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need to manage a list in jQuery

Posted on 2011-09-22
7
339 Views
Last Modified: 2012-05-12
I need to create a list, fill it with unique 8 digit numbers, then query the list to see if the next 8 digit number is unique. If it's already in the list my function will return true. If it gets through the whole list and does not find a match, I will return false.

What is the syntax to:

1) create the list
2) add an item to the list
3) check the list to see if a given 8 digit number is found.

Thanks.
0
Comment
Question by:newbieweb
  • 3
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7 Comments
 
LVL 13

Expert Comment

by:haloexpertsexchange
ID: 36581310
well for a list you would normally use an array.
var uniqueNumber = new Array();
uniqueNumber[0] = number1;
uniqueNumber[1] = number2;
and so on.
0
 

Author Comment

by:newbieweb
ID: 36581322
how do I query the list to see if number1 is found?
0
 
LVL 13

Expert Comment

by:haloexpertsexchange
ID: 36581340
If you are using jquery as your title suggests then you can use jquery inarray for part of it.
http://api.jquery.com/jQuery.inArray/
Let me put something together to test.
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Author Comment

by:newbieweb
ID: 36581349
thanks.
0
 
LVL 63

Assisted Solution

by:Zvonko
Zvonko earned 200 total points
ID: 36581351
Check this:
<script>

var myList=[];
for(var i=1;i<9;i++){
  myList['_'+(12345670+i)]=(12345670+i);
}


function checkList(theList,theNum){
  return theList['_'+theNum]!=undefined;
}

alert(checkList(myList,12345678)); // returns: true
alert(checkList(myList,12345687)); // returns: false

</script>

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0
 
LVL 13

Accepted Solution

by:
haloexpertsexchange earned 300 total points
ID: 36581634
This works, I just filled in the array with any old numbers, with this array you will get false returned.
var uniqueNumber = new Array();
uniqueNumber[0] = 3082969;
uniqueNumber[1] = 3082963;
uniqueNumber[2] = 3082963;
uniqueNumber[3] = 3082965;
uniqueNumber[4] = 3082966;
uniqueNumber[5] = 3082967;

var checkNumbers = function(uniqueNumber) {

while(uniqueNumber.length!=0){
var searchFor = uniqueNumber[0];
uniqueNumber.shift();
if(jQuery.inArray(searchFor, uniqueNumber)!=-1)
return false;
}
return true;
};

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0
 

Author Closing Comment

by:newbieweb
ID: 36593862
Thanks!
0

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