Populating php drop down with additional values

Question

I have a drop down list that is generated from values in a table, however, there are two additional values I'd like to add that are not in this table.

$sql = mysql_query("select category_id,category_name from categories"); 
    $selection="";
    while ($row=mysql_fetch_array($sql)){ 
        $id=$row["category_id"]; 
        $category_name = $row["category_name"]; 
        $selected = ($id == $idlookup) ? 'selected="selected"' : '';
        $selection.="<OPTION " . $selected . " VALUE=\"$id\">".$category_name."</OPTION>";
    }

<tr><td><input type = "text" name ="search_value"></input></td><td><select name ="category"><Option >Choose Category<? echo $selection; ?></Select></td>

This works fine, but I'm not sure how to add two more values to the beginning of the list. How could I add the values "User" and "Group" to the beginning of this list? I do not want them in the categories table because this will only be used on one page.

1) Stop using mysql_*, switch to PDO. 2) just echo those 2 options before while. Or add them to DB — E_p
– E_p, Commented Nov 23, 2012 at 21:18
just open this page and read red box php.net/manual/en/function.mysql-query.php — E_p
– E_p, Commented Nov 23, 2012 at 21:29

Wes Cossick · Accepted Answer · 2012-11-23 21:19:28Z

1

Instead of instantiating the variable with no content ($selection="";), instantiate it with this:

$selection="<OPTION VALUE=\"id\">User</OPTION><OPTION VALUE=\"id\">Group</OPTION>";

answered Nov 23, 2012 at 21:19

Wes Cossick

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2 Comments

Wes Cossick Over a year ago

Naturally, you'll want to replace id with the appropriate text/number

user1406951 Over a year ago

Still have a few minutes before it'll let me.

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