Using PHP to run a query and place it in a javascript object

Question

I am trying to run a stored procedure using PHP. I want to put the results of the query into a javascript object. I cannot find how to accomplish this. I have run this code but I get one of the results I am returning is undefined. Here is my code:

$result = mysql_query("call sp_getGenre()");
    if($result === FALSE){
            die(mysql_error());
        }
        while($row = mysql_fetch_array($result)){
    ?>
            <script type="text/javascript">
                console.log(<? $row['Type'] ?>);
                var genreObj = new Object();
                genreObj.name = "<? echo $row['Type'] ?>";
                genreObj.level = <? echo $row['Level'] ?>;
    <?
                $parentID = $row['PrevID'];
                if($parent == null){ $parent = "0"; }
    ?>
                genreObj.parent = <? $parent ?>;
                arrGenre.push(genreObj);
            </script>
    <?
        }
    ?>

I am fairly new to the PHP world but would greatly appreciate a point in the right direction. Thanks in advance.

Can you paste here the output, why you don't use echo or = to display variable content in genreObj.parent = <? $parent ?>; You can easily create the object using Ajax — Hamza
– Hamza, Commented Feb 9, 2013 at 1:31
Thank you for teh fast replies. I'm not fully understanding. Can you give me an example of what genreObj.name = "<? echo $row['Type'] ?>"; should look like please. — seroth
– seroth, Commented Feb 9, 2013 at 1:34

Ayman Safadi · Accepted Answer · 2013-02-09 01:45:12Z

3

Build an associative array in PHP resembling the JavaScript object you want, then use json_encode() to convert it into a JSON object, you can then easily pass on to JavaScript.

You may even be able to do something like:

<?php
// stuff
$resultObj = json_encode($row);
?>

<script type="text/javascript">
var genreObj = <?= $resultObj ?>;
</script>

This is not supposed to be a perfect solution to your problem, but hopefully it will give you some ideas.

answered Feb 9, 2013 at 1:45

Ayman Safadi

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3 Comments

seroth Over a year ago

Is it possible to build a PHP object and pass that object to a javascript object?

Barmar Over a year ago

Not by default, your object class has to provide encoding methods. See stackoverflow.com/questions/4697656/…

seroth Over a year ago

Is using that method the only way?

Leeish · Accepted Answer · 2013-02-09 01:45:14Z

1

I think

<? $row['Type'] ?>

Should be

<?= $row['Type'] ?> or
<? echo $row['Type'] ?>

Same goes for

<? $parent ?>;

answered Feb 9, 2013 at 1:45

Leeish

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4 Comments

seroth Over a year ago

I added the echo and still get no results. Firebug says one of the first returned value "bob" is undefined. genreObj.name = "<? echo $row['Type'] ?>";

Leeish Over a year ago

You sure all of your PHP variables have values?

seroth Over a year ago

yes, I ran the stored procedure in HeidiSQL and saw all the results.

seroth Over a year ago

This did the trick I had to get rid of echo and just use ?= Thanks for the help.

swbullis · Accepted Answer · 2013-02-09 02:03:17Z

Code below is not fully tested but the basic Idea is to Write all values into an array, then convert that array to a json object. From there assign the json object to the jsonresults variable. Now you can pull up all the info you need via javascript.

<?php


$result = mysql_query("call sp_getGenre()");
if($result === FALSE){
    die(mysql_error());
}
while($row = mysql_fetch_array($result)){
    $resultsarray = array("type"=> $row['type'], "level" => $row['Level'}, "previd" => $row['PrevID']);
    $jsonifiedresults = json_encode($resultsarray);
    ?>
            <script type="text/javascript">
                jsonresult = <? echo $jsonifiedresults; ?>;

                console.log(jsonresult.type);
                var genreObj = new Object();
                genreObj.name = jsonresult.type;
                genreObj.level = jsonresult.level;
                genreObj.parent = jsonreult.prevID;
                arrGenre.push(genreObj);
            </script>
<?php
        }
?>

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