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jQuery - What am i doing wrong. if - else statements. wrap() if element is parent

开发者 https://www.devze.com 2023-03-08 23:19 出处:网络
Basic idea of my if else statements is this: If .wrap is parent element: do something Else: do something else

Basic idea of my if else statements is this:

  • If .wrap is parent element: do something
  • Else: do something else

This might not make much sense but if you look at the code here: http://jsfiddle.net/C6NQM/ ..it might

if( $('.wrap:parent'); ) 
    { $('.wrap').children().wrap('<div class="new" />'); }
else {开发者_运维问答 $('.wrap').wrap('<div class="new" />'); }

Things work separately if i remove if and else, so i came to the conclusion that there's something wrong with my if condition.

Any ideas about what's wrong here?


I think you want:

$('.wrap').each(function() {
    if($(this).children().length > 0) {
        $(this).children().wrap('<div class="new" />'); 
    }
    else { 
        $(this).wrap('<div class="new" />'); 
    }
});

Updated fiddle.

Comments on your code:

if( $('.wrap:parent')) will always evaluate to true. A call to jQuery always returns a jQuery object. You could do if( $('.wrap:parent').length > 0), but even then you would select all elements in your example, as :parent selects also nodes that have text nodes as children.

But even with that, the whole statement will work as follows: If there are any .wrap elements that are parents, wrap their children in a new element. If not, wrap the elements themselves.
So it will not treat every .wrap element individually.

Update: To cover every case (children, text content, empty), you probably have to do:

$('.wrap').each(function() {
    if($(this).children().length > 0) {
        $(this).children().wrap('<div class="new" />'); 
    }
    else if($(this).text().length > 0) { 
        $(this).html('<div class="new">' + $(this).text() + '</div>'); 
    }
    else {
       $(this).wrap('<div class="new" />'); 
    }
});


I tried and you code will work if you remove the ';' within the if condition.

Update:

PS: If you use jfiddle.net, general error in javascript will also be printed out in the error console of FireFox and Safari.

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