Help me please

[nasirr]

[code]<tr>
      <?
      echo "<td>";
         $link = mysql_connect( "localhost", 'root', 'root' );
if ( ! $link )
die( "Couldn't connect to MySQL" );
mysql_select_db( 'project', $link )
or die ( "Couldn't open user: ".mysql_error() );
$sql="SELECT * FROM labels";
$result = mysql_query($sql);
    echo "<select name=project>";
       echo "<option value='1'>-Select Project-</option>";
while($nt=mysql_fetch_row($result))
       {
//echo "<option value=".$nt[id]."></option>";
echo "<option value=".$nt["id"].">".$nt["wht"]."</option>";
}
echo "</select>";
?>
</td>[/code]

what i m trying to do is that get the values from the database for a drop down menu
and i got nothing there is only one option "select project" which is outside from the while statement of fetch intstruction please help me what is the problem with this code

[freakus_maximus]

[!--quoteo(post=363779:date=Apr 11 2006, 04:11 PM:name=Nasir)--][div class=\'quotetop\']QUOTE(Nasir @ Apr 11 2006, 04:11 PM) [snapback]363779[/snapback][/div][div class=\'quotemain\'][!--quotec--]
[code]<tr>
      <?
      echo "<td>";
         $link = mysql_connect( "localhost", 'root', 'root' );
if ( ! $link )
die( "Couldn't connect to MySQL" );
mysql_select_db( 'project', $link )
or die ( "Couldn't open user: ".mysql_error() );
$sql="SELECT * FROM labels";
$result = mysql_query($sql);
    echo "<select name=project>";
       echo "<option value='1'>-Select Project-</option>";
while($nt=mysql_fetch_row($result))
       {
//echo "<option value=".$nt[id]."></option>";
echo "<option value=".$nt["id"].">".$nt["wht"]."</option>";
}
echo "</select>";
?>
</td>[/code]

what i m trying to do is that get the values from the database for a drop down menu
and i got nothing there is only one option "select project" which is outside from the while statement of fetch intstruction please help me what is the problem with this code
[/quote]


Try this, should work as long as your query is actually returning what it should.

[code]while($nt=mysql_fetch_row($result))
       {
echo '<option value="'. $nt['id']. '">"' .$nt['wht'] ."</option>\n";
}
echo '</select>';[/code]

[nasirr]

by using this what it return in the option is just "
mean there are two enteries in the table
and its return " for both
help me please

[wildteen88]

change [b]mysql_fetch_row[/b] to [b]mysql_fetch_array[/b] instead. As mysql_fetch_row doesnt return the results with the column names as the indexes of the array, as currently mysql_fetch_row is returning your results like so:
$nt[0], $nt[1]

However mysql_fetch_array will retrun the results with the column headers as the array index, ie:
$nt['id'], $nt['wht'] etc.

So the following should now work:
[code]while($nt = mysql_fetch_array($result))
{
    echo '<option value="'. $nt['id']. '">"' .$nt['wht'] ."</option>\n";
}
echo '</select>';[/code]