我已經和在工作只是在firefoex沒有錯誤罰款,但IE6加載頁面的狀態欄上一個醜陋的錯誤圖標一個頁面一個AJAX腳本。什麼是我可以去解決/調試這個最好的方法?我怎樣才能調試這個Ajax腳本的IE瀏覽器?
以下是錯誤報告:
我已經檢查線路323多次下面是函數:
function checkAvailability(){
var card_select = document.getElementById('card_select').value;
var price_select = document.getElementById('price_select').value;
var num_of_cards = document.getElementById('num_of_cards').value;
var url = 'checkAvailability.php?cardName=' + card_select + '&value=' + price_select + '&amount=' + num_of_cards;
var xmlhttp;
if (window.XMLHttpRequest)
{
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
}
else if (window.ActiveXObject)
{
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
else
{
alert("Your browser does not support XMLHTTP!");
}
xmlhttp.onreadystatechange = function()
{
if(xmlhttp.readyState == 4 && xmlhttp.responseText) /**** line 323 ****/
{
document.getElementById('submit_button').className = 'hidden';
document.getElementById('div_error_massage').className = 'anounce_div';
document.getElementById('error_massage').innerHTML = xmlhttp.responseText;
document.getElementById('num_of_cards').className = 'red_inputs';
}
else if(isNaN(num_of_cards))
{
document.getElementById('submit_button').className = 'hidden';
document.getElementById('num_of_cards').className = 'red_inputs';
document.getElementById('div_error_massage').className = 'hidden';
}
else if(num_of_cards != "" && !xmlhttp.responseText)
{
document.getElementById('submit_button').className = '';
document.getElementById('error_massage').innerHTML = 'Total: $' + document.getElementById('price_select').value * document.getElementById('num_of_cards').value + '.00';
document.getElementById('div_error_massage').className = 'anounce_div';
}
else
{
document.getElementById('submit_button').className = 'hidden';
document.getElementById('num_of_cards').className = 'red_inputs';
}
}
xmlhttp.open("GET",url,true);
xmlhttp.send(null);
}
您可能希望除了檢查(xmlhttp.status == 200),以檢查的readyState – jdigital 2009-05-30 23:50:02