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Home/ Questions/Q 7695311
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T21:31:10+00:00 2026-05-31T21:31:10+00:00

I have this table: <tbody> <tr> <th></th> <th></th> <th></th> <th></th> </tr> <tr> <th></th> <th></th>

  • 0

I have this table:

<tbody>
<tr>
    <th></th>
    <th></th>
    <th></th>
    <th></th>
</tr>

<tr>
    <th></th>
    <th></th>
    <th></th>
    <th></th>
</tr>

<tr>
    <th></th>
    <th></th>
    <th></th>
    <th></th>
</tr>

<tr>
    <th></th>
    <th></th>
    <th></th>
    <th></th>
</tr>

...
</tbody> 

and it keeps going…
I want on doc ready to add class odd and even depends on their position. To end up something like this:

<tbody>
    <tr class="odd">
        <th></th>
        <th></th>
        <th></th>
        <th></th>
    </tr>

    <tr class="even">
        <th></th>
        <th></th>
        <th></th>
        <th></th>
    </tr>

    <tr class="odd">
        <th></th>
        <th></th>
        <th></th>
        <th></th>
    </tr>

    <tr class="even">
        <th></th>
        <th></th>
        <th></th>
        <th></th>
    </tr>

    ...
    </tbody>

What is the way to achieve this? Do I have to count how many tables I have and add classes depends on their position? It doesn’t sound safe/proper way to me.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-31T21:31:11+00:00Added an answer on May 31, 2026 at 9:31 pm

    If you don’t need a pure CSS solution, go with jQUery and use :odd / :even selectors. Otherwise Sirko’s solution is better. (Note that nth-child method would not run on IE8 and below)

    ​$("table tr:odd").addClass("odd");
    ​$("table tr:even").addClass("even");
    
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