javascript 从触发事件的元素返回属性 id
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Return the attribute id from the element that triggerd the event
提问by Elian ten Holder
I have an element, this element calls the function calcTotal via the following code:
我有一个元素，该元素通过以下代码调用函数 calcTotal：
$('.pause').change(function(e) {
window.alert("pause changed");
calcTotal(e);
The code of calcTotal(e) is as follows:
calcTotal(e)的代码如下：
function calcTotal(event)
{ alert('calcTotal called');
var myId = event.currentTarget.attr('id');
myId = myId.replace(/[^0-9]/g, '');
var timeRegex = /^[0-9]{1,2}:[0-9]{2}$/;
if($('#start'+myId).val().match(timeRegex) && $('#end'+myId).val().match(timeRegex) && $('#pause'+myId).val().match(timeRegex))
{
var minutes = 0;
var n = $('#end'+myId).val().split(':');
minutes = parseInt(n[0])*60 + parseInt(n[1]);
var n = $('#start'+myId).val().split(':');
minutes -= parseInt(n[0])*60 + parseInt(n[1]);
var n = $('#pause'+myId).val().split(':');
minutes -= parseInt(n[0])*60 + parseInt(n[1]);
var hours = Math.floor(minutes/60);
minutes = minutes % 60;
alert(hours + ':' + minutes);
$('#total' + myId).val(hours + ':' + minutes);
}
else
{
$('#total' + myId).val('00:00');
}
}
It doesn't work as I had exceptected and when I debug with firebug it says the following:
它无法正常工作，当我使用 firebug 进行调试时，它会显示以下内容：
TypeError: event.currentTarget.attr is not a function
var myId = event.currentTarget.attr('id');
I would like to store the id of the element in myId. How would I do this?
我想将元素的 id 存储在 myId 中。我该怎么做？
回答by Niet the Dark Absol
event.currentTargetis not a jQuery object, it's a DOM node.
event.currentTarget不是一个 jQuery 对象，它是一个 DOM 节点。
var myIf = event.currentTarget.id;
回答by Ricardo Alvaro Lohmann
event.currentTargetisn't a jQuery object, so attris undefined.
event.currentTarget不是一个 jQuery 对象，所以attr是undefined.
You can fix by the following ways:
您可以通过以下方式修复：
$(event.currentTarget).attr('id');
event.currentTarget.id;
回答by David says reinstate Monica
Assuming the rest of your code works as intended, I'd suggest:
假设您的其余代码按预期工作，我建议：
var myId = event.currentTarget.id;
Because it's a plain DOM node, not a jQuery object, the attr()method won't, and can't, work.
因为它是一个普通的 DOM 节点，而不是一个 jQuery 对象，所以该attr()方法不会也不能工作。
References:
参考：
