Moving data from SQL database to an array in JavaScript via PHP

Question

Ive been trying this for hours now but it wont quite get there.

I have a database which amongst other things contains geocodes, lat and lon. I have accessed these using the following PHP

<?php
 mysql_connect("localhost", "tompublic", "public") or die(mysql_error()); 
 mysql_select_db("first_section") or die(mysql_error()); 

$data = mysql_query("SELECT geo_lat, geo_lon FROM first_page_data")
or die(mysql_error());
while ($row = mysql_fetch_assoc($data)){
$lat[] = $row['geo_lat'];
$lon[] = $row['geo_lon'];
}
?>

These values in $lat and $lon then need to be put into an array in a javascript function like so:

var latit = [];
var longi = [];
latit = '<?php echo $lat[]; ?>';
longi = '<?php echo $lon[]; ?>';

But it wont work! Any ideas?

Wouldn't latit just equal Array?

Leeish
– Leeish

2014-01-10 22:51:42 +00:00
Commented Jan 10, 2014 at 22:51 — Leeish
– Leeish, Commented Jan 10, 2014 at 22:51

johnnycardy · Accepted Answer · 2014-01-10 23:10:00Z

1

Try:

var latit = <?php echo json_encode($lat); ?>;
var longi = <?php echo json_encode($lon); ?>;

Edit: Also, the mysql_ functions are deprecated.

edited Jan 10, 2014 at 23:10

answered Jan 10, 2014 at 22:51

johnnycardy

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Comments

ManfredP · Accepted Answer · 2014-01-10 22:53:58Z

1

You could try this:

var latit = [<?php echo implode(",",$lat) ?>];
var longi = [<?php echo implode(",",$lon) ?>];

answered Jan 10, 2014 at 22:53

ManfredP

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Comments

Tomas · Accepted Answer · 2014-01-10 22:56:24Z

1

First thing is first try to switch to MySQLi due to the fact that Mysql is depreciated.

But try

var latit = <?php echo json_encode($lat); ?>;
var longi = <?php echo json_encode($lon); ?>;

answered Jan 10, 2014 at 22:56

Tomas

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user1607528 · Accepted Answer · 2014-01-10 22:59:48Z

0

You can either do as @Robbert stated

OR Using php json_encode convert it to JSON string and you need to JSON.parse() it to convert it to javascript object

 var latit = JSON.parse("<?php echo json_encode($lat); ?>");
 var longi = JSON.parse("<?php echo json_encode($lon); ?>");

answered Jan 10, 2014 at 22:59

user1607528

Comments

Community · Accepted Answer · 2023-11-17 20:33:26Z

0

JavaScript arrays are created using a comma separated list surrounded by brackets

var latit = [];

To use your PHP values

var latit = [<?PHP echo implode(",",$lat); ?>];

This assumes your values are numbers. If not, you'll need to include quotes.

var latit = ['<?PHP echo implode("','",$lat); ?>'];

Finally, json_encode is a good option as many of the other answers indicate.

edited Nov 17, 2023 at 20:33

CommunityBot

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answered Jan 10, 2014 at 22:51

Robbert

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2 Comments

Giovanni Over a year ago

what does 'implode' do?

Robbert Over a year ago

Implode is the same as join, it takes an array and joins the items together with the glue, in this case a comma: us3.php.net/function.implode

Alex Shilman · Accepted Answer · 2014-01-10 23:00:43Z

0

Try this:

<?php
 mysql_connect("localhost", "tompublic", "public") or die(mysql_error()); 
 mysql_select_db("first_section") or die(mysql_error()); 

$data = mysql_query("SELECT geo_lat, geo_lon FROM first_page_data")
 or die(mysql_error());
while ($row = mysql_fetch_assoc($data)){
$lat[] = $row['geo_lat'];
$lon[] = $row['geo_lon'];
}
 echo '
 <script type="text/javascript">    
  var latit  = '.json_encode($lat).';
  var longi  = '.json_encode($lon).';
 </script>
';
?>

answered Jan 10, 2014 at 23:00

Alex Shilman

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