up vote 3 down vote favorite

Below is my code where the php array $keys assigned to a JS array var keysArr, But the value display on the alert box is not correct.

Thus, there is something wrong on assign Php array to Js array.

Can anyone help me? Thanks in advance!

<?php
$keys = array(1, 2, 3, 4);
?>
<html>
    <head>
    <script type="text/javascript">
        var keysArr = <?php print $keys?>;
        for (var i = 0; i < keysArr.length; ++i){

                    alert(keysArr[i]);

        }
    </script>
    </head>
    <body>
    </body>
</html>
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3 Answers

up vote 0 down vote accepted

Try this

<?php
$keys = array(1, 2, 3, 4);
?>
<html>
    <head>
    <script type="text/javascript">
        var keysArr = Array(<?php echo implode(',',$keys);?>);
        for (var i = 0; i < keysArr.length; ++i){

                    alert(keysArr[i]);

        }
    </script>
    </head>
    <body>
    </body>
</html>

whith print you're just showing Array, and not proper array code for js. Look inside the HTML you generate.

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4  
That will break horribly for any array containing non-Number data or any array consisting of just one value. – Quentin Nov 8 '11 at 12:38
yes, youre right. It will only work for arrays containg more than one number. – Kokers Nov 8 '11 at 12:44
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up vote 10 down vote 
var keysArr = <?php json_encode($keys) ?>;

requires PHP >= 5.2.0, but third-party json_encode() implementations are available for older versions.

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up vote 1 down vote

Try this, it doesn't need JSON extension:

var keysArr = [ <?php print implode(',', $keys); ?> ];
    for (var i = 0; i < keysArr.length; ++i){
        alert(keysArr[i]);
    }
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2  
That will break horribly for any array containing non-Number data – Quentin Nov 8 '11 at 12:39
I'm sorry, my mistake. – fonini Nov 8 '11 at 12:54
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