如何在jQuery中显示所有子节点

Question

<a href="javascript:(void);" id="lnkP">show all children</a>
<a href="javascript:(void);" id="lnkC1">hide child 1</a>
<a href="javascript:(void);" id="lnkC2">hide child 2</a>

<div id="p" style="display:none;">
<div id="c1">child 1</div>
<div id="c2">child 1</div>...
</div>​

$("#lnkP").click(function(){
    $("#p").children().show(); //seems there's a problem here...
});
$("#lnkC1").click(function(){
   $("#c1").hide(); 
});
$("#lnkC2").click(function(){
   $("#c2").hide(); 
});​

jsFiddle: http://jsfiddle.net/CBGsF/1/

What I am trying to do is:

1. p is a parent container
2. click show all children link, display all child divs under p
3. click lnkC1 or lnkC2 to hide individual child div

But it seems that I didn't get .children() working correctly. So how to fix it? Any ideas?

Answer 1

Since the parent ( #p in your case) has a display:none , it's children won't be visible.

You'll need to show the parent first,

$("#p")
.show()
.children().show();

(jQuery's chaining, very helpful)

Please try and get rid of the inline styling (it gets unmanageable after a while), use classes as much as possible.

You can have a class in css,

.displayNone
{
    display: none;
} 
.displayBlock
{
   display: block;
}

And then use jquery methods .removeClass() , .addClass() or .toggleClass() to show/hide your elements.

This is just a recommendation :)

Test link: http://jsfiddle.net/CBGsF/8/

Answer 2

You need to show the #p also

Updated fiddle: http://jsfiddle.net/CBGsF/7/

$("#lnkP").click(function(){
    $("#p").show().children().show(); //Add show() before children.show call
});
$("#lnkC1").click(function(){
   $("#c1").hide(); 
});
$("#lnkC2").click(function(){
   $("#c2").hide(); 
});​

Answer 3

Updated fiddle : http://jsfiddle.net/CBGsF/5/

$("#lnkP").click(function(){
$("#p").show();
$("#p").children().show();
});
  $("#lnkC1").click(function(){
   $("#c1").hide(); 
});
$("#lnkC2").click(function(){
   $("#c2").hide(); 
});​

Answer 4

Parent element is set to "display":"None" That is the problem

$("#p").css("display","block"); //is required in show all anchor click

Check the fiddle

http://jsfiddle.net/CBGsF/6/

Thanks

Answer 5

(Posted solution on behalf of the question author) .

I thought .children() would search for invisible nodes as well. Well, I was wrong on that.