Creating a Drop Down Menu with images using PHP and HTML

Question

I am attempting to create a drop-down menu that links to user profile pages and includes user images next to it. I am a bit new to PHP so bear with me if this is a simple question. Here is the code I've created:

 function dropdown() {



    $query = "SELECT * FROM usersbadges";
    $result = mysql_query($query);


    while($row = mysql_fetch_array($result)){
        $name=$row['name'];
        $image=$row['image'];

    }
    mysql_close(); 
}   

$options.=<option value=\$name\>.$image </option>;
<select name=name>
<option name=0>
print $options
</select>

It tells me that the < brackets are unexpected but I'm at a loss as to how to put them. If those were fixed, would this code create me a drop down menu with user names accompanied by their images? I'd just like to know if I'm on the right track/what the next step is. All help is greatly appreciated

You need to either enter and exit PHP mode or construct the HTML as a string and echo it all at once. Currently, your code makes use of totally invalid syntax. — Nightfirecat
– Nightfirecat, Commented Oct 4, 2013 at 0:39

Osama Jetawe · Accepted Answer · 2013-10-04 00:33:40Z

0

use this

  $query = "SELECT * FROM usersbadges";
    $result = mysql_query($query);


    while($row = mysql_fetch_array($result)){
        $name=$row['name'];
        $image=$row['image'];

    }
    mysql_close(); 
}   

$options.="<option value='$name'>$image </option> <select name=name> <option name=0>";
echo $options;
echo '</select>';

answered Oct 4, 2013 at 0:33

Osama Jetawe

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Osama Jetawe Over a year ago

tell me if this help you or not

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