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up vote 2 down vote favorite

This question already has an answer here:

Referring to the title. I have an array which I coded like this:-

$query = "SELECT * FROM server";
$result = mysql_query($query);
$dServer = array();

     while($row = mysql_fetch_assoc($result))
      {
              $dServer[] = $row['model'];
      }

Now, how do I pass the $dServer array into a Javascript array?

For example, this array:

var a = new Array();
share|improve this question

marked as duplicate by Benjamin Gruenbaum May 19 '14 at 15:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
    
What are you trying to do? –  wkm Apr 19 '12 at 2:00
    
pass all the array from $dServer into a. –  Hafiz Abdullah Apr 19 '12 at 2:01
3  
Please use the search function. About 1/4 of the questions in the "Related" sidebar answer your question. –  jprofitt Apr 19 '12 at 2:02
1  
@zerkms If they used quotes it would make a a string rather than a variable defined by JSON. –  jprofitt Apr 19 '12 at 2:04
    
@jprofitt: oops, missed that it is not an ajax response, but inline JS –  zerkms Apr 19 '12 at 2:19

4 Answers 4

up vote 7 down vote accepted 
$query = "SELECT * FROM server";
$result = mysql_query($query);
$dServer = array();

while($row = mysql_fetch_assoc($result)){
    $dServer[] = $row['model'];
}    

?>
<script type="text/javascript">
    var a = <?php echo json_encode($dServer); ?>;
</script>
share|improve this answer
up vote 2 down vote

Encode it as a json object.

<?
    $arr = array('entry' => 'content');
?>

<script>
var data = <?=json_encode($arr);?>;
alert(data['entry']);
</script>
share|improve this answer
up vote 1 down vote

Try to get use of ajax request and json_encode.

Second variant

<?php
$query = "SELECT * FROM server";
$result = mysql_query($query);
$dServer = array();

     while($row = mysql_fetch_assoc($result))
      {
              $dServer[] = $row['model'];
      }    
?>
var a = <?php echo json_encode($dServer);?>;
share|improve this answer
up vote 0 down vote

In addition to the ajax / json methods mentioned, you can directly print out the values:

<?php
  $query = "SELECT * FROM server";
  $result = mysql_query($query);
?>

<script type="text/javascript">
  var a = new Array();

<?php
  while($row = mysql_fetch_assoc($result)){
    echo "a['model'] = " . $row['model'] . ";"; 
    echo "a['nextField'] = " . $row['nextField'] . ";"; 
  }
?>
</script>
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