function ajaxFunction(id){
var ajaxRequest;
var response;
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
} catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e){
alert("Ajax Failed");
return false;
}
}
}
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
response = ajaxRequest.responseText;
}
}
ajaxRequest.open("GET", "http://priest/getpic.php?id="+id, true);
ajaxRequest.send(null);
return response;
}
function lightbox(id) {
var image;
var imageArr;
document.write(image);
image = ajaxFunction(id);
imageArr = image.split('|');
imageSrc = imageArr[0];
imageWidth = imageArr[1];
imageHeight = imageArr[2];
getElementById('lightbox').visibility=visible;
getElementById('lightboximg').src=imageSrc;
if(imageWidth > 700) {getElementById('lightboximg').width=700;}
if(imageHeight > 500) {getElemetnById('lightboximg').height=500;}
}
我遇到的問題是在我的代碼調用ajaxFunction()到圖像變量ajaxFunction()不返回任何東西到變量,使我得到返回值以下錯誤。的Javascript:從我的AJAX功能
遺漏的類型錯誤:無法調用未定義 收藏 (匿名函數)
任何幫助,將不勝感激的「分裂」。
這是越來越與「解析HTML與正則表達式」儘可能多的頻率它被問到......愚蠢的http://stackoverflow.com/questions/9286045/get-json-response-var - jQuery的外部功能,http://stackoverflow.com/questions/562412/return-value-from-function-with-an-ajax-call,還有更多。 – 2012-03-08 23:46:09