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Conflict between .show() and .append() in JQuery

开发者 https://www.devze.com 2023-01-26 09:00 出处:网络
I have a div with id=\"poidiv\" whose display is \'none\' initially. Now I want to l开发者_开发技巧oad it more than once with a loop (the max value of the loop is dynamic). I tried it with JQuery .app

I have a div with id="poidiv" whose display is 'none' initially. Now I want to l开发者_开发技巧oad it more than once with a loop (the max value of the loop is dynamic). I tried it with JQuery .append().clone().

Here is the example code--

$(document).ready(function(){  

    $("#levelnext").click(function(){
        for(i=1; i<=level; i++){  
            $("#leveldiv").append($("#poidiv").clone().removeAttr("id"));
        }
    });
});

But because the display of "poidiv" was initially 'none', it does not appear with this piece of code. Now if I want to show it with .show() before the loop starts, the loop is not working in a good manner. What might be a good solution in this situation?


You can .show() in the chain, like this:

$("#leveldiv").append($("#poidiv").clone().removeAttr("id").show());
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