PHP variable falls outside the scope of Javascript, how do I get it in the scope?

Question

I'm trying to make a dropdown in HTML, but I want to have it's default value to be determined by the value of a PHP variable. I wanted to do it with JavaScript, but I end up with the famous "Uncaught ReferenceError: $ is not defined" error :/.

PHP:

while(list($ArtikelID, $Type) = mysql_fetch_array($result))
{
  $arid = $ArtikelID;
  $te = $Type;
}

HTML + JS:

<form name="send" method="post" action="editartticlesdef.php">
    articlenumber: </br>
    <input readonly="readonly" name="ArticleID" value=<?php echo $arid ?>></br>
    Type: </br>
    <select name="Type">
        <option value="Article">Article</option>
        <option value="Code">Code</option>
        <option value="News">News</option>
        <option value="Project">Project</option>
    </select></br>
    <script>
        document.getElementById("Type").selectedIndex = <?php echo '$te'; ?>;
    </script>
</form>

I tried to place the script on other places on the page but had no success :/. I suspect the variable falls outside the scope, but I don't know how to get it back in :/.

Is PHP or JavaScript throwing the error? What is the resulting JavaScript created by this code? — David
– David, Commented Oct 14, 2015 at 12:32

Sougata Bose · Accepted Answer · 2015-10-14 12:40:52Z

4

It should be -

<select name="Type" id="Type">

And

document.getElementById("Type").selectedIndex = '<?php echo $te; ?>';

answered Oct 14, 2015 at 12:40

Sougata Bose

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1 Comment

BadAtPHP Over a year ago

Well that worked, but it turns out I'm also an idiot. While this solves the error I got it also shows me that I'm trying to select the value of the dropdown by id and not the value of the variable $te. facepalm

Steve · Accepted Answer · 2015-10-14 12:42:18Z

1

Provided $te is an integer between 0 and 3, then you need to:

Give your select an id attribute:

<select name="Type" id="Type"> <option value="Article">Article</option> <option value="Code">Code</option> <option value="News">News</option> <option value="Project">Project</option> </select>
remove the quotes areound the php variable itsself and place them in the html:

<script> document.getElementById("Type").selectedIndex = "<?php echo $te; ?>"; </script>

answered Oct 14, 2015 at 12:42

Steve

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Comments

thomas_inckx · Accepted Answer · 2015-10-14 12:53:55Z

0

Sorry for not answering the question, but you should not use mysql_, it's deprecated and unsafe (See PHP Manual). Use MySQLi or PDO.

answered Oct 14, 2015 at 12:53

thomas_inckx

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