Question

MySQL syntax query problems and PHP

Asked by: -ZugZug-

Hey thanks for taking a look, I am trying to to access data through a form and i am incountering a couple problems. This is the error i am getting when i pass it to the php script:

"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 'SELECT sa.AgencyID FROM AgencyAreas aa"

here is the code for the query:

$result = mysql_query("SELECT ag.AgencyID, ag.CompanyName, ag.Address, ag.StateOrProvince, ag.City, ag.PostalCode
FROM Agencies ag
WHERE ag.StateOrProvince ='$province'

AND ag.AgencyID IN ( SELECT sa.AgencyID
FROM AgencyAreas aa
INNER JOIN Areas ar ON (aa.AreaServedID=
ar.AreaServedID)
WHERE a.AreaServed LIKE '%$city')

AND ag.AgencyID IN ( SELECT as.AgencyID
FROM AgencyServices
WHERE ServiceID IN ($services))"
) or die("Invalid query: " . mysql_error() );
// $services is a check box array

My second problem is that $services is an array from check boxes and i have no clue how to include it in the query, WHERE ServiceID IN ($services))" just seems way to easy for the luck i have been having with this.
what i am trying to do is to return information based on the areas the agency serves based on a text field input from the user  and services provided based on a check box query. I hope this makes sense, if not sorry my brain is mush right now. Thanks again

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Asked On
2003-07-28 at 22:04:18ID20693112
Topic

MySQL Server

Participating Experts
1
Points
20
Comments
7

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Answers

 

by: pizzametsalamiPosted on 2003-07-29 at 06:38:52ID: 9027871

You might want to place a comma between the two tables from which you read; like:
SELECT ag.AgencyID, ag.CompanyName, ag.Address, ag.StateOrProvince, ag.City, ag.PostalCode FROM Agencies, ag WHERE ag.StateOrProvince ='$province' AND ag.AgencyID IN ( SELECT sa.AgencyID FROM AgencyAreas aa INNER JOIN Areas ar ON (aa.AreaServedID= ar.AreaServedID) WHERE a.AreaServed LIKE '%$city') AND ag.AgencyID IN ( SELECT as.AgencyID FROM agencyServices WHERE ServiceID IN ($services));

That would get you closer.

The $services array is a harder part.

$sql = "You might want to place a comma between the two tables from which you read; like:
SELECT ag.AgencyID, ag.CompanyName, ag.Address, ag.StateOrProvince, ag.City, ag.PostalCode FROM Agencies, ag WHERE ag.StateOrProvince ='$province' AND ag.AgencyID IN ( SELECT sa.AgencyID FROM AgencyAreas aa INNER JOIN Areas ar ON (aa.AreaServedID= ar.AreaServedID) WHERE a.AreaServed LIKE '%$city') AND ag.AgencyID IN ( SELECT as.AgencyID FROM agencyServices WHERE ";

for ($i=0; $i<count($services); $i++)
{
  if ($i>0)
    $sql.= " AND ";

  $sql.= "ServiceId IN (" & $services[$i] & ")";
}
$sql.=");"

mysql_query($sql, $link) or die("OOPS!")


I think this will help you out. Good Luck!

 

by: -ZugZug-Posted on 2003-07-29 at 08:36:07ID: 9028784

for the "mysql_query($sql, $link) or die("OOPS!")" line,  what is the $link for?

 

by: pizzametsalamiPosted on 2003-07-29 at 09:32:56ID: 9029263

$link is the connection to the mysql server. It is some kind of pointer. It points to the mysql connection.

use
$link = mysql_connect("server:port", "username", "password");
to connect to the server.

If you have only one connection it will take that connection by default.

But if you have more than one it will use the connection specified.

 

by: pizzametsalamiPosted on 2003-07-29 at 09:33:44ID: 9029273

mysql_query($sql, $link) or die("OOPS!");

Don't forget the ; in PHP, like I did :-(

20120131-EE-VQP-002

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