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
> > >>
> > >> --
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