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up vote 0 down vote favorite

I'm wondering for some time now why this fails:

if (longest.length >= 3) {
    for ( var i = 0; i < longest.length-1; i++) {
      var $last[i] = longest[i].splice(-1).toString();
      //if ( $( $last[i] ).closest(':jqmData(role="panel")') == "popover" ) { 
      //var gotoPage = $last[i];
    // }
      }
    }

longest is an array, which contains array elements.

I want to loop through the number of arrays in longest and create variables from each nested arrays' last element. The .splice(-1).toString() fails telling me "missing ; before statement"

What am I doing wrong?

EDIT: longest will look something like this:

[[#menu, #menuSub1, #menuSub2], [#main, yield, yield], [#pop1, #pop1-1, #pop1-2]]

It's within a function to track the browser-history in a mobile-app with different panels.

EDIT2: Finished code (Thx Mic):

if (longest.length >= 3) {
   var $last = [];
   for ( var i = 0; i < longest.length; i++) {
      $last.push( longest[i][ longest[i].length - 1 ]);
      if ( $( $last[i] ).closest(':jqmData(role="panel")').jqmData('panel')  == "popover" ) { 
         var gotoPage = $last[i]; }
      }
   }
share|improve this question
    
what is $last?? –  Baz1nga Oct 21 '11 at 19:18
    
should be the variable name. $last0, $last1, $last2... –  frequent Oct 21 '11 at 19:18
    
I believe splice() requires two parameters, both being positive integers or zero. –  Diodeus Oct 21 '11 at 19:21
    
You have one extra "}" out there, you probably forgot to comment out. –  Birey Oct 21 '11 at 19:21
    
@Diodeus: I'm using it quite a lot, always like this, so I think that won't help. I think either chainging splice().toString() doesn't go or I can't declare variables like this: $last[i] –  frequent Oct 21 '11 at 19:24

1 Answer 1

up vote 5 down vote accepted

You should write something like:

if (longest.length >= 3) {
    var $last = [];
    for ( var i = 0; i < longest.length-1; i++) {
      $last.push( longest[i][ longest[i].length - 1 ] );
    }
  }

and get the values with $last[0]

Or use a hash like:

if (longest.length >= 3) {
    var hash = {};
    for ( var i = 0; i < longest.length-1; i++) {
      hash['$last'+ i] = longest[i][ longest[i].length - 1 ];
    }
  }

and get the content using for instance hash.$last0

share|improve this answer
    
may be give a sample of longest, it would help –  Mic Oct 21 '11 at 19:27
    
edited my code above –  frequent Oct 21 '11 at 19:31
    
I edited the answer, is it better? This is ok if you have just 2 levels of arrays, if you have more, it needs to recurse –  Mic Oct 21 '11 at 19:31
    
longest.length-1 is a bit strange as you will never reach the last element of longest –  Mic Oct 21 '11 at 19:33
    
Your first snippet works perfectly! THANKS. I'll probably be dreaming "array-style" tonight... –  frequent Oct 21 '11 at 19:49

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