php - 使用 cURL 时数据未发布到 php 文件

Question

我正在尝试使用 PHP 创建一个简单的 API，但数据未发布到文件中。

$url = "http://localhost/api.php";
$session = curl_init();
curl_setopt($session, CURLOPT_URL, $url);
curl_setopt($session, CURLOPT_REFERER, "http://".$_SERVER['HTTP_HOST'].$_SERVER['PHP_SELF']);
curl_setopt($session, CURLOPT_POST, true);
curl_setopt($session, CURLOPT_RETURNTRANSFER, true);
//curl_setopt($session, CURLOPT_TIMEOUT, 200);
curl_setopt($session, CURLOPT_HEADER, true);
curl_setopt($session, CURLOPT_HTTPHEADER, array('Content-Type: multipart/form-data'));
curl_setopt($session, CURLOPT_POSTFIELDS, http_build_query(array("process"=>"login","user"=>$_POST['user'],"pass"=>$_PO    ST['pass'])));
printArr2($session);
$result = curl_exec($session);
$httpCode = curl_getinfo($session, CURLINFO_HTTP_CODE);
curl_close($session);

以下是API

if(isset($_RESPONSE['process']))
{

    if(!strcmp($_RESPONSE['content']['process'],"login"))
    {
        $con = dbConnect();
        $str = userLogin($con,$_POST['content']['user'],$_POST['content']['pass']);
        if(is_bool($str))
        {
            $jsonData = json_encode($str);
            header("HTTP/1.1 200 OK");
            header("Content-type: application/json");
            echo $jsonData;
        }
        else
        {
            $jsonData = json_encode($str);
            header("HTTP/1.1 401 Unauthorised access");
            header("Content-type: application/json");
            header("Location: ".$_SERVER['HTTP_REFERER']);
            echo $jsonData;
        }
        dbClose($con);
        exit;
    }

    if(!strcmp($_POST['process'],"plagiarism"))
    {
        $con = dbConnect();
        $user = $_POST['user'];
        $text = $_POST['text'];

    }
}
else
{
    $jsonData = json_encode(array("Error"=>"No methods called"));
    if(isset($_SERVER['HTTP_REFERER']))
    {
        header('HTTP\1.1 400', true, 400);
        header("Content-type: application/json");
        header("Location: ".$_SERVER['HTTP_REFERER']);
        echo $jsonData;
        exit;
    }
    else
    {
        echo "Invalid entry";
    }
}

无论我做什么，输出始终是 isset($_RESPONSE['process']) 的“else”部分。

我尝试将“process”作为 get 附加到 URL 中。

$url = "http://localhost/checkapi.php?process=login";

Answer 1

您在 $_RESPONSE 和 $_POST super 全局变量中使用了无效的数组索引(['content'])。

if(!strcmp($_RESPONSE['content']['process'],"login"))
    {
        $con = dbConnect();
        $str = userLogin($con,$_POST['content']['user'],$_POST['content']['pass']);

相反，您需要:

if(!strcmp($_POST['process'],"login"))
    {
        $con = dbConnect();
        $str = userLogin($con,$_POST['user'],$_POST['pass']);