Jquery inArray always returns -1 in loop

Question

I'm having problem with Jquery inArray.

//Get selected devices
selectedDevices = $('.selectDevices').val();
console.log(selectedDevices);

//If nothing selected, do nothing
if(selectedDevices == null || selectedDevices == ""){}

    //Else remove devices not selected
    else{ 
        //Loop thru table rows with class "enhet"
        $.each( $('td.enhet'), function() {

        thisDevice = $('.enhet').text();

        var found = $.inArray(thisDevice, selectedDevices);

        console.log(found);

        if (found > -1) {
            console.log(thisDevice);
        }
        else {
            console.log('nope');
        }


    })
}

Console.log(selectedDevices) give:
["GulAvformning", "RosaAvformning"]

Console.log(found) give: (everytime!!)
-1

Console.log(thisDevice) give:
``

Console.log('nope') give:
nope

Even if thisdeviceexists in selectedDevices i get the else case.
What am i missing?

selectedDevices is an array. I checked it by: console.log(selectedDevices[1]); And it gave me one of the devices.

thisDevice = $('.enhet').text(); what is the value of thisDevice — gurvinder372
– gurvinder372, Commented Feb 16, 2016 at 12:18

Marcos Pérez Gude · Accepted Answer · 2016-02-16 12:18:44Z

2

Use the each() method to the elements and then use $(this) to identify the current element:

selectedDevices = $('.selectDevices').val();
console.log(selectedDevices);

//If nothing selected, do nothing
if(selectedDevices == null || selectedDevices == ""){}

//Else remove devices not selected
else{ 
  //Loop thru table rows with class "enhet"
  $('td.enhet').each(  function() {

    thisDevice = $(this).text(); // use $(this) here

    var found = $.inArray(thisDevice, selectedDevices);

    console.log(found);

    if (found > -1) {
      console.log(thisDevice);
    }
    else {
      console.log('nope');
    }


  })
}

answered Feb 16, 2016 at 12:18

Marcos Pérez Gude

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2 Comments

Björn C Over a year ago

Yes of corse, that was my problem! Thank you!

Marcos Pérez Gude Over a year ago

Well, my pleasure, check this answer as correct when you can make it. Thank you!

Lior Shitrit · Accepted Answer · 2016-02-16 12:51:33Z

1

Use:

thisDevice = $(this).text();

Instead of:

thisDevice = $('.enhet').text();

edited Feb 16, 2016 at 12:51

answered Feb 16, 2016 at 12:28

Lior Shitrit

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Comments

Rustin Cohle · Accepted Answer · 2016-02-16 12:20:48Z

0

You are iterating but not selecting the iterator. Change your lines to this: ...

$.each( $('td.enhet'), function(i) {

        thisDevice = $('td.enhet').eq(i).text();

...

answered Feb 16, 2016 at 12:20

Rustin Cohle

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4 Comments

charlietfl Over a year ago

better to use the element itself returned within the each callback...otherwise you are searching the DOM each iteration to look for that same element. DOM searches are expensive

Rustin Cohle Over a year ago

I agree. I was not sure with "this".

charlietfl Over a year ago

Use $('td.enhet').each(... and this is current element

Rustin Cohle Over a year ago

I know. but I wasn't sure of it, never tried, so I didnt use "this". I stand corrected

Community · Accepted Answer · 2017-05-23 11:59:25Z

0

If console.log(selectedDevices) gives ["GulAvformning", "RosaAvformning"], and console.log(thisDevice) gives "", then $.inArray(thisDevice, selectedDevices) will return -1, since there's no "" in ["GulAvformning", "RosaAvformning"]. See Marcos Pérez Gude's solution for more info about how to properly get the correct value for thisDevice.

edited May 23, 2017 at 11:59

CommunityBot

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answered Feb 16, 2016 at 12:24

falsarella

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