Variable between php and Javascript

Question

code:

<?php

    mysql_connect("localhost", "bikemap", "pedalhard") or die(mysql_error()); 
    mysql_select_db("test") or die(mysql_error()); 
    $data = mysql_query("SELECT * FROM gpsdata");     
    $aData = array();
    while($row = mysql_fetch_assoc($data))
        $aData[$row['idgpsdata']] = array($row);
        $json = json_encode($aData);
?>

    var json = "<?php echo $json; ?>";
    document.write(json.length);
    alert('half done');

    for(var x=0; x < json.length; x++) {
        document.write("<li>"+ x + " " + json[x]+ "</li>");
    }

Output of the code:

20

0 <
1 ?
2 p
3 h
4 p
5
6 e
7 c
8 h
9 o
10
11 $
12 j
13 s
14 o
15 n
16 ;
17
18 ?
19 >

if you take out the line numbers is: "<?php echo $json; ?>"

Looks suspiciously like a line where I'm trying to transfer the php variable $json to javascript variable json.

I've tried every type of bracketry and quote to get this to work. Anyone see any mistakes or have any suggestions?

Are you sure that the PHP code is being executed? That is, are you requesting the page from a web server with PHP enabled or are you just opening the page from the file system in the browser? — Vincent Ramdhanie, Nov 16 '11 at 04:41
@VincentRamdhanie I am running the code from a site hosted with PHP enabled. — Loren Zimmer, Nov 16 '11 at 04:43
@LorenZimmer, why don't you "view source" on the output of your script and show us that. — Brad, Nov 16 '11 at 04:44
http://stackoverflow.com/questions/168214/pass-a-php-string-to-a-javascript-variable-including-escaping-newlines — Shomz, Nov 16 '11 at 04:44
@Brad this kind of development is new to me where would I "view source" from? — Loren Zimmer, Nov 16 '11 at 04:51
from mozilla: [23:56:39.850] An unbalanced tree was written using document.write() causing data from the network to be reparsed. For more information https://developer.mozilla.org/en/Optimizing_Your_Pages_for_Speculative_Parsing @ http://98.169.0.140:8080/Map/trials/testii.html:5 is this what you were mentioning Brad? — Loren Zimmer, Nov 16 '11 at 04:57
The code is doing exactly what you programmed it too. What did you expect to see? — James Anderson, Nov 16 '11 at 05:05
@Shomz I did try it without success thanks for the find though — Loren Zimmer, Nov 16 '11 at 05:06
@JamesAnderson I thought that it would pass the contents of the variable $json from the php script to var json in javascript. I did not expect it to pass the text of the code to json in javascript. — Loren Zimmer, Nov 16 '11 at 05:08

score 1 · Accepted Answer · answered Nov 16 '11 at 11:00

1

Try

var json = {<?php echo $json; ?>};

or maybe

var json = eval("{<?php echo $json; ?>}");

Be sure your code is interpreted as php by the server. The file extension '.js' is usually not configured in this way.

If you want to embed your php file as a js, you'd better to rename it as .php and add the required header like that :

<?php
header("Content-type","text/javascript");
?>

var json = {<?php echo $json; ?>};

answered Nov 16 '11 at 11:00

Naypa

182
6

I needed just the php file to pass the variables – Loren Zimmer Nov 17 '11 at 06:04

score 0 · Answer 2 · answered Nov 16 '11 at 05:10

0

You could try:-

var json = <? echo $json ?>;

You cannot execute php code inside the Javascript. Instead you must get php to write the relevent javascript code;

answered Nov 16 '11 at 05:10

James Anderson

26,221
7
45
76

score 0 · Answer 3 · answered Nov 16 '11 at 05:44

You could do it other way, that is, put your javascript inside php:

echo "<script>
var json = $json; 
document.write(json.length);
alert('half done');

for(var x=0; x < json.length; x++) {
    document.write(\"<li>\"+ x + \" \" + json[x]+ \"</li>\");
}
</script>";

Variable between php and Javascript

3 Answers3

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