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up vote 1 down vote favorite

So I want to store all of my MySQL results in an array I can manipulate with javascript/jQuery.

Here is my current code:

<?php
    $sql = "SELECT * FROM potentials";
    $result = mysql_query($sql) or die(mysql_error());
    $potential = mysql_fetch_array($result);
    echo json_encode($potential);
?>

And my Javascript:

<script type="text/javascript">
    $(document).ready(function(){
        var myArray = "<?php print(json_encode($potential)); ?>";
        console.log(myArray)
    )};
</script>

I keep getting "Unexpected number" or "unexpected identifier". Whats going on?

share|improve this question
 
it echos out fine, but doesnt console.log in the javascript. –  Rick Bross Apr 14 '13 at 15:31
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3 Answers

up vote 2 down vote accepted

json_encode() returns an object using JSON notation. Strangely, you surround it with quotes. Try removing them :

<script type="text/javascript">
$(document).ready(function(){
    var myArray = <?php print(json_encode($potential)); ?>;
    console.log(myArray)
)};
</script>

(and it's not an array, it's an object, so you might want to rename your variable ;) )

share|improve this answer
 
That did it thanks! –  Rick Bross Apr 14 '13 at 15:36
 
Its now spitting out only the first row of potentials. How to I get all rows? a multidemensional array? what would that look like? maybe a foreach statement? –  Rick Bross Apr 14 '13 at 15:39
 
I would go for something like this : <?php $sql = "SELECT * FROM potentials"; $result = mysql_query($sql) or die(mysql_error()); $potential = array(); while($row = mysql_fetch_assoc($result)) { $potential[] = $row; } echo '<script type="text/javascript"> $(document).ready(function(){ var myArray = "' . json_encode($potential) . '"; console.log(myArray) )}; </script>' . "\n"; ?> –  Über Apr 14 '13 at 16:34
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up vote 0 down vote

You are using json_encode() twice, however if you want to use it, you need to parse it. In jQuery this can be done via jQuery.parseJSON like this

var myArray = jQuery.parseJSON('<?php print json_encode($potential); ?>');

Also if you want all the results, you need to loop the query (e.g. with while) and then save it to an array

$array = array();
while ($row = mysql_fetch_array( mysql_query( $query ) )
{
    $array[] = $row;
}
share|improve this answer
 
That gives me this in the console: Uncaught TypeError: Object function (E,F){return new o.fn.init(E,F)} has no method 'parseJSON' –  Rick Bross Apr 14 '13 at 15:34
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up vote 0 down vote 
<script>
    var data = '<?php echo $data; ?>';
    var json = JSON.parse(data);
    console.log(json[0]); //etc
</script>

Notice that var data = ... is SINGLE QUOTED, so you catch the echo from php as a String

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