var office = $('#gameid') || $('#netid');
//$("#gameid" && "#netid").on('change', function(){
office.on('change', function(){
I want that if there is a change between one of the var office
, then the function will be there is one event.
我希望如果其中一个var办公室之间有变化,那么该函数将有一个事件。
How to do that?
怎么做?
3 个解决方案
#1
6
Try using .add()
function to club the both,
尝试使用.add()函数来支持两者,
var office = $('#gameid').add($('#netid'))
office.on('change', function(){
Apart from the solution, i just need to comment on what you tried,
除了解决方案,我只需要评论你的尝试,
var office = $('#gameid') || $('#netid');
The above code of yours always return $('#gameid')
to the variable office, because Jquery object is basically a collection. And it won't be undefined/falsy at any point of time.
你的上述代码总是将$('#gameid')返回给变量office,因为Jquery对象基本上是一个集合。并且在任何时候它都不会是不确定/错误的。
#3
0
Try with -
尝试 -
$('#gameid, #netid').on('change', function() { ... })
#1
6
Try using .add()
function to club the both,
尝试使用.add()函数来支持两者,
var office = $('#gameid').add($('#netid'))
office.on('change', function(){
Apart from the solution, i just need to comment on what you tried,
除了解决方案,我只需要评论你的尝试,
var office = $('#gameid') || $('#netid');
The above code of yours always return $('#gameid')
to the variable office, because Jquery object is basically a collection. And it won't be undefined/falsy at any point of time.
你的上述代码总是将$('#gameid')返回给变量office,因为Jquery对象基本上是一个集合。并且在任何时候它都不会是不确定/错误的。
#2
#3
0
Try with -
尝试 -
$('#gameid, #netid').on('change', function() { ... })