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Re: mySQL query in PHP - Parse error

Posted by noone on 02/04/06 03:37

JT wrote:
> I am working on a basic webpage in PHP connecting to a mySQL database,
> here is my query...
>
> //query 1
> $query_rs_staff = "SELECT * FROM jtStaff WHERE
> locationID=".$_GET['myDesk'];
> $rs_staff = mysql_query($query_rs_staff);
> $row_rs_staff = mysql_fetch_assoc($rs_staff);
>
>
> //query 2
> $query_rs_section = "SELECT * FROM jtSection WHERE
> sectionID=$row_rs_staff['sectionID']";
> $rs_section = mysql_query($query_rs_section);
> $row_rs_ssection = mysql_fetch_assoc($rs_section);
> ?>
>
>
> I had it working with query 1. then I added query 2 to get from another
>
> table, some additional data. So i have no problems with query 1. When
> ran thru my server, it throws back a parse error on line 20, which is
> the first line of query 2. can someone tell me the correct syntax for
> what i am trying to do here?
>
>
> thanks
>


why not do this in one query.. It saves coding and reduces
network/system traffic by only returning columns and rows that are
necessary instead of "give me everything and let the program sort it
out". That method is considered VERY bad coding practices, but taught
in way too many institutions of "higher lerarning".

(you will need to fix the php exact syntax for this...)

select a.field1,a.field2,,,,b.field1,b.field2,,,,
from jtstaff a, jtsection b
where a.locationID= $_GET['myDesk']
and b.sectionid=a.sectionid

this will return all rows where locationid = mydesk and the
jsstaff.sectionid is in jtsection.stationid

depending upon what you are actually trying to retreive, you may also
need a locationid column (field) in the jtsection table to make them
more unique - then just add:

and a.locationid=b.locationid

to the end of the sql statement above.


Michael Austin
DBA.

 

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