Wk7xr9iE7Vatk_5vJimEbHveGSqX52B00QsBXqbLh9mGZxMHcjThQ_ITsCZ3JxKOY88WOVsFTLPrGtHRkK0E9ZDVh_Wz326QZlNlwx2

我想将此代码编写在php文件中。 我不知道如何开始编写此代码。 谁能说出如何在php中编写curl代码？

Answer 1

尝试这个：

$url = 'API URL';

try {
    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    $APIresponse = curl_exec($ch);
    curl_close($ch);
    $res = json_decode($APIresponse, true);
    // $res contains the API response as an array
}
catch(Exception $e)
{
    // Error mesage
}

参考

Answer 2

您需要使用PHP客户端URL库 。

您的代码的解决方案：

<?php
$username = "60CF3Ce97nRS1Z1Wp5m9kMmzHHEh8Rkuj31QCtVxjPWGYA9FymyqsK0Enm1P6mHJf0THbR";
$password = "API-P4ss";
$process = curl_init("https://api.sandbox.ewaypayments.com/Transaction/44DD7aVwPYUPemGRf7pcWxyX2FJS-0Wk7xr9iE7Vatk_5vJimEbHveGSqX52B00QsBXqbLh9mGZxMHcjThQ_ITsCZ3JxKOY88WOVsFTLPrGtHRkK0E9ZDVh_Wz326QZlNlwx2");
curl_setopt($process, CURLOPT_HTTPHEADER, array('Content-Type: application/json'));
curl_setopt($process, CURLOPT_HEADER, 1);
curl_setopt($process, CURLOPT_USERPWD, $username . ":" . $password);
curl_setopt($process, CURLOPT_TIMEOUT, 30);
curl_setopt($process, CURLOPT_POST, 1);
curl_setopt($process, CURLOPT_RETURNTRANSFER, TRUE);
$return = curl_exec($process);
curl_close($process);
var_dump($return);

API的响应位于$return变量中。

Answer 3

首先使用file_get_contents php函数。

如果找不到解决方案，请尝试以下代码

//  Initiate curl
$ch = curl_init();
// Disable SSL verification
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
// Will return the response, if false it print the response
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
// Set the url
curl_setopt($ch, CURLOPT_URL,$url);
// Execute
$result=curl_exec($ch);
// Closing
curl_close($ch);

// Will dump a beauty json :3
var_dump(json_decode($result, true));