Find item in array with indexOf

Question

I have a problem with 'indexOf'. When i pass a numerical value in argument, it works, but not with a variable:

$.ajax({
    url: 'bdd.php',
    type: 'POST',
    data: {'jsonNomCommune': jsonNomCommune, 'client': client},
    success: function(data) {
        data = JSON.parse(data);
        listClients= data.clients;
        listProspects = data.prospects;

        $('#tableCSV tbody tr ').each(function(){
            var id = $(this).attr('id');
            id=id.substring(4);

            if (listClients.indexOf(id) > -1){
                $(this).css("background-color","red");

            }

            if (listProspects.indexOf(id) > -1){
                $(this).css("background-color","blue");
            }


        });
    }
});

listProspects and listClients are arrays of numericals values. For example, 27 is in listClients, so when id=27, "listClients.indexOf(id) > -1" should works, but it doesn't. And when i write: "(listClients.indexOf(27) > -1)" it works.

Where is the problem??

can you add a console.log(id) right before the indexOf, to check it has the expected value? — richardtz
– richardtz, Commented May 22, 2015 at 10:18
I reckon id might be a string. If so, you can use parseInt(id) to make sure it's an int. — evolutionxbox
– evolutionxbox, Commented May 22, 2015 at 10:20
@evolutionxbox When using parseInt() you should always pass the radix parameter. — Phylogenesis
– Phylogenesis, Commented May 22, 2015 at 10:21

score 1 · Accepted Answer · 2015-05-22 10:23:05Z

1

27 and "27" are different.

Try parsing id to integer:

listClients.indexOf(+id) //note, its parsing

OR:

listClients.indexOf(parseInt(id, 10))

edited May 22, 2015 at 10:23

answered May 22, 2015 at 10:19

user2575725

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3 Comments

evolutionxbox Over a year ago

Wouldn't +id turn id into a string? Surely it needs to be an int?

evolutionxbox Over a year ago

"When using parseInt() you should always pass the radix parameter." Thanks to @Phylogenesis.

Barmar Over a year ago

@evolutionxbox Unary + always returns a number. This is a common idiom for converting a numeric string to a number.

qtgye · Accepted Answer · 2015-05-22 10:19:47Z

0

You should cast the value of $(this).attr('id') into a number.
So it should be var id = Number($(this).attr('id'));

answered May 22, 2015 at 10:19

qtgye

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Comments

Joey Dias · Accepted Answer · 2015-05-22 10:28:36Z

0

cast id to integer by using parseInt() method

answered May 22, 2015 at 10:28

Joey Dias

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