Question

我有这个脚本

<?php
include_once "fungsi.php";
include_once "conf/conf_Star04.php";;
header('Access-Control-Allow-Origin: *');

if($_SERVER['REQUEST_METHOD'] == "POST"){
        if(!isset($_POST['card'])){
            $json = array("status" => 0,"Desc" => "Error", "message" => "Please insert CardNumber");
        }else{
            $card = $_POST['card'];
            $email = $_POST['email'];
            $address = $_POST['address'];
            $hp = $_POST['hp'];
            // QUERY HERE
            $sql = "MY Query";
            $result = sqlsrv_query( $conn, $sql );
            if ($result){
                $json = array("status" => 1,"Desc" => "Done");
            }else{
                $json = array("status" => 0,"Desc" => "Error");
            }
        }
    }else{
        $json = array("status" => 0,"Desc" => "Error");
    }
?>

我制作API POST方法。该脚本有什么问题？

当我访问时，它总是给出错误未定义的变量：json

Answer 1

您没有正确初始化 $ json 变量，请尝试以下代码，这可能有所帮助......

 <?php
    include_once "fungsi.php";
    include_once "conf/conf_Star04.php";;
    header('Access-Control-Allow-Origin: *');

    $json = array("status" => 1);

    if($_SERVER['REQUEST_METHOD'] == "POST"){
    if(!isset($_POST['card'])){
        echo "Please insert CardNumber";
    }else{
        $card = $_POST['card'];
        $email = $_POST['email'];
        $address = $_POST['address'];
        $hp = $_POST['hp'];
        // QUERY HERE
        $sql = "MY Query";

        $result = sqlsrv_query( $conn, $sql );
        if ($result){

            $json["status"] = 1;
            $json["Desc"] = "Done";
            echo json_encode($json);
        }else{

            $json["status"] = 0;
            $json["Desc"] = "Error";
            echo json_encode($json);
        }
    }
    }else{

    $json["status"] = 0;
    $json["Desc"] = "Error";
    echo json_encode($json);
    }

    ?>

Answer 2

你的问题就在这一行 - ＆gt; echo "Please insert CardNumber";

在代码的最后，您正在执行echo json_encode($json);，因此期望$json变量。

因此，您应该更改此行：

echo "Please insert CardNumber";

到

$json = array("status" => 0,"Desc" => "Error", "message" => "Please insert CardNumber");

如何在PHP中进行Web API方法POST

2 个答案: