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Home/ Questions/Q 6155525
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T20:27:04+00:00 2026-05-23T20:27:04+00:00

Hi im trying to replace a link if i click this other link. but

  • 0

Hi im trying to replace a link if i click this other link. but i cant get click to work when i add the link with jquery replace() or html();. what im i doing wrong?

this is the code i try to use

$(document).ready(function() {
        var countTotal = $('#myTable tr').length;
        $("#hideTotalCount").val(countTotal);

        $('#clickMeAll').click(function() {
            var totalCount = $("#hideTotalCount").val() - 1;
            $('#myTable').paginateTable({ rowsPerPage: totalCount, pager: '.pageNumbersOnly', maxPageNumbers: 4 });
            $('#placeLink').replaceWith("<a href='#' id='clickMeRegular'>Back to regular</a>");
        });

        $('#clickMeRegular').click(function() {
            var totalCount = $("#original").val();
            $('#myTable').paginateTable({ rowsPerPage: totalCount, pager: '.pageNumbersOnly', maxPageNumbers: 4 });
            $('#placeLink').replaceWith("<a href='#' id='clickMeAll'>ViewAll</a>");
        });

        $('#myTable').paginateTable({ rowsPerPage: 5, pager: '.pageNumbersOnly', maxPageNumbers: 4 });
        $('#placeLink').html("<a href='#' id='clickMeAll'>ViewAll</a>");
    });

and #placeLink is just a regular <span>

maybe somebody know something that can push me in the right direction? thanks

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  1. Editorial Team
    Editorial Team
    2026-05-23T20:27:04+00:00Added an answer on May 23, 2026 at 8:27 pm

    The problem that the replacement DOM node has no click event attached. When you call $(“#clickMeAll”).click(…), jQuery searches for an element with id clickMeAll that is present right now and assigns an event to it.

    If you replace it with some other html code later on, the event is not automatically re-assigned to the new DOM node.

    Two solutions:

    • Use jQuery live events
    • Do not replace but just hide / show #clickMeAll and #clickMeRegular. They both exist all the time, but only one of them is visible. (better solution).
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