Re: Compare and inserting with php

>> > > 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()); }