On Tue, Jun 30, 2009 at 2:19 PM, Miller, Terion<tmil...@springfi.gannett.com> wrote: > > > > >>> >> >> Does this list of inspections exist in the db? Could you not use an >> INSERT INTO SELECT * FROM TABLE WHERE type statement...much less >> processing overhead then >> >> -- >> >> Bastien >> >> Cat, the other other white meat >> >> Hi Bastien, >> >> Something like this is what you mean? >> >> if (!empty($Go)) { $query = "SELECT * FROM restaurants WHERE name = >> '$ucName' AND address = '$ucAddress' AND inDate ='$inDate' AND inType = >> '$inType' "; $result = mysql_query ($query); $row = mysql_fetch_object >> ($result); If (mysql_num_rows($result) == 0) { $sql = "INSERT INTO >> `restaurants` (name, address, inDate, inType, notes, critical, cviolations, >> noncritical) VALUES ("; $sql .= " '$ucName', '$ucAddress', '$inDate', >> '$inType', '$notes', '$critical', '$cleanViolations', '$noncritical')"; >> $result = mysql_query($sql) or die(mysql_error()); } So if all >> three things are met, that entry is in there if not insert right? >> > > More like > > sql = "insert into restaurants select * from restaurants where name = > '$ucName' AND address = '$ucAddress' AND inDate ='$inDate' AND inType > = '$inType' "; > > http://dev.mysql.com/doc/refman/5.0/en/insert-select.html > > > > -- > > Bastien > > Cat, the other other white meat > > > Well I didn't get anywhere, now I just keep getting this error.... > > > Warning: mysql_num_rows(): supplied argument is not a valid MySQL result > resource in > /var/www/vhosts/getpublished.news-leader.com/httpdocs/ResturantInspections/compare.php > on line 119 > You have an error in your SQL syntax; check the manual that corresponds to > your MySQL server version for the right syntax to use near 's Roast Beef > Restaurant #9459', ' 1833 W Republic Rd ', '3/2/09', '' at line 1 > > Using this code: > > $query = "SELECT * FROM restaurants WHERE name = '$ucName' AND address = > '$ucAddress' AND inDate ='$inDate' AND inType = '$inType' "; $result = > mysql_query ($query); // $row = mysql_fetch_object ($result); echo > $result; If (mysql_num_rows($result) == 0) { $sql = "INSERT INTO > `restaurants` (name, address, inDate, inType, notes, critical, cviolations, > noncritical) VALUES ("; $sql .= " '$ucName', '$ucAddress', '$inDate', > '$inType', '$notes', '$critical', '$cleanViolations', '$noncritical')"; > $result = mysql_query($sql) or die(mysql_error()); } > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > >
For one thing, that error message shows that you are not properly escaping the strings you are sending to mysql_query() (specifically, $ucName in this case). Even if you do get it to work, you'll be vulnerable to SQL injection. Andrew -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php