How to change an attribute of a variable in Jquery?

Question

I'm struggeling with something basic and I can't seem to work it out.

The following works perfectly:

switch (url_param) {
    case '80':
        $('input[name=power][value=80]').prop('checked', true);
        break;
    case '90':
        $('input[name=power][value=90]').prop('checked', true);
        break;
    case '99':
        $('input[name=power][value=99]').prop('checked', true);
        break;
}

However, I have to interact with many different inputs on the page so I would like to store them into variables first, for example

$input_power = $('input[name=power]');

How can I achieve the same result, using a variable?

Satpal · Accepted Answer · 2016-06-10 12:09:02Z

2

You can use filter() method to filter out the matched element.

Reduce the set of matched elements to those that match the selector or pass the function's test.

var $input_power = $('input[name=power]');
switch (url_param) {
    case '80':
    case '90':
    case '99':
        $input_power.filter(function(){
            return parseInt(this.value, 10) == parseInt(url_param, 10);
        }).prop('checked', true);
        break;
}

In the above example I have improved the code and used parseInt(str, radix) for integer.

answered Jun 10, 2016 at 12:09

Satpal

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Comments

Ansh Takalkar · Accepted Answer · 2016-06-10 12:20:08Z

1

Remove switch case statements and use only one line this

$('input[name=power][value='+ url_param +']').prop('checked', true);

edited Jun 10, 2016 at 12:20

answered Jun 10, 2016 at 12:15

Ansh Takalkar

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1 Comment

NicoF Over a year ago

this indeed would work, however I do need the switches to check the values. In addition I have to use a variable for all inputs, as I'm using them multiple times in the code. Thanks for answering!

Furhan S. · Accepted Answer · 2016-06-10 12:07:55Z

0

Try this:

var myInput = $('input[name=power][value=80]');

Somewhere later in your code:

$(myInput).prop('checked', true);

Hope this helps.

answered Jun 10, 2016 at 12:07

Furhan S.

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1 Comment

NicoF Over a year ago

I don't want to create a new variable for each new value...this was my problem in the first place. Thanks for answering nevertheless!

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