>>
>
> 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()); }
--
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