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Home/ Questions/Q 6209149
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T05:55:58+00:00 2026-05-24T05:55:58+00:00

I’ve inherited another developer’s code for a project I’m working on. Here’s an example

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I’ve inherited another developer’s code for a project I’m working on. Here’s an example of their code – notice the first comment…

function disableSuggestButton(themePath) {
    // could not see a better way of doing this :(
    var row = $("input[type='text']:inputnum:inputnumsmall");
    $(row).each(function () {
        if ($(this).attr("disabled") == true) {
            var btn = $(this).closest("tr").find(".btnSuggest")
            $(btn).hover(function () {
                $(this).css({ backgroundImage: 'url(' + themePath + '/images/buttons/btnSuggest.png)' }).css('cursor', 'default');
            }, function () {
                $(this).css({ backgroundImage: 'url(' + themePath + '/images/buttons/btnSuggest.png)' }).css('cursor', 'default');
            });
            $(btn).removeAttr("title").removeAttr("href");
        }
    });
}

I needed to upgrade the version of jQuery (from 1.3.2 to 1.4) however when doing this the following line:

var row = $("input[type='text']:inputnum:inputnumsmall");

produces the following error:

Error: uncaught exception: Syntax error, unrecognized expression:
inputnum

How can I improve on the previous devloper’s selector? inputnum & inputnumsmall are CSS classes.

Thanks

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-24T05:55:59+00:00Added an answer on May 24, 2026 at 5:55 am

    Had the previous developers written some custom selectors? (maybe right into the jquery file?).

    You should just be able to do:

    var row = $("input[type='text'].inputnum.inputnumsmall");
    

    Example: http://jsfiddle.net/BqHBf/

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