Array in jQuery does not work

Question

I have an array in jQuery like this

var matches = [];
  jQuery(".block").each(function() {
  matches.push(this);
});

Now I want to use the matches array like this

var len = jQuery(matches).length;
for (var i = 0; i < len; i++){
  function slider() {
    jQuery(matches[i]).show("slow");
    jQuery(matches[i]).animate({left:'+=730'},3000);
    jQuery(matches[i]).show("normal", slider);
  }
}

When I put number in the place of I, it works, but the for loop is not working. Please help me where the mistake.

Why do you have a function declaration in a loop? Is that even valid? — IsisCode
– IsisCode, Commented Feb 3, 2012 at 13:13
What you seem to be trying to do is highly unlikely to be necessary. jQuery objects are array-like containers for a set of DOM elements. You can iterate over a jQuery object (you already do, to build your array) so there shouldn't be any need to build an array just so you can iterate over it! Why not just do all the necessary work in the original each? But maybe I'm wrong... what are you actually trying to achieve? — James Allardice
– James Allardice, Commented Feb 3, 2012 at 13:15
i am trying to call the the function like jQuery(matches[1]).show("slow"); then jQuery(matches[2]).show("slow"); — Rain
– Rain, Commented Feb 3, 2012 at 14:29

LordZardeck · Accepted Answer · 2012-02-03 13:33:46Z

Ss the commenter said, you have it wrapped in a functon thats not called. Also, i tested it, wrapping jquery around the array has no effect, so that's not the problem. Also, is there any reason you are not executing that code in the each function? Try this instead:

jQuery(".block").each(function() {
    jQuery(this).show("slow");
    jQuery(this).animate({left:'+=730'},3000);
    jQuery(this).show("normal", slider);
});

You could even chain them together like so:

jQuery(".block").each(function() {
    jQuery(this).show("slow", function(){jQuery(this).animate({left:'+=730'},3000, function(){jQuery(this).show("normal", slider);});});
});

One final note, if this is an effect you are using multiple different times, you could wrap the effect in a function and call it in the iteration:

jQuery(".block").each(function() {
    slider(this);
});
function silder(el)
{
     jQuery(el).show("slow", function(){jQuery(el).animate({left:'+=730'},3000, function(){jQuery(el).show("normal", slider);});});
}

Reinstate Monica Cellio · Accepted Answer · 2012-02-03 13:16:47Z

1

Try this...

for (var i= 0; i< matches.length; i++){
    jQuery(matches[i]).show("slow");
    jQuery(matches[i]).animate({left:'+=730'},3000);
    jQuery(matches[i]).show("normal", slider);
}

It's more likely that you want to replace the whole of your example code with just this...

jQuery(".block").each(function() {
    jQuery(this).show("slow");
    jQuery(this).animate({left:'+=730'},3000);
    jQuery(this).show("normal", slider);
});

With that you don't need to declare or use an array. Obviously if there's a reason you're doing it that way then fair enough, but it does seem unnecessary.

There will be conflicts between the 2 show functions though.

answered Feb 3, 2012 at 13:16

Reinstate Monica Cellio

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