JSON 解析不起作用且属性未出现

Question

我仍在学习 JSON，所以当我尝试这样做时，名称没有出现。

index.php :

$myObj->name = "John";
$myObj->age = 30;
$myObj->city = "New York";
$myJSON = json_encode($myObj);
echo $myJSON;
header("Content-Type: text/html");

index.html :

$(document).ready(function() {
            $("input").blur(function() {
                var data = "pName=" + $(this).val();
                //alert($(this).val());
                $.ajax({
                    type: "POST",
                    url: "index.php",
                    data: data,
                    success: function (respond) {
                        var JQuery = JSON.parse(respond);
                        $("#result").html('');
                        $("#result").append(JQuery.name);
                    }
                });
            });
        });

1：

Answer 1

试试这个代码。

试试这个代码。你忘记声明对象了。 $myObj = new stdClass();

<?php
$myObj = new stdClass();

$myObj->name = "John";
$myObj->age = 30;
$myObj->city = "New York";
$myJSON = json_encode($myObj);
header("Content-Type: text/html");
echo $myJSON;
?>

Answer 2

改变

SELECT 
(select ( select Sum(T_No) where Transactions = R) - (select Sum(T_No) where Transactions = D) as C_T ) 
FROM CustomerTrans WHERE C_T > 0 ;

到

header("Content-Type: text/html");

Answer 3

谢谢大家，终于成功了<3

$myObj = new stdClass();
$myObj->name = "John";
$myObj->age = 30;
$myObj->city = "New York";
$myJSON = json_encode($myObj);
header("Content-Type: text/html");
echo $myJSON;
exit();