""Dan Joseph"" <dmjoseph@xxxxxxxxx> wrote in message
news:a20394490809171140u436b0d40yad003e31592a70f9@xxxxxxxxxxxxxxxxx
On Wed, Sep 17, 2008 at 2:30 PM, Vinny Gullotta
<vgullotta@xxxxxxxxxxxxxx>wrote:
""Dan Joseph"" <dmjoseph@xxxxxxxxx> wrote in message
news:a20394490809171120o63691c31idc356aeae78681f3@xxxxxxxxxxxxxxxxx
On Wed, Sep 17, 2008 at 2:17 PM, Vinny Gullotta
<vgullotta@xxxxxxxxxxxxxx
>wrote:
What I want to do is find the top 10 servers where the column steps =
iisreset. The following code works great except that the page is not
displaying the servername in the 'Server Name' column of my results
(nothing
appears, the column is just blank).
servername and steps are the important columns in the database table.
$_POST[time1] and $_POST[time2] come from a form submitted.
When I copy and paste the entire select statement into the SQL tab in
phpmyadmin (and replace the time variables with actual times
corresponding
to the timestamp column), it displays the correct results including
servername. Everything works in the php page's results except for the
servername. I feel like it's right in front of my face and that's why I
can't see it lol. Any help would be greatly appreciated. Thanks in
advance
=)
My code...
$query = "SELECT servername, COUNT(steps) FROM monitoring WHERE steps
LIKE
'iisreset' AND timestamp <= '$_POST[time2]' AND timestamp >=
'$_POST[time1]'
GROUP BY servername ORDER BY COUNT(*) DESC LIMIT 10";
$result = mysql_query($query) or die(mysql_error());
# display column titles
echo "<center><table class='table'><tr>";
echo "<td
class='tableHeader'><center><small><b>Count</b></small></td>";
echo "<td class='tableHeader'><center><small><b>Server
Name</b></small></td>";
echo "</tr>";
#display results
while($i = mysql_fetch_row($result))
{
echo "<tr><td><small><center>", $i[COUNT('steps')],
"</center></small></td>";
echo "<td><small><center>", $i[servername]
,"</center></small></td></tr>";
}
echo "</table></center><br>";
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Change that "COUNT(steps)" to "COUNT(steps) AS CountSteps", that might
be
the issue. Then you're using $i['CountSteps']. That seems a bit more
normal looking to me atleast.
Also, try echoing out your query on the screen to see that its formating
properly in the PHP code. You may have something wrong in there,
although
I
don't see any off hand.
--
-Dan Joseph
www.canishosting.com - Plans start @ $1.99/month.
"Build a man a fire, and he will be warm for the rest of the day.
Light a man on fire, and will be warm for the rest of his life."
Adding as CountSteps $i['CountSteps'] still leaves the column blank.
echo $result; gives me an output of:
Resource id #3
and
echo $query;
just gives me an error.
One thing I don't understand is why echo $result; gives me Resource id #3
as an output. What does that mean?
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That's basically your result set ID number inside PHP.
as for $query, what error are you getting? Does this $query echo out:
$query = "SELECT servername, COUNT(steps) AS CountSteps FROM monitoring
WHERE steps LIKE 'iisreset' AND timestamp <= '" . $_POST['time2'] . "' AND
timestamp >= '" . $_POST['time1'] . "' GROUP BY servername ORDER BY
COUNT(*)
DESC LIMIT 10";
--
-Dan Joseph
www.canishosting.com - Plans start @ $1.99/month.
"Build a man a fire, and he will be warm for the rest of the day.
Light a man on fire, and will be warm for the rest of his life."
It's actually not an error, it's the select statement that is echo'd
echo $query;
yields
SELECT servername, COUNT(steps) as CountSteps FROM monitoring WHERE steps =
'IISRESET' AND timestamp <= '2008-09-17 11:40:34' AND timestamp >=
'2008-08-17' GROUP BY servername ORDER BY COUNT(*) DESC LIMIT 10
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