SQL While and If statement error

[Chris_Evans]

Hi there,
I have this little :rolleyes: 😃 bit of code underneath this, it basically retrieves the menu... but it appears the if statement doesn't work as I added part to the url and the $ variable is assigned but it still gives the second option. Is there anything obvious that could be fixed in this as to why it is chosing the second option when it should be the first?

			$lmenu = mysql_query("SELECT name,link FROM menu WHERE menu='left' AND part='$part'") 
				or die(mysql_error()); 
			While ($row = mysql_fetch_array($lmenu))
				{
				If($row['part'] != "")
					{
						echo '<tr><td><a href="';
						echo $siteurl;
						echo $row['link'];
						echo '&part=';
						echo $part;
						echo '">';
						echo $row['name'];
						echo '</a></td></tr>';
					}
					else
					{
						echo '<tr><td><a href="';
						echo $siteurl;
						echo $row['link'];
						echo '">';
						echo $row['name'];
						echo '</a></td></tr>';
					}
				}
				mysql_free_result($lmenu);

Thanks,

Chris

[kevinsequeira]

change query to
SELECT name,link,part FROM menu WHERE menu='left' AND part='$part'"

or

SELECT * FROM menu WHERE menu='left' AND part='$part'"

reg
kevin

[The_Chancer]

Why don't you echo out $row[part] and see what is there, does it then make any more sense ? (Where is it defined though, as I can only see name and link in the select query....?)

[Chris_Evans]

Thanks both... echoed and changed it to:

$lmenu = mysql_query("SELECT name,link,part FROM menu WHERE menu='left' AND part='$part'")

...it worked then. 🙂 Thanks.

[laserlight]

The problem is that you arent selecting the 'part' column as well.

$lmenu = mysql_query("SELECT name, link, part FROM menu WHERE menu='left' AND part='$part'")
	OR die(mysql_error());
while ($row = mysql_fetch_array($lmenu))
{
	if ($row['part'] != '')
	{
		echo '<tr><td><a href="';
		echo $siteurl;
		echo $row['link'];
		echo '&part=';
		echo $part;
		echo '">';
		echo $row['name'];
		echo '</a></td></tr>';
	}
	else
	{
		echo '<tr><td><a href="';
		echo $siteurl;
		echo $row['link'];
		echo '">';
		echo $row['name'];
		echo '</a></td></tr>';
	}
}
mysql_free_result($lmenu);

[Chris_Evans]

It's always the simple stuff I miss in php. 🙂 Things like:

;
and

" "
and

' '
and you get the idea. 🙂