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How to use JQuery to select the nth option

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借酒劲吻你
借酒劲吻你 2021-02-05 01:15

I have the following HTML
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  • 被撕碎了的回忆
    2021-02-05 01:28

    I'd not advice setting value of select by using .attr('selected', 'selected'). It's bad practice. If two options have 'selected' attribute - then what is the value of select?

    Here is the simple and easy version that will work:

     $('select[name="name"]').val('admin');

    DEMO

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  • 后悔当初
    2021-02-05 01:36

    I think the nth-child pseudo-class is what you're looking for:

    $('select option:nth-child(3)').attr('selected', 'selected');
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  • 执念已碎
    2021-02-05 01:38

    Like this:

    $("select[name='name'] option:eq(2)").attr("selected", "selected");

    EDIT:

    To do what you want in your new edit, that is, select the option that has the value of admin:

    $("select[name='name'] option:contains('admin')").attr("selected", "selected");

    See it working.

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  • 野性不改
    2021-02-05 01:42

    Just updating in case it helps someone. I had a similar requirement but in my case the dropdown was stored as a JavaScript variable. Also I liked how Denis Matafonov suggested to make the option in question selected. So here's the final block of code:

    var target_dd = $('select[name=name]');
$(target_dd).val($(target_dd).find('option:eq(2)').html());
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  • 我在风中等你
    2021-02-05 01:52

    If you want to select nth option and trigger onchange event then:

    function selectOption(nr) {
    var select = $("select");
    if (select.children().length >= nr) {
        let value = select.find('option:eq(' + nr + ')').val();
        select.val(value).change();
    }
}
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