Re: Search Query on two tables not working
| From: | Miller, Terion | Date: | Tue, 21 Jul 2009 16:41:22 +0000 |
| Subject: | Re: Search Query on two tables not working | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-295629@lists.php.net to get a copy of this message | ||
Turned off the redirects on the whole script and tried to the the query to echo and these are the
errors I got:
Notice: Undefined offset: 1 in
/var/www/vhosts/getpublished.news-leader.com/httpdocs/ResturantInspections/processRestaurantSearch.php
on line 89
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in
/var/www/vhosts/getpublished.news-leader.com/httpdocs/ResturantInspections/processRestaurantSearch.php
on line 119
On 7/21/09 11:32 AM, "Kevin Smith" <kevin@netsmith.ltd.uk> wrote:
Can you supply the actual generated SQL, I can potentially see a
problem, but need to see the final SQL statement.
Miller, Terion wrote:
> Why isn't this working for searching?
>
> // Run query on submitted values. Store results in $SESSION and redirect to restaurants.php
> $sql = "SELECT name, address, inDate, inType, notes, critical, cviolations, noncritical
> FROM restaurants, inspections WHERE restaurants.name<> '' AND restaurant.ID =
> inspection.ID"; if ($searchName) { $sql .= "AND restaurants.name LIKE
> '%". mysql_real_escape_string($name) ."%' "; if(count($name2) == 1)
> { $sql .= "AND restaurants.name LIKE '%".
> mysql_real_escape_string($name2[1]) ."%' "; } else { foreach($name2
> as $namePart) { $sql .= "AND restaurants.name LIKE '%".
> mysql_real_escape_string($namePart) ."%' "; } } } if ($searchAddress)
> { $sql .= "AND restaurants.address LIKE '%".
> mysql_real_escape_string($address) ."%' "; } $sql .= "ORDER BY
> restaurants.name;"; $result = mysql_query($sql);
>
>