If you look at the query, there's no column name between 'WHERE' and 'LIKE'
which is a syntax error. That's the problem.
Looks like your sql generator has some issues, or you didn't specify the WHERE
column properly. I'm not familiar with the system you're using, but keep in
mind, that mysql_error will usually not steer you wrong.
-Micah
On Friday 10 September 2004 10:03 am, Stuart Felenstein wrote:
> I had not, my apologies. I think your post slipped
> by. Anyway, yes I have now inserted mysql_error() and
> got my return. Though I'm not entirely sure how to
> fix it.
> Here is the error.
>
> 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 'Like 'ACCFIN'' at line 1
> SELECT `VendorJobs`.`JobID`,
> `VendorSignUp`.`CompanyName`,
> `StaIndTypes`.`CareerCategories`,
> `StaUSCities`.`City`, `USStates`.`States`,
> `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`,
> `VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`,
> `VendorJobDetails`.`Details`,
> `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM
> `VendorJobs` INNER JOIN `VendorSignUp` ON
> (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`)
> INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` =
> `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN
> `StaUSCities` ON (`VendorJobs`.`LocationCity` =
> `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON
> (`VendorJobs`.`LocationState` = `USStates`.`StateID`)
> LEFT OUTER JOIN `staTaxTerm` ON
> (`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`)
> INNER JOIN `staTravelReq` ON (`VendorJobs`.`TravelReq`
> = `staTravelReq`.`TravelReqID`) INNER JOIN
> `VendorJobDetails` ON (`VendorJobs`.`JobID` =
> `VendorJobDetails`.`JobID`) where Like 'ACCFIN' limit
> 0,1
>
> --------------------------------------------------
>
> --- Micah Stevens <micah@xxxxxxxxxxxxxxxxxx> wrote:
> > did you make the change to the code I suggested?
> > What does MySQL say the error
> > is?
> >
> > -Micah
> >
> >
> > On Friday 10 September 2004 07:49 am, Stuart
> >
> > Felenstein wrote:
> > > As I said this is a code generator (dbqwiksite).
> >
> > So,
> >
> > > describing the process for creating the code is
> > > different. The $sql is fine, as far as typos or
> > > incorrect characterrs. I've gone through those
> > > statement very carefully.
> > > I've also tried to run a debug with no luck.
> > > But I do know that there is something wrong with
> >
> > the
> >
> > > statement, since I can get a complete display of
> >
> > all
> >
> > > records, but can't search using criteria.
> > >
> > > Below is the $sql, one I know works , though I've
> > > tried others. VendorJobs is the table I'm
> >
> > querying ,
> >
> > > and since it's made up of values from other static
> > > type tables I have all the joins in place.
> > >
> > > I've also taken the tick marks out, thinking maybe
> >
> > it
> >
> > > was a style issue. I'm not sure where to go with
> >
> > it.
> >
> > > Getting a response from the company is like pretty
> > > difficult.
> > >
> > > `VendorJobs`.`JobID`,
> > > `VendorSignUp`.`CompanyName`,
> > > `StaIndTypes`.`CareerCategories`,
> > > `StaUSCities`.`City`,
> > > `USStates`.`States`,
> > > `VendorJobs`.`AreaCode`,
> > > `staTaxTerm`.`TaxTerm`,
> > > `VendorJobs`.`PayRate`,
> > > `staTravelReq`.`TravelReq`,
> > > `VendorJobDetails`.`Details`,
> > > `VendorJobs`.`PostStart`,
> > > `VendorJobs`.`JobTitle`
> > > FROM
> > > `VendorJobs`
> > > INNER JOIN `VendorSignUp` ON
> > > (`VendorJobs`.`VendorID` =
> >
> > `VendorSignUp`.`VendorID`)
> >
> > > INNER JOIN `StaIndTypes` ON
> >
> > (`VendorJobs`.`Industry`
> >
> > > = `StaIndTypes`.`CareerIDs`)
> > > LEFT OUTER JOIN `StaUSCities` ON
> > > (`VendorJobs`.`LocationCity` =
> >
> > `StaUSCities`.`CityID`)
> >
> > > LEFT OUTER JOIN `USStates` ON
> > > (`VendorJobs`.`LocationState` =
> >
> > `USStates`.`StateID`)
> >
> > > LEFT OUTER JOIN `staTaxTerm` ON
> > > (`VendorJobs`.`TaxTerm` =
> >
> > `staTaxTerm`.`TaxTermID`)
> >
> > > INNER JOIN `staTravelReq` ON
> > > (`VendorJobs`.`TravelReq` =
> > > `staTravelReq`.`TravelReqID`)
> > > INNER JOIN `VendorJobDetails` ON
> > > (`VendorJobs`.`JobID` =
> >
> > `VendorJobDetails`.`JobID`)
> >
> > > Stuart
> > >
> > > --- Philip Thompson <prthomp@xxxxxxxx> wrote:
> > > > I think everyone knows that $sql is a statement.
> >
> > But
> >
> > > > what people are
> > > > asking is: what is that statement?! Because if
> >
> > there
> >
> > > > are "incorrect"
> > > > characters or 's in that statement, then that
> >
> > can
> >
> > > > break your
> > > > code/statement.
> > > >
> > > > ~Philip
> > > >
> > > > On Sep 9, 2004, at 5:49 PM, Stuart Felenstein
> >
> > wrote:
> > > > > Just getting back to this thing. There are 3
> > > >
> > > > files
> > > >
> > > > > involved, search.php, connections.php and
> > > > > functions.php.
> > > > >
> > > > > I searched through all and couldn't find the
> > > >
> > > > meaning
> > > >
> > > > > of the $sql_ext. $sql is just the sql
> >
> > statement
> >
> > > > I
> > > >
> > > > > inserted into the code.
> > > > >
> > > > > Great that I can't get a response from the
> > > >
> > > > company.
> > > >
> > > > > If anyone thinks it would be useful to post
> >
> > the
> >
> > > > code I
> > > >
> > > > > will. That is if anyone would want to see and
> >
> > look
> >
> > > > at
> > > >
> > > > > it.
> > > > >
> > > > > The weird thing is searching on one table is
> > > >
> > > > great.
> > > >
> > > > > The list (all records) works great. The
> >
> > problem
> >
> > > > > starts when I insert all my joins. Now I've
> > > >
> > > > tested
> > > >
> > > > > and retested the query and I know it's fine.
> > > > >
> > > > > Thank you ,
> > > > > Stuart
> > > > >
> > > > > --- Steve Davies <steve@xxxxxxxxxxxxxxxxxx>
> >
> > wrote:
> > > > >> What's contained in $sql and $sql_ext ???
> > > > >>
> > > > >> Stuart Felenstein wrote:
> > > > >>> I'm using a product called dbqwiksite pro.
> >
> > PHP
> >
> > > > >>> generator for PHP - MySQL
> > > > >>>
> > > > >>> The code seems to be working fine except in
> >
> > my
> >
> > > > >> search
> > > > >>
> > > > >>> page where I receive an "invalid query"
> > > > >>>
> > > > >>> $result = mysql_query($sql . " " . $sql_ext
> >
> > . "
> >
> > > > >> limit
> > > > >>
> > > > >>> 0,1")
> > > > >>> or die("Invalid query");
> > > > >>>
> > > > >>> This is the place I where the code is taking
> >
> > the
> >
> > > > >> die
> > > > >>
> > > > >>> path.
> > > > >>> I'm guessing something belongs between the
> > > > >>
> > > > >> quotation
> > > > >>
> > > > >>> marks , just not sure.
> > > > >>>
> > > > >>> Anyone ?
> > > > >>>
> > > > >>> Thank you
> > > > >>> Stuart
> > > > >>
> > > > >> --
> > > > >> PHP Database Mailing List
> >
> > (http://www.php.net/)
> >
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