Displaying return value of a function in JQuery in another function

Question

I have a JQuery which is retunring an array of LockedProjects [] which is decalred as global variable.

var LockedProjects = [];

I wanted to display the values of this array LockedProjects in another JQuery I m accesing this array of values in an alert box to display and separating them with comma. Before displaying that I am checking the length of it.

if(LockedProjects.length>0)
alert(LockedProjecrs.join(,));

'if' condition is passed But It is not displaying any values. Can any one help on this?

syntax error , isn't a string. "," is.

Matt Burland
– Matt Burland

2014-04-09 18:32:34 +00:00
Commented Apr 9, 2014 at 18:32 — Matt Burland
– Matt Burland, Commented Apr 9, 2014 at 18:32

George · Accepted Answer · 2014-04-09 18:31:41Z

1

join() expects string, you've given it a comma. Also, 'LockedProjects' is spelt incorrectly. I believe it should be:

alert(LockedProjects.join(","));

answered Apr 9, 2014 at 18:31

George

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2 Comments

user3356020 Over a year ago

I am using like this only. I kept within double quotes still it did nt work

George Over a year ago

Can you re-produce the issue? Perhaps on jsfiddle.net? See this

Claudio Redi · Accepted Answer · 2014-04-09 18:34:07Z

0

The default separator for join is , so you could simply do

LockedProjects.join()

The separator is converted to a string if necessary. If omitted, the array elements are separated with a comma.

answered Apr 9, 2014 at 18:34

Claudio Redi

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