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Home/ Questions/Q 8059243
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Editorial Team
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Editorial Team
Asked: June 5, 20262026-06-05T09:36:33+00:00 2026-06-05T09:36:33+00:00

If I do $var = $(‘[height > 100]’); It is throwing error runtime error

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If I do $var = $('[height > 100]');

It is throwing error runtime error

Microsoft JScript runtime error: Syntax error, unrecognized expression: [height > 100]

But equals(=) works

$var = $('[height = 100]');

So how can I filter for the first case? (using attribute selector only just because I am learning it)

I have <table height="200">

Thanks

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-05T09:36:35+00:00Added an answer on June 5, 2026 at 9:36 am
    $var = $(target).filter(function() {
      return $(this).height() >= 100;
    });
    

    Here target will be replaced with any valid jQuery selectore

    Note

    If you use height with style css then your way is not possible.

    According to edit

    <table height="200">
    

    Then try

    $('table[height=200]');
    

    DEMO

    OR

    $var = $('table').filter(function() {
      return $(this).attr('height') == 100;
    });
    

    DEMO

    According to commebt

    Q. Does it work for !=?

    A. YES

    $var = $('table').filter(function() {
      return $(this).attr('height') != 100;
    });
    

    DEMO

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