PHP脚本中的错误，未定义的变量

Question

<?PHP
include_once("connection.php");

$username = $_POST['txtUsername'];

$query = "SELECT * FROM cart_management WHERE UserName = '$username' ORDER BY Cart_Name ASC"; 

$result = mysqli_query($conn, $query);

while($row = mysqli_fetch_assoc($result)){
        $data[] = $row;
}
echo json_encode($data)
?>

请帮我解决这个问题，我无法解决这个问题？

Answer 1

您需要检查您发送的数据是否具有必须相同的索引txtUsername。 （查看你在哪里打电话cartRetrive.php api）

，第二个错误是未定义的变量。

您在宣布之前尝试使用$data[]。

<?PHP
include_once("connection.php");

$username = $_POST['txtUsername']; // check the index from where you are calling  cartRetrive.php

$query = "SELECT * FROM cart_management WHERE UserName = '$username' ORDER BY Cart_Name ASC"; 

$result = mysqli_query($conn, $query);

car data[] = array();// declare variable here

while($row = mysqli_fetch_assoc($result)){
        $data[] = $row;
}
echo json_encode($data)
?>

Answer 2

您应首先检查txtUsername，这是您在屏幕截图中的首次通知。其次，在$data循环中使用它之前，您应该将while声明为空数组，否则会导致屏幕截图中的第二个通知：

if($_POST['txtUsername'] && $_POST['txtUsername'] != '') {
    $username = $_POST['txtUsername'];
    $data = [];
    $query = "SELECT * FROM cart_management WHERE UserName = '$username' ORDER BY Cart_Name ASC"; 

    $result = mysqli_query($conn, $query);

    while($row = mysqli_fetch_assoc($result)){
        $data[] = $row;
    }
    echo json_encode($data);
}