Displaying 0 rows

crawfd

After my while statement I want to suggest that if their are no rows that match a query (which I haven't pasted but is above the script below), then display in printed form that their are no rows that match my sort criteria. How do I accomplish this?
I'm assuming I have to use mysql_num_rows somehow, thanks.


function listmenu() {

return "<form name=\"form2\" method=\"post\" action=\"home.php?page=manage\"><select name=\"sort\" style=\"font-size: 11px; font-weight: bold; color: 000000;\" onchange=\"document.forms['form2'].submit()\"> 
<option selected>Sort by Category</option>
<option value=\"1\">Sort by Date</option> 
<option value=\"2\">Sort by Awareness(Low)</option> 
<option value=\"3\">Sort by Awareness(Medium)</option> 
</select></form>";
}

$p .= "<table align=\"center\" width=\"95%\" border=\"0\" cellspacing=\"1\" cellpadding=\"2\" class=\"table_border\">\n";
$p .= "<tr><td colspan=\"6\">" . listmenu() . "</td></tr>";
$p .= "<tr class=\"header_cell\">";

$p .= "<td>Date</td><td>Title</td><td>Past or Recent</td><td>Awareness</td><td>Modify</td><td>Delete</td></tr>\n";

echo $p;
while(list($id,$date,$username,$title,$dreamstatus,$entry,$awareness,$pagination) = mysql_fetch_row($result)){
		$d = explode("-",$date);
		$titletrimmed = substr($title, 0, 20);
		$thedate = "$d[2]/$d[1]/$d[0]";
		$print  = "\n<tr class=\"cells\">\n";
		$print .= "<td><li>$date</td><td><b>$titletrimmed...</b></td>\n";
		$print .= "<td>$dreamstatus</td>";
		$print .= "<td>$awareness</td>";
		$print .= "<td align=\"center\"><a href=\"home.php?page=manage&action=modify&entry=$id\">";
		$print .= "Modify</a></td>\n";
		$print .= "<td align=\"center\"><a href=\"home.php?page=manage&action=remove&entry=$id\">";
		$print .= "Delete</a></td>\n";
		$print .= "</tr>\n";
		echo $print;
	}
	echo"</table><br></center>";
}

superwormy

if (mysql_num_rows ($result) == 0)
{print ("no rows found");

www.php.net/mysql_num_rows

crawfd

Well I tried adding this below the while statement with an else statement but it still doesn't display the print message "no rows found'.


while(list($id,$date,$username,$title,$dreamstatus,$entry,$awareness,$pagination) = mysql_fetch_row($result)){

if (mysql_num_rows ($result) == 0) {
print ("no rows found"); 

} else {

$d = explode("-",$date);
$titletrimmed = substr($title, 0, 20);
$thedate = "$d[2]/$d[1]/$d[0]";
$print  = "\n<tr class=\"cells\">\n";
$print .= "<td><li>$date</td><td><b>$titletrimmed...</b></td>\n";
$print .= "<td>$dreamstatus</td>";
$print .= "<td>$awareness</td>";
$print .= "<td align=\"center\"><a href=\"home.php?page=manage&action=modify&entry=$id\">";
$print .= "Modify</a></td>\n";
$print .= "<td align=\"center\"><a href=\"home.php?page=manage&action=remove&entry=$id\">";
$print .= "Delete</a></td>\n";
$print .= "</tr>\n";
echo $print;
		}
	}
echo"</table><br></center>";

tekky

try more like....

if (mysql_num_rows ($result) == 0) {
print ("no rows found"); 

} else {


while(list($id,$date,$username,$title,$dreamstatus,$entry,$awareness,$pagination) = mysql_fetch_row($result)){

$d = explode("-",$date);
$titletrimmed = substr($title, 0, 20);
$thedate = "$d[2]/$d[1]/$d[0]";
$print  = "\n<tr class=\"cells\">\n";
$print .= "<td><li>$date</td><td><b>$titletrimmed...</b></td>\n";
$print .= "<td>$dreamstatus</td>";
$print .= "<td>$awareness</td>";
$print .= "<td align=\"center\"><a href=\"home.php?page=manage&action=modify&entry=$id\">";
$print .= "Modify</a></td>\n";
$print .= "<td align=\"center\"><a href=\"home.php?page=manage&action=remove&entry=$id\">";
$print .= "Delete</a></td>\n";
$print .= "</tr>\n";
echo $print;
		}
	}
echo"</table><br></center>";

quaTTro

I have a similar problem, at least with mysql_num_rows.
While doing this

$Query = "SELECT * FROM Table WHERE Estado=Venta ORDER BY id DESC";
$result	= mysql_query($Query, $Link);
@$num_rows	= mysql_num_rows($result);

I have that $num_rows is 0. But I know that $query finds three rows so, why is $num_rows saying "0"?

If I take out the @, it says it´s "not a valid resource".
Thanks in advance!

tuf-web_co_uk

try changing

$result = mysql_query($Query, $Link);

to $result = mysql_query($Query, $Link) or die(mysql_error());

and if it returns an error you know something is wrong with your query.
if you get an error post it here for more assstance

mrhappiness

WHERE Estado = 'Venta'

you forgot the ' i think

quaTTro

Originally posted by tuf-web.co.uk
try changing

$result = mysql_query($Query, $Link);

to $result = mysql_query($Query, $Link) or die(mysql_error());

and if it returns an error you know something is wrong with your query.
if you get an error post it here for more assstance

Great! It was "No Database Selected", so now It´s solved. Thanks, tuf-web.co.uk and mrhappiness too! 🙂