How can I use a JSON array generated in PHP in a javascript function?

Question

I'm trying to use a JSON array generated in a php function in a javascript function. My code is this:

$query = "SELECT lat, lng FROM Eventi";
$result = mysql_query($query) or die(mysql_error() . "<br/><br/>" . $query);
if (mysql_affected_rows() != 0) {
     while($r = mysql_fetch_array($result)) {
        $rows =  array(
           "latitudine" => $r['lat'],
           "longitudine" => $r['lng'],
       );}
       $risultato_rows = json_encode($rows);

Now I want to recover them in a subroutine did javascript to use them, and I tried so:

var res = JSON.parse($risultato_rows);
    alert var prova = res.[latitudine];

This code doesn't work; what can I do to make it function properly?

Is the first snippet PHP and the second snippet JavaScript? If so, you can't use the same variables directly. — pimvdb
– pimvdb, Commented Nov 13, 2012 at 18:35
are you echo? var res = JSON.parse(<?php echo $risultato_rows?>); console log after its, console.log(res); — jd_7
– jd_7, Commented Nov 13, 2012 at 18:37

Sushanth -- · Accepted Answer · 2012-11-13 18:35:05Z

2

res.[latitudine];  // You seem to mix up both the
                   // dot and bracket notation..

supposed to be either

res.latitudine; OR res["latitudine"];

answered Nov 13, 2012 at 18:35

Sushanth --

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Comments

Denys Séguret · Accepted Answer · 2012-11-13 18:42:56Z

1

A PHP variable isn't directly visible in Javascript. Supposing you're not doing AJAX but just trying to embedd JSON in your script, you might do this :

?><script>
var res = JSON.parse('<?php echo $risultato_rows; ?>');
var prova = res.latitudine;
alert (prova);
</script><?php

edited Nov 13, 2012 at 18:42

answered Nov 13, 2012 at 18:36

Denys Séguret

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3 Comments

pimvdb Over a year ago

One may just echo the JSON directly since JSON is a subset of JavaScript.

Musa Over a year ago

@Asad The keys are always quoted, how does that prevent you from just echoing it.

Stefania Over a year ago

I try with: var res = JSON.parse('<?php echo $risultato_rows; ?>'); var prova = res.latitudine; alert (prova); and var res = JSON.parse('<?= $risultato_rows; ?>'); var prova = res.latitudine; alert (prova); but in both case don't work!

score 0 · Accepted Answer · 2012-11-13 18:48:52Z

0

There are a few problems with your code:

You need to echo the JSON string that is contained in $risultato_rows. That is:

var res = JSON.parse('<?= $risultato_rows; ?>');
When accessing properties, you put the key name as a string in the brackets. To access the latitudine property, you would use res.["latitudine"]
The var keyword is used when declaring variables, not when accessing them. Use separate statements to assign the value to prova and alert it.

edited Nov 13, 2012 at 18:48

answered Nov 13, 2012 at 18:42

user1726343

1 Comment

Stefania Over a year ago

I try with: var res = JSON.parse('<?php echo $risultato_rows; ?>'); var prova = res.latitudine; alert (prova); and var res = JSON.parse('<?= $risultato_rows; ?>'); var prova = res.latitudine; alert (prova); but in both case don't work!

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