javascript jquery无法读取未定义的属性“完成” - 避免这种情况
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/23115928/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
jquery Cannot read property 'done' of undefined - avoid this
提问by Linesofcode
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
我有一个返回结果（或不返回）的函数。问题是当它不返回任何值时我在控制台中收到消息
cannot read property 'done' of undefined
无法读取未定义的属性“完成”
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?
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？
The function in ajax is:
ajax中的函数是：
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}
});
}
}
And to call it I use:
并称之为我使用：
getDelivery().done(function(data){ // the problem is here
if(data == false){
return;
}
});
So, is there any way of avoid the error? I have tried without success the following:
那么，有没有什么办法可以避免这个错误呢？我尝试了以下方法但没有成功：
if(items.length > 0){
return $.ajax({
url: 'response.php?type=getDelivery',
type: 'POST',
data: {content: items}
});
}else{
return false;
}
And I get the error:
我得到错误：
Uncaught TypeError: undefined is not a function
未捕获的类型错误：未定义不是函数
回答by adeneo
You could just return a deferred, that way the done()callback won't generate errors, and you can choose to resolve it or not
你可以只返回一个延迟，这样done()回调就不会产生错误，你可以选择是否解决它
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;
}
