通过PHP将数据从SQL数据库移动到JavaScript中的数组

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?

Answer 1

Try:

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

Edit: Also, the mysql_ functions are deprecated.

Answer 2

You could try this:

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

Answer 3

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); ?>;

Answer 4

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.

Answer 5

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); ?>");

Answer 6

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>
';
?>