issue using .hide() when passing a variable from an array

Question

I have the following code which works fine:

 $(function mvp() {
    var theMvp = ['#mvpWtd', '#mvpStd'];
    $.each(theMvp, function (index, value) {
        $(value + ' .budTySales').hide();
        $(value + ' .lySales').hide();
        $(value + ' .budReceipts').hide();
        $(value + ' .lyReceipts').hide();
    });
})

According to jquery documentation I should be able to pass multiple elements in followed by a comma instead of doing it line by line (less code too!). I tried changing my code to the below but it fails...

$(function mvp() {
    var theMvp = ['#mvpWtd', '#mvpStd'];
    $.each(theMvp, function (index, value) {
        $(value + ' .budTySales',value + ' .lySales',value + ' .budReceipts',value + ' .lyReceipts').hide();
    });
})

You are wrongly concatening string here. Commas should be part of former string. Anyway, you would have better to use $(theMvp.toString()).find('.budTySales, .lySales, .budReceipts, .lyReceipts').hide(); or better just use a specific common class... — A. Wolff
– A. Wolff, Commented Jul 19, 2016 at 18:40

imtheman · Accepted Answer · 2016-07-19 18:48:03Z

6

This line:

$(value + ' .budTySales',value + ' .lySales',value + ' .budReceipts',value + ' .lyReceipts').hide();

should be:

$(value + ' .budTySales,' + value + ' .lySales,' + value + ' .budReceipts,' + value + ' .lyReceipts').hide();

Notice that the commas are inside the quotes. This is because jQuery expects a single parameter to be passed into it, not multiple, which is what you were doing.

edited Jul 19, 2016 at 18:48

answered Jul 19, 2016 at 18:42

imtheman

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Comments

larz · Accepted Answer · 2016-07-19 18:43:49Z

2

As stated in the comments, comma should be part of the string.

$(value + ' .budTySales, ' + value + ' .lySales, ' + value + ' .budReceipts, '+ value + ' .lyReceipts').hide(); should work

answered Jul 19, 2016 at 18:43

larz

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