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I want to perform javascript functions on a page after I retrieve some values from a local database. However once the PHP code is in my javascript the javascript function won't even run in the first place. If I separate the PHP code out and run the PHP code alone, it works fine, as well as the javascript code too. Here's my code:

<?
mysql_connect("127.0.0.1:3307", "username", "password") or die ("Error fetching value from database.");
mysql_select_db("ccmalumni");
$result = mysql_query("SELECT DISTINCT value FROM ccm_bp_xprofile_data WHERE field_id = 16") or die ("What da heck");
$states = db_result_array_values($result);
echo $states[0];
mysql_close();

function db_result_array_values($result) { 
for ($array = array(); $row = mysql_fetch_row($result); isset($row[1]) ? $array[$row[1]] = $row[0] : $array[] = $row[0]); 
return $array; }
?>

var present = <? echo json_encode($states); ?>;

Please, what am I doing wrong? You can see the full code here.

Thanks.

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declare your vars between <script> & </script> tags –  Yellow Bird Jun 28 '12 at 7:33
 
are you using a javascript console in your browser to debug this? is it showing any parsing errors when loading the page? –  mikegreiling Jun 28 '12 at 7:36
 
also what is being echoed in the line echo $states[0];? is it valid javascript? –  mikegreiling Jun 28 '12 at 7:37
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3 Answers

up vote 4 down vote accepted

json_encode will out put json string in the form that javascript can understand so basically fix like below;

var darr = <? echo json_encode($states); ?>;  //Already readable JSON
var present = darr ;
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1  
You can try to view source on browser and see what is printed out –  Trinh Hoang Nhu Jun 28 '12 at 7:34
1  
Or simply var present = <?php echo json_encode... –  Salman A Jun 28 '12 at 7:47
 
Actually, that was a more recent modification, my original code simply had var present = <? echo json_encode($states); ?> (who wants to do extra coding). I solved the problem by viewing the source, I discovered all my PHP code was not being processed. The problem was because I started my code with <? instead of <? php. Thanks for your help! –  Chibueze Opata Jun 28 '12 at 9:57
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up vote 2 down vote

$.parseJSON() parses a string to javascript object. var darr is already an object , is not a string, and therefore should not be passed to $.parseJSON()

See $.parseJSON() API Docs

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up vote 0 down vote

if you want to get data from server to client, i think the best way is to do a ajax request.

Because you dont have to mix server with client side
(makes your code more readable, easy to understand etc.)

here you get a short description how to do it with jQuery: http://stackoverflow.com/a/415894/1067061

hope it helps, good luck!

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