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Home/ Questions/Q 8292937
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Editorial Team
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Editorial Team
Asked: June 8, 20262026-06-08T13:39:24+00:00 2026-06-08T13:39:24+00:00

Because I am creating DOM using Jquery it was difficult to copy the output

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Because I am creating DOM using Jquery it was difficult to copy the output so i am adding one image of code that i have captured using one tool

enter image description here

i have attached hover and mouseout event to id='nf1' using this code

$("#nf"+n).hover(function(){
    $("#nf"+$(this).attr("post_id")+"post_delete").show();
});
$("#nf"+n).mouseout(function(){
    $("#nf"+$(this).attr("post_id")+"post_delete").hide();
});

Here n is post_id and i am looping all post_id got from response.This attach events but not giving expected behaviour Like when mouse over to id='nf1post_delete' it is hide

Please ask if any doubts

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-08T13:39:26+00:00Added an answer on June 8, 2026 at 1:39 pm

    The way you’re describing this, you will actually want to pass two functions to .hover(), one for the action on mouseenter and one for the action on mouseleave. You can pass only one function to .hover(), but it will run that function when you roll over and when you roll out.

    http://api.jquery.com/hover/

    So, try this instead:

    $("#nf"+n).hover(function(){
        $("#nf"+$(this).attr("post_id")+"post_delete").show();
    },function(){
        $("#nf"+$(this).attr("post_id")+"post_delete").hide();
    });
    

    The .mouseout() function isn’t needed at all.

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