jquery无法读取未定义的属性'done' - 避免这种情况 [英] jquery Cannot read property 'done' of undefined - avoid this
问题描述
我有一个返回结果的函数(或不返回)。问题是当它没有返回我在控制台中得到的任何值时消息
I have a function which returns results (or not). The problem is when it does not return any value I get in the console the message
无法读取未完成的属性'done'
cannot read property 'done' of undefined
这是真的,我确实理解这个问题。此外,此错误不会使我的代码停止工作,但我想知道是否有机会避免这种情况?
Which is true and I do understand the problem. Also, this error doesn't make my code stop working, but I would like to know if there's any chance of avoiding this?
ajax中的函数是:
The function in ajax is:
function getDelivery(){
var items = new Array();
$("#tab-delivery tr").each(function(){
items.push({"id" : $(this).find('.form-control').attr('id'), "id_option" : $(this).find('.form-control').val()});
});
if(items.length > 0){
return $.ajax({
url: 'response.php?type=getDelivery',
type: 'POST',
data: {content: items}
});
}
}
为了调用它,我使用:
getDelivery().done(function(data){ // the problem is here
if(data == false){
return;
}
});
那么,有什么方法可以避免错误吗?我试过以下没有成功:
So, is there any way of avoid the error? I have tried without success the following:
if(items.length > 0){
return $.ajax({
url: 'response.php?type=getDelivery',
type: 'POST',
data: {content: items}
});
}else{
return false;
}
我收到错误:
Uncaught TypeError:undefined不是函数
Uncaught TypeError: undefined is not a function
推荐答案
你可以只返回一个延迟,这样 done()
回调不会产生错误,你可以选择解决它
You could just return a deferred, that way the done()
callback won't generate errors, and you can choose to resolve it or not
if(items.length > 0){
return $.ajax({
url: 'response.php?type=getDelivery',
type: 'POST',
data: {content: items}
});
}else{
var def = new $.Deferred();
def.resolve(false);
return def;
}
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