Multidimensional Array Javascript

Question

I can't seem to get anything to go into my ['count'] key. Any help would be appreciated.

<script>
 $.getJSON( "someAddress", function( data ) {
 $( ".result" ).html( data );
 var totalUse = new Array();
 var totalLen = data.stats.length;
 for(x = 0; x < totalLen; x++){
 var user = data.stats[x].userId;
 if(totalUse.indexOf(user) > -1){ // yes it does have it
 totalUse[user]["count"] += data.stats[x].count;
 }else{
 totalUse[user] = data.stats[x].userId;
 totalUse[user]['count'] = data.stats[x].count;
 console.log(totalUse[user]['count']);
 }
 }
 console.log(totalUse[1]['count']);
 });
 </script>

This line is giving me grief: totalUse[user]['count'] = data.stats[x].count; I know that data.stats[x].count; contains data but it comes out as undefined when I do console.log(totalUse[user]['count']);.

• jameslafferty: Do you have an example of the JSON you get back?
• Brian S: "I know that data.stats[x].count; contains data but it comes out as undefined." -- If the value is undefined, then what you "know" is wrong. Either count is undefined, stats does not have an index x, stats is null/undefined, or data is null/undefined.
• Austin Lovell: { "stats":[ { "userId":13, "user":"AM Mobile", "service":"total", "month":"05/2013", "count":19694 }, { "userId":13, "user":"AM Mobile", "service":"total", "month":"06/2013", "count":4008 },

Answer 1

instead of:

totalUse[user] = data.stats[x].userId;
totalUse[user]['count'] = data.stats[x].count;

try :

totalUse[user] = { count : data.stats[x].count };

Answer 2

Use this:

if (totalUse[user]){ // yes it does have it
 totalUse[user].count += data.stats[x].count;
}else{
 totalUse[user] = {
 userId: user,
 count: data.stats[x].count
 };
}

If you want to save both the userID and count in each element of totalUse, you have to put them in different properties of the object. You can't assign the userID directly to the array element.

• Austin Lovell: Thank you for helping me understand. I really appreciate you taking the time to answer my question accurately.