Cannot set multiple value of bootstrap select from array

Question

So I have this string: 93, 94 which are the references of the values of a select.

I'm trying to set these values using:

let values = "93, 94";
$('#property-categories').selectpicker('val', values);

but not working, none of the value is selected in the select control.

If I write: $('#property-categories').selectpicker('val', [93,94]);

works, any idea?

Rory McCrossan · Accepted Answer · 2019-10-12 14:24:17Z

2

To turn a comma delimited string in to an array, which is the type required here, use split().

let values = "93, 94";
$('#property-categories').selectpicker('val', values.split(', '));

answered Oct 12, 2019 at 14:24

Rory McCrossan

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6 Comments

jishan siddique Over a year ago

please test your code before posting.link

Rory McCrossan Over a year ago

@jishansiddique It already works. jsfiddle.net/g4y02nfa

Rory McCrossan Over a year ago

Because you used a very outdated version of selectpicker, for some reason

sfarzoso Over a year ago

@RoryMcCrossan could I ask a little more help? The issue happen again if I provide non numerical values, eg: modern,historic,seaside,minimalist, could you please take a look at it?

Rory McCrossan Over a year ago

Text values work too. The difference with that string is that you're using just , to delimit the string instead of , (note the trailing space after the comma). Fix that and it works fine jsfiddle.net/rgkmwd9v.

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jishan siddique · Accepted Answer · 2019-10-12 14:27:23Z

1

Please check this one for simple select control

var values="93,94";
$.each(values.split(","), function(i,e){
    $("#strings option[value='" + e + "']").prop("selected", true);
});

For selectpicker you need to set below code

$('.selectpicker').selectpicker();
$('.selectpicker').selectpicker('val', ['93','94']);

Check here working demo - > Click here

answered Oct 12, 2019 at 14:27

jishan siddique

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3 Comments

Rory McCrossan Over a year ago

Your first example won't update the selectpicker UI and the second calls selectpicker() twice for no reason.

Rory McCrossan Over a year ago

Also please don't plagiarise other answers

jishan siddique Over a year ago

Yes, I understood the problem thanks for sharing your response.

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