php array to 2d Javascript array

Question

I'm working on passing 2 fields of SQL through php to javascript. I have read many tutorials on how to create a multidimensional javascript array. Where I get confused is how to code from php to javascript. I have seen a couple of tutorials on how to get the php data to javascript, but none on how to do this with 2 dimensions.

My first hangup is that if I'm creating a multidimensional array I need to count the number of records in my sql data before I declare the java array right?

update:

I got the data to JSON format as suggested below. Is there a way for me to get all of the contents printed to the web page so that I can see them and then narrow down what is displayed?

update III:

code:

mysql_connect("localhost", "bikemap", "pedalhard") or die(mysql_error()); 
 mysql_select_db("test") or die(mysql_error()); 
 $data = mysql_query("SELECT * FROM gpsdata");
 $new_row = array();
$new_column = array();
 while($info = mysql_fetch_array($data)){

   foreach( $info as $row_num => $row)
{
      $thisItem = $row;
      $new_row[] = $thisItem;
   }
   array_push($new_column = $new_row);

 }
 $json = json_encode($new_column);
echo $json;    
?>

Working code:

 $data = mysql_query("SELECT * FROM gpsdata");     
$aData = array();
while($row = mysql_fetch_assoc($data))
   $aData[$row['idgpsdata']] = array($row['userID'],$row['date'],$row['lat'], $row['longi'], $row['alt']);


 $json = json_encode($aData);
echo $json;

You can shorten the current code: var scriptAr = <?php echo json_encode($lines); ?>; — ThiefMaster
– ThiefMaster, Commented Nov 13, 2011 at 19:32
@LorenZimmer Yea, you're using mysql_fetch_array incorrectly. This only gets one row at a time. Programming languages don't know about tables, so you grab a row at a time. You need to actually build the 2D array. — Matt
– Matt, Commented Nov 14, 2011 at 11:08
update... the code above returns all of the values but not the keys. The weird thing for me is that it creates duplicate values when plugged into @Laith Shadeed 's java routine. — Loren Zimmer
– Loren Zimmer, Commented Nov 15, 2011 at 2:19

Marco Mariani · Accepted Answer · 2011-11-13 19:46:31Z

0

Fill a PHP array first, it's easier than building the string for a javascript array. Then - as ThiefMaster said as comment - use JSON to make the javascript array.

edited Nov 13, 2011 at 19:46

Marco Mariani

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answered Nov 13, 2011 at 19:43

Yogu

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If I add the varable $json between <?php and ?> can it be used outside as well?

Loren Zimmer
– Loren Zimmer

11/13/2011 20:20:11
Commented Nov 13, 2011 at 20:20
Yes, it can be used in any PHP section of the script.

Yogu
– Yogu

11/13/2011 21:02:42
Commented Nov 13, 2011 at 21:02

Add a comment |

Community · Accepted Answer · 2017-05-23 11:47:59Z

In PHP, you can use JSON PECL extension

<?php
$arr = array( 1=>array(3,4),
              2=>array(4,5));
$json = json_encode($arr);
echo $json;
?>

Output

{"1":[3,4],"2":[4,5]}

In Javascript

var obj = JSON.parse('{"1":[3,4],"2":[4,5]}');
for(var i in obj){
    for(var j in obj[i]) {
        console.log(obj[i][j]);
    }
}

JSON.Parse is for Gecko (Firefox alike), for cross-browser javascript JSON parse check jQuery.parseJSON or https://stackoverflow.com/search?q=json+parse+javascript

Sample implementation in PHP/jQuery, would be something like this:

json.php

$arr = array( 1=>array('Name', 'Age'),
              2=>array('Fares','18'));
$json = json_encode($arr);
echo $json;

json.html

<html><head></head>
<body>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.0/jquery.min.js"></script>
<script>
$.getJSON('json.php', function(data) {
  var items = [];

  $.each(data, function(key, val) {
    items.push('<li id="' + key + '">' + val + '</li>');
  });

  $('<ul/>', {
    'class': 'my-new-list',
    html: items.join('')
  }).appendTo('body');
});
</script>
</body>
</html>

thanks for your suggestions. In my original question you will see that I have substituted some code to get sql results converted to json but I have not been completely successful. Any thoughts what I have entered incorrectly. I have a feeling that I am not passing the information into an array correctly. — Loren Zimmer
– Loren Zimmer, Commented Nov 14, 2011 at 11:01

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