Assigning array variables in Javascript

Question

I am having a small issue with variable assignment. For some reason my line variable doesn't properly assign.

var records = new Array();
var recid = -5
var subrecid = 6495;
var line = new Array();
line['recid'] = recid;
line['subrecid'] = subrecid;
if (subrecid > 0) records.push(line);

• Ry-: Actually, it does. Instead of checking what you think is the solution, can you provide a jsFiddle showing the error?
• Gareth: How do you know it doesn't "assign"? What result are you expecting to get that you aren't getting?
• Nanne: mind you that javascript does NOT have associative arrays. You can use the 1line['syntax'] because you are actually calling an object there, but there is associative array in javascript. It might be the fact you just init it as an array?
• Ry-: JavaScript doesn't need arrays for that, by the way; there's no such thing as a hash or associative array. You can use var line = {}; line.recid = recid; line.subrecid = subrecid; instead.
• Ry-: @MДΓΓБДLL: Yeah, it depends on your definition.

Answer 1

Don't use an array for non-integer indexing. Use an object. Also, it's generally better to use [] instead of new Array(). Oh yeah, and there's a line missing a semicolon.

var records = [];
var recid = -5;
var subrecid = 6495;
var line = {}; // object, not array
line.recid = recid;
line.subrecid = subrecid;
if (subrecid > 0) records.push(line);

Even more concise:

var records = [];
var recid = -5;
var subrecid = 6495;
var line = {
 recid: recid,
 subrecid: subrecid
};
if (subrecid) records.push(line);

• RobG: Ha! You did that while I was typing. I considered the different test but it's not clear whether or how subrecid is being modified so it might become -ve at some point.

Answer 2

Matt's answer is fine, but you could take greater advantage of object literal syntax:

var records = [];
var line = {recid: -5, subrecid: 6495 };
if (line.subrecid > 0) records.push(line);

• Jason Sundram: or for maximum terseness: var records = [{recid: -5, subrecid: 6495 }];

Answer 3

var val1=$('#Accountstype_account_c').val();

• Trisha: It would be helpful to add a bit more explanation as to why this is a solution.