Question

php工作有奇怪的行为。我通过POST传递json数据并期望数据出现在php://input中。而不是它使用奇怪的键/值对填充$ _POST。这是curl call

$process = curl_init("https://www.myurl.com/script");
curl_setopt($process, CURLOPT_SSL_VERIFYPEER, FALSE);
curl_setopt($process, CURLOPT_SSL_VERIFYHOST, FALSE);


$params = '{"Response":{"Data":{"RMID":"0910403545","QID":"965102499460","RspCode":"000","RspDesc":"Successful Transaction Complete","TrxID":"61801","TrxStatus":"COMPLETE","BID":"61801","TrxRC":"4201","TrxTime":"2015-06-15 14:53:51","TrxValue":"9"}},"Signature":"5bf094adb23e40e1de135c055684dd2098ab18d0","Certificate":"-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1

5bf094adb23e40e1de135c055684dd2098ab18d0
-----BEGIN PGP SIGNATURE-----
Version: GnuPG v1.4.5 (GNU/Linux)

iD8DBQFVfoSPQ8sozXgiBRURArdkAKCwx5ggf5dE+djAAMIDsHaapLRgOACePyys
a5scG9GhRFDefGz5CLHrqfI=
=AWpR
-----END PGP SIGNATURE-----
"}';

curl_setopt($process, CURLOPT_POST, true);
curl_setopt($process, CURLOPT_POSTFIELDS, $params);
curl_setopt($process, CURLOPT_RETURNTRANSFER, 1);

$return = curl_exec($process);

另一方面，我得到了

echo "<pre>";
var_export($_POST);
---------------------------------------------
array (
  '{"Response":{"Data":{"RMID":"0910403545","QID":"965102499460","RspCode":"000","RspDesc":"Successful_Transaction_Complete","TrxID":"61801","TrxStatus":"COMPLETE","BID":"61801","TrxRC":"4201","TrxTime":"2015-06-15_14:53:51","TrxValue":"9"}},"Signature":"5bf094adb23e40e1de135c055684dd2098ab18d0","Certificate":"-----BEGIN_PGP_SIGNED_MESSAGE-----
Hash:_SHA1

5bf094adb23e40e1de135c055684dd2098ab18d0
-----BEGIN_PGP_SIGNATURE-----
Version:_GnuPG_v1_4_5_(GNU/Linux)

iD8DBQFVfoSPQ8sozXgiBRURArdkAKCwx5ggf5dE_djAAMIDsHaapLRgOACePyys
a5scG9GhRFDefGz5CLHrqfI' => '
=AWpR
-----END PGP SIGNATURE-----
"}',
)

为什么这样做？

Answer 1

关于curl_setopt($process, CURLOPT_POSTFIELDS, $params);

此参数既可以作为urlencoded字符串传递，例如'para1 = val1＆amp; para2 = val2＆amp; ...'，也可以作为数组传递。（ref）

cURL期待一个参数/值对，并且您只提供了一个字符串，因此cURL假定参数是第一个=符号和值是之后的所有内容，直到遇到第一个&符号，没有任何符号。这就是为什么$_POST在转储它时会这样看的原因。

缓解此问题的一种方法是对$params字符串进行urlencode并将其传递到curl_setopt "data=".rawurlencode($params)，然后使用rawurldecode($_POST["data"]);检索它。

php填充$ _POST数据而不是php：// input

1 个答案: