我有這個片斷這是在事實上正常工作的更新我的數據庫,只有一次對其做它沒有執行我的代碼的其餘AJAX代碼...阿賈克斯文件停止執行
$(".getPoint").click(function()
{
var theid = $(this).attr("id");
var onlyID = theid.split("_");
var onlyID = onlyID[1];
var credoff = parseInt($(this).children('input.credoff:hidden').val());
$.ajax({
url: 'do.php',
type: 'POST',
data: "userID=" + onlyID + "&credoff=" + credoff,
success: function(data) {
if(data != "success1" && data != "success5") {
$("#" + theid).text(data);
}else{
$("#thediv_" + onlyID).fadeOut("slow");
$('#creditsBalance').fadeOut("slow");
newbalance = parseInt($('#creditsBalance').text());
if(data != "success5") {
newbalance = newbalance+credoff;
}else{
newbalance = newbalance+5;
}
alert ('hi');
$('#creditsBalance').text(newbalance);
$('#creditsBalance').fadeIn("slow");
$("#" + theid).text("Done");
}
},
beforeSend: function()
{
$("#" + theid).text("Working...");
},
error: function()
{
$("#" + theid).text("Failed...Click to Retry");
}
});
});
線
if(data != "success5") {
newbalance = newbalance+credoff;
}else{
newbalance = newbalance+5;
}
alert ('hi');
更新我的數據庫,但然後我沒有收到警報,這是足夠的代碼,讓任何人看到我哪裏會出錯?
有你在瀏覽器中發射了一個JavaScript控制檯(螢火蟲爲例)並看看是否有腳本錯誤? – 2011-05-29 22:52:14
@傑夫,是的,我只有它的所有運行良好... – Liam 2011-05-29 22:54:09