CODEWITHSUNDEEP
jquery
Adam Skiba
Adam Skiba

Reputation: 633

jQuery.inArray keeps returning -1

EDIT jQuery.inArray kept failing to do string != int comparisons, and this was resolved with the help of ShankarSongoli. Originally it was thought to be a variable scope issue, so please remember to look at all aspects and not just one in any given problem.

   omit_all_user_contacts = new Array();
    add_contacts = new Array();

    function is_added_contact(id) {
        if ( jQuery.inArray(id, add_contacts) == -1 )
            return false;
        return true;
    }

    function selectcontacts(userid){
      document.getElementById("ic"+userid).checked = true;
      omit_all_user_contacts.push(userid);
      var buf = '<table class="contactlist" width="100%">';
      buf = buf + "<tr class='contactlistheader'><td></td><td>Contact Name</td><td>Company</td><td>Email</td><td>Profession</td></tr>";
      var tmp;
      jQuery.getJSON('/user/get_user_contacts/'+userid, function(data) {

          jQuery.each(data.contacts, function(key, value) {

            buf = buf + "<tr><td><input type='checkbox' name='manualcontact[]' onclick='manualcontact(" + value.id + ",this.checked)' ";
                alert(is_added_contact(value.id)); 

            if ( jQuery.inArray(value.id, add_contacts) > -1 ) {

                buf = buf + " checked ";
            }
            buf = buf + "/></td>";
            buf = buf + "<td>" + value.firstname + " " + value.lastname + "</td>";
            buf = buf + "<td>" + value.company + "</td>";
            buf = buf + "<td>" + value.email + "</td>";
            buf = buf + "<td>" + value.profession + "</td></tr>";
          });
          buf = buf + '</table>';
          iBox.show(buf);
      }); 
    }

    function manualcontact(id,val) {
        if ( val == true ) {
            add_contacts.push(id);
        } else {
            add_contacts.splice( jQuery.inArray(val, add_contacts), 1 );
        }   
    }

Upvotes: 0

Views: 2102

Answers (2)

Roger Cable
Roger Cable

Reputation: 19

It is important that the type of the variable is the same. I needed to compare integers, but since my original text contained numbers split by commas then the items in my array were being seen as strings.

Here was a simple function I wrote to check for integers only. It works the same as jquery.inArray but it only checks a specific data type.

function in_array(val,arr) {
    cnt=0;
    index=-1;
    $(arr).each(function () {
        if(parseInt(this) == parseInt(val)) { index=cnt; }
        cnt++;
    });
    return index;
}

Upvotes: 0

ShankarSangoli
ShankarSangoli

Reputation: 69905

You should change the title of the question first and also since the issue is resolved through our comments accept this answer and close it.

Solution:

The array values were of string type and comparison was done with integers, so it was failing.

var tmpContacts = ["2"];
jQuery.inArray(2, tmpContacts) was always returning `-1`

Change it to jQuery.inArray("2", tmpContacts)

Upvotes: 2

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