Its a version problem, UNION was not implempted till V4, and my host has V3.23 Anybody know of a work around with my code? I have found some work arounds on google but its too advanced for me...i dont really understand left joins, right joins etc
Cheers, -Ryan We will slaughter you all! - The Iraqi (Dis)information ministers site http://MrSahaf.com ----- Original Message ----- From: "esctoday.com | Wouter van Vliet" <[EMAIL PROTECTED]> To: "'Ryan A'" <[EMAIL PROTECTED]> Cc: <[EMAIL PROTECTED]> Sent: Monday, September 15, 2003 1:09 AM Subject: RE: [PHP] Program works only when there are records :-( > This looks like a pretty decent query to me... To figure out more about the > errors mysql is producing, you can try to put a > > print mysql_error(); > > After ever call to a mysql function .. From there you should be able to > learn more about the errors, and if not: don't bother posting what > mysql_error() produces. Ow, and one question: does it give the errors if > there's no rows in any table, or also if there's no rows in one of the > tables? And in the second option, does it give you the errors less often? > > Wouter > > -----Original Message----- > From: Ryan A [mailto:[EMAIL PROTECTED] > Sent: Monday, September 15, 2003 12:49 AM > To: esctoday.com | Wouter van Vliet > Cc: [EMAIL PROTECTED] > Subject: Re: [PHP] Program works only when there are records :-( > > Hey, > Thanks for replying. > > sorry, i thought i explained what $tt was, but here it is in program code if > you want to be clear on it. > > $tt = "SELECT COUNT(*), 'Count1' FROM shared WHERE user ='".$mmmy_user."' > and ccno=".$mmmy_ccno." UNION SELECT COUNT(*), 'Count2' FROM dedicated > WHERE user='".$mmmy_user."' and ccno=".$mmmy_ccno." UNION SELECT COUNT(*), > 'Count3' FROM reseller WHERE user ='".$mmmy_user."' and ccno=".$mmmy_ccno." > UNION SELECT COUNT(*), 'Count4' FROM colocated WHERE user ='".$mmmy_user."' > and ccno=".$mmmy_ccno." UNION SELECT COUNT(*), 'Count5' FROM freehosting > WHERE user ='".$mmmy_user."' and ccno=".$mmmy_ccno; > > I'll try those things you outlined till then. > > Cheers, > -Ryan > > > > Start by giving us the value of $tt, ... That'll probably tell more about > > the problem. Or you can do something like this: > > > > if ($res = mysql_query($tt)) { > > $num_rows = mysql_num_rows($res); //This is line 25 > > $x=1; > > > > for($i=0; $i<5; $i++) { > > $a = mysql_result($res, $i); //This is line 30 > > $bb[$x]=$a; > > $x++; > > } > > } else { > > print "sorry, no results found"; > > }; > > > > Also, for: > > ===== > > for($i=0; $i<5; $i++) { > > $a = mysql_result($res, $i); //This is line 30 > > ===== > > > > I'd suggest: while($a = mysql_fetch_array($res)) { > > > > Wouter > > > > > > -----Original Message----- > > From: Ryan A [mailto:[EMAIL PROTECTED] > > Sent: Sunday, September 14, 2003 11:45 PM > > To: [EMAIL PROTECTED] > > Subject: [PHP] Program works only when there are records :-( > > > > Hi, > > I have a select statment with a UNION so that i am getting 5 COUNT(*)'s > from > > a table. > > > > After that I have this code: > > ***************************** > > $res = mysql_query($tt); > > $num_rows = mysql_num_rows($res); //This is line 25 > > $x=1; > > > > for($i=0; $i<5; $i++) > > { > > $a = mysql_result($res, $i); //This is line 30 > > $bb[$x]=$a; > > $x++; > > } > > > > $one = $bb[1]; > > $two = $bb[2]; > > $thr = $bb[3]; > > $fou = $bb[4]; > > $fiv = $bb[5]; > > > > $tot=$one + $two + $thr + $fou + $fiv; > > ***************************** > > > > When i have some records in the database this works fine and i get all the > > values in $one,$two etc but when a new account is created and i the > > COUNT(*) gets back just 0 I get these ugly errors/warnings on my page: > > > > Warning: mysql_num_rows(): supplied argument is not a valid MySQL result > > resource in /usr163/home/b/e/bestweb/public_html/co.details.php on line 25 > > > > Warning: mysql_result(): supplied argument is not a valid MySQL result > > resource in /usr163/home/b/e/bestweb/public_html/co.details.php on line 30 > > > > Warning: mysql_result(): supplied argument is not a valid MySQL result > > resource in /usr163/home/b/e/bestweb/public_html/co.details.php on line 30 > > > > Warning: mysql_result(): supplied argument is not a valid MySQL result > > resource in /usr163/home/b/e/bestweb/public_html/co.details.php on line 30 > > > > Warning: mysql_result(): supplied argument is not a valid MySQL result > > resource in /usr163/home/b/e/bestweb/public_html/co.details.php on line 30 > > > > Warning: mysql_result(): supplied argument is not a valid MySQL result > > resource in /usr163/home/b/e/bestweb/public_html/co.details.php on line 30 > > > > > > > > What should i do and wheres the problem? I searched on google but just > found > > some sites outputting this, no real help. > > Totally confused, please help. > > > > Thanks, > > -Ryan > > > > -- > > PHP General Mailing List (http://www.php.net/) > > To unsubscribe, visit: http://www.php.net/unsub.php > > > > -- > > PHP General Mailing List (http://www.php.net/) > > To unsubscribe, visit: http://www.php.net/unsub.php > > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
