Preset data base data in drop down menu ?

[developer]

Hello! 🙂

I do have this:

<INPUT TYPE="text" size="20" NAME="category" value="<? echo $row['category'] ?>">

How can I please change it to a drop down menu, so the data base value is preset in the menu:

<select name="category">
<option value="KC" selected>Kansas City
<option value="LA" selected>Los angelses
<option value="NY" selected>New York
</select>

[petitduc]

For example :

<select name="category">

while($row=mysql_fetch_array($result)) {
echo "<option value=\"$row['whatever']\">$row['category']</option>";
}

</select>

[developer]

Hello! 🙂

Actually I get this error message using that code:

Parse error: parse error, expecting T_STRING' orT_VARIABLE' or `T_NUM_STRING'

😕

[petitduc]

Ok, there is a small stupid syntax error somewhere.
If you can't figure it out, post the relevant part of your code.

[Sxooter]

If you want something preselected you need this:

<form ...>
<select name="test">
<option value="1">Number 1</option>
<option value="2" SELECTED>Number 2</option>
<option value="3">Number 3</option>
</select>
</form>

See that SELECTED in there, that's what you need. So, you do something like this:

print '<select name="test">';
$val=2;
for ($i=0;$i<4;$i++){
  print '<option value="'.$i.'"';
  if ($val==$i) print ' SELECTED';
  print '>Number '.$i.'</option>';
}
print '</select>';

[petitduc]

I always forget that array variables don't nest in strings. This is the error.

The 2nd line should be :

echo "<option value=\"".$row['whatever']."\">".$row['category']."</option>"; 

No matter how many times I make this mistake myself, I will always make it again.
Sorry...

[Sxooter]

Yeah, that's why a little while back I just started using ' instead of " since no vars inside ' marks are parsed, I'd be force to just do it with '.$var.' from then on. Once you get used to just doing them all that way it's pretty easy.

[developer]

Hello! 🙂

Thanks a lot for all the help in this thread so far.

I have for hours been trying to get it right, this is the nearest I have came, but there is no output att all, just a small drop down that is emty.

<?
echo '<select name="category">'; 

while($row=mysql_fetch_array($Result)) { 
	echo  '<option value="'.$row['category'].'"';
	if ($category==$row['category']) print ' SELECTED'; 

echo  ">";
echo  $row['category'];
echo "</option>"; 
} 

echo "</select>"
?>

[Sxooter]

What does view source of the delivered document show? Any options at all? Empty ones, malformed ones?