php - SQL 错误或 PHP 错误在我的 ajax [object object] 中返回

Question

当模式窗口内的按钮被点击时，ajax被调用。所以我有一个按钮，然后它会弹出一个带有"is"或“否”按钮的模式窗口，如果单击"is"，它将触发此ajax并将数据发送到我的php。它实际上在我的数据库中保存或更新我的表，但当它返回时，它在我的ajax中返回一个[对象对象]。

这里是php

// get data
$selGuest = $_POST["selGuest"];

include("openDB.php");

//3.) insert a record

$insertintoCanceled = "insert into tbl_canceled "
        ."(reserved_id, guest_id, checkin, checkout, type_id, numAdults, numChildren, transacstatus, amountDue)"
        ."("
        ."SELECT * FROM tbl_bookings where reserved_id = " .$selGuest
        .")";
if(!mysql_query($insertintoCanceled, $con))//if it fails
    {
    echo json_encode(array('msg'=>'Error')) //error msg goes here
    die('Error: ' . mysql_error() . "\n");//show the mysql error

    }
echo json_encode(array('msg'=>'Successfully updated')) //success msg goes here



include("closeDB.php");
?>

这是我的ajax

var canceldata_json = {
            'selGuest': selGuest,
    };
    $.ajax({
            type: "POST",
            data: canceldata_json,
            dataType:'json',
            url: "./php/cancelBooking.php",
            success: function(msg) {
                    alert("guest information updated real")
                    $('#confirmDialog').fadeOut('slow');

            },
            error:function(msg){
            alert(msg)
            }
    });

Answer 1

它正在解析您返回的 JSON。改为这样做:

alert(msg.msg);