javascript - 删除功能不起作用 像这样的错误 JSON 解析错误 : Unexpected identifier "Param"

Question

这是 Angular 代码。我喜欢这个删除功能代码。运行时出现这样的错误:

JSON Parse error: Unexpected identifier "Param"

$scope.delete = function(cart_ID, index) {
var params = $.param({"cart_ID":cart_ID});
console.log(cart_ID);
 $http({
        headers: {'Content-Type': 'application/x-www-form-urlencoded'},
        url: 'http://localhost/test1/shopCartProductDelete.php?cart_ID'+ cart_ID, 
        method: "GET",
        data: params


}).success(function(data){
$scope.data.splice(index, 1);


 });

shopCartProductDelete.phpt

<?php

header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json;charset=UTF-8");
$con = mysqli_connect("localhost","root","","look4com_lk");
if(isset($_GET['cart_ID'])){
$cart_ID = $_GET['cart_ID'];
$res = "DELETE FROM l4wlk_cart WHERE cart_ID='".$cart_ID."'";
mysqli_query($con, $res) or mysqli_error($con);
}else{
die("Param value not set up"); 
}

Answer 1

您的 jQuery AJAX 请求中缺少 =(可能不相关，但您最好检查 $_GET['cart_ID'] 实际上是通过 data 发送到您的 PHP 脚本的):

url: 'http://localhost/test1/shopCartProductDelete.php?cart_ID'+ cart_ID

应该是:

url: 'http://localhost/test1/shopCartProductDelete.php?cart_ID='+ cart_ID

---

现在，如果您尝试将其作为 json 使用，则需要发送 application/json，而不是 text/plain:

echo json_encode(['error' => true, 'message' => 'Param not set']);