how do I loop a php variable within a javascript script

Question

I created a form with a text field that has Spry Validation (ie javascript). The user can select the number of rows in the form from 1 to 10. I need the code below to also expand but I'm not familiar enough with javascript to make it work.

$divkey is the variable that controls how many rows are in the form.

Original

<script type="text/javascript">
var sprytextfield1 = new Spry.Widget.ValidationTextField("sprytextfield1", "none", {validateOn:["change"], maxChars:20});
var sprytooltip1 = new Spry.Widget.Tooltip("sprytooltip1", "#sprytrigger1");
</script>

so I need the line 'var sprytextfield1...' to repeat based on $divkey with the next line being 'var sprytextfield2...' and so on. Can someone please rewrite this so it will work?

Trying to use php

<script type="text/javascript">
<?php   for ($i = 0; $i < $divkey; $i++) { $num=$i+1; ?>  
var sprytextfield<?php echo $num;?> = new Spry.Widget.ValidationTextField("sprytextfield<?php echo $num;?>", "none", {validateOn:["change"], maxChars:20});
<?php }?>
var sprytooltip1 = new Spry.Widget.Tooltip("sprytooltip1", "#sprytrigger1");
</script>

Trying to use javascript

<script type="text/javascript">
var numwrestler = <?php echo $wrestlerkey; ?>;
var sprytextfield = [];
for (var i = 0; i < numwrestler; i++) {
    var num = i+1;  
var sprytextfield[num] = new Spry.Widget.ValidationTextField("sprytextfield"+num, "none", {validateOn:["change"], maxChars:20});
}
var sprytooltip1 = new Spry.Widget.Tooltip("sprytooltip1", "#sprytrigger1");
</script>

In either attempt, what was the JavaScript which was emitted to the browser? — David
– David, Commented Jul 11, 2013 at 1:27
It looks to me like the only error in your JavaScript is the var in line beginning var sprytextfield[num] = ..., though you probably want to use index i, not num and it would be best practice to var num; outside of the loop and just set num inside (if you feel this variable necessary). — Paul S.
– Paul S., Commented Jul 11, 2013 at 1:38

maxton · Accepted Answer · 2013-07-11 01:39:38Z

1

I'd recommend that you use a Javascript array for this type of task. Your code is mostly correct, but the var in your for loop is incorrect, and the creation of the num variable instead of just using i is redundant.

<script type="text/javascript">
var sprytextfield = new Array();
var numwrestler = <?php echo $wrestlerkey; ?>;
for(var i = 0; i < numwrestler; i++){
    sprytextfield[i] = new Spry.Widget.ValidationTextField("sprytextfield"+(i+1), "none", {validateOn:["change"], maxChars:20});
}

var sprytooltip1 = new Spry.Widget.Tooltip("sprytooltip1", "#sprytrigger1");
</script>

Be sure that your PHP variable(s) are defined in the file before you include them in your script.

answered Jul 11, 2013 at 1:39

maxton

3921 silver badge6 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Paul S. Over a year ago

This still gives sprytextfield form [undefined, Object, Object, ..]

Matt Keel Over a year ago

very weird, anytime I make 'sprytextfield1 =' an array (ie sprytextfield[i] or sprytextfield[divnum] the javascript validation doesn't work. however this DOES work:

Matt Keel Over a year ago

var sprytextfield1 = new Spry.Widget.ValidationTextField("sprytextfield"+divnum, "none", {validateOn:["change"], maxChars:20}); ------- This is in the loop this way and the first sprytextfield1 doesn't mess it up... is that ok?

eebbesen · Accepted Answer · 2013-07-11 01:55:33Z

0

In your PHP code, you never define the variable divkey, by deafault the value will be 0. Try:

<script type="text/javascript">
<?php  $divkey = 10; for ($i = 0; $i < $divkey; $i++) { $num=$i+1; ?>  
var sprytextfield<?php echo $num;?> = new Spry.Widget.ValidationTextField("sprytextfield<?php echo $num;?>", "none", {validateOn:["change"], maxChars:20});
<?php }?>
var sprytooltip1 = new Spry.Widget.Tooltip("sprytooltip1", "#sprytrigger1");
</script>

Please, note that the variable that you are using as index $num in every iteration of the loop will be increasing 2, because of the i++ and the $num=$i+1

edited Jul 11, 2013 at 1:55

eebbesen

5,1558 gold badges52 silver badges73 bronze badges

answered Jul 11, 2013 at 1:34

gryphes

272 bronze badges

3 Comments

Fallen Over a year ago

nope. $num is not increasing twice. It's getting $i+1 and $i is increasing once every iteration. So if $i=1 then $num=2, if $i=2, $num=3 etc :)

gryphes Over a year ago

That's true, my fault. Sorry :\

Matt Keel Over a year ago

$ divkey is defined further up the page. This not the complete page.

Collectives™ on Stack Overflow

how do I loop a php variable within a javascript script

2 Answers 2

3 Comments

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related