Javascript Insert into Array

Question

I have javascript code some thing like this --

var count=3;
var pl=new Array(count);
var il=new Array(count); 
pl.push('<?php echo $var1; ?>');
il.push('<?php echo $var2; ?>');

Why is this not working?

But if I try some thing like this

pl[0]='<?php echo $var1; ?>';

It works. Can some one point out the problem?

Thanks

• mario: You're aware where PHP runs, and where Javascript runs? Explain how / from where your "javascript code" is delivered, not just what it contains.
• Amadan: Also, please define "doesn't work". Both of those should work, but have wildly different results. pl.push(10) results in [undefined, undefined, undefined, 10]; while pl[0] = 10 results in [10, undefined, undefined].

Answer 1

By using the Array constructor you are creating an array with a number of "slots" already defined. In your case, the arrays both have 3 elements when you create them (the value of each element is undefined.)

When you use push, you add an extra element to that array. When you set the element at a specific index, you just set the value of that element.

I would suggest not using the Array constructor, and instead creating array literals:

var pl = [];
var il = [];
pl.push('<?php echo $var1; ?>');
il.push('<?php echo $var2; ?>');

This time, the arrays are created with no elements, so when you use push you are pushing an element into index 0, instead of 4.

• Fox: Thank You for providing the explanation too. My code is working .. :)
• Rob Grant: This isn't safe; @anttihaapala's is correct. And shorter and quicker to run.

Answer 2

Try this:

var pl=[];
var il=[]; 
pl.push('<?php echo $var1; ?>');
il.push('<?php echo $var2; ?>');

Answer 3

The new Array(count) creates a new Array object with length set to count, but no indices set. When you print the array, you get array.length times undefined. Now,

array.push(value);

is the same as

array[array.length] = value;
array.length ++;

Which after new Array(3) would set the element number 3 (the 4th element);

pl[0] = '<?php echo $var1; ?>';
pl[1] = '<?php echo $var2; ?>';
pl[2] = '<?php echo $var3; ?>';

does work "correctly"; that is according to the expectations; but you still do not need to initialize the length of array; even doing

var pl = [];
pl[0] = 1;
pl[1] = 2;
pl[2] = 3;

Would still cause the pl.length to be 3.

---

However your PHP code has a serious flaw that can undermine the security of both your website and the users using it. If $var1 contains a ' character (because it comes from the user or you insert it by accident), you will shoot yourself in the foot - say if $var1 was set to:

$var1 = "'; $.ajax('/user/delete', { method: 'post' }); '"

Please do yourself a favor and use the json_encode function instead to properly quote the values:

var pl = [],
 il = []; 
pl.push(<?php echo json_encode($var1); ?>);
il.push(<?php echo json_encode($var2); ?>);

or even better notice that you can use the json_encode to readily create arrays, as

var pl = <?php echo json_encode(array(1, 'a', 'b', 5.7)) ?>;

This would produce JavaScript code:

var pl = [1,"a","b",5.7];

---

You must be careful with the unicode characters U+2028 and U+2029, as these are proper JSON but not proper JavaScript. json_encode does seem to encode them as \u2028 and \u2029 which is safe.