-5

I tried the following code to retrieve values from the database and store them in a javascript array using php array. I tried using the following code, But it is returning me a Reference Error array is not defined.The code is as follows.

<?php
$con = mysql_connect("localhost","root","root");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("cerebra", $con);
$sql="select name from details order by download desc limit 20";
if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
$query=mysql_query($sql,$con);
$names=array();
$index=0;
while($row=mysql_fetch_array($query)){
    $names[$index]=$row[0];
    $index++;
}
?>
<script>
var comp=new array();
<?php
 $i=0;
 foreach($names as $a){
        $i++;
        echo "comp[$i]='".$a."';\n";

        }
?>
for(i=0;i<comp.length;i++)
            alert(comp[i]);

</script>
Improve this question
2
  • An possibly unrelated note: there's no need for $index in your while loop you may as well use $names[] = $row[0]. Commented Mar 14, 2013 at 9:30
  • 2
    Your browser's developer tools will be able to point you to what exactly is not defined; then you will have to determine why and fix that. Commented Mar 14, 2013 at 9:30

4 Answers 4

Reset to default
1

It is var comp = new Array() not array(). Skip that anyhow and use var comp = [] right away.

Improve this answer
Sign up to request clarification or add additional context in comments.

3 Comments

but the first value am getting is undefined. How come?
See why you get issues with undefined in you Javascript in my updated answer.
yes stupid me...sorry..i placed the increment there to debug the code. And that ended up giving me an error.
1

I think you're going about this the wrong way. Here is a much better way of going about it:

<?php
  $phpArray = array("foo", "bar", "baz");
  //....
?>

<script type="text/javascript">
var jsArray = <? echo json_encode($phpArray); ?>;
</script>

Taken from here: How to use an array value from php to javascript?

Improve this answer

Comments

1 
new array();

should be

new Array();

in your javascript.

EDIT: you should also stop using mysql_* functions.

Improve this answer

Comments

1

First of all, why do you run your query twice?

...

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
$query=mysql_query($sql,$con);

...

Second in Javascript the array object is called Array, not array. So try with 

var comp = new Array();

Update:

but the first value am getting is undefined. How come? – user2129868

Because you increment $i before getting the value. So $i is 1 and not 0 on your first iteration in your Javascript.

So change

<script>
var comp=new array();
<?php
 $i=0;
 foreach($names as $a){
        $i++;
        echo "comp[$i]='".$a."';\n";

        }
?>
for(i=0;i<comp.length;i++)
            alert(comp[i]);

</script>

to

<script>
var comp=new array();
<?php
 $i=0;
 foreach($names as $a){
        echo "comp[".$i++."]='".$a."';\n";
        }
?>
for(i=0;i<comp.length;i++)
            alert(comp[i]);

</script>

Otherwise when you try to fetch the elements with

for(i=0;i<comp.length;i++)
            alert(comp[i]);

0 will be undefined since 1was the first index you added.

Improve this answer

Comments

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.