[Resolved] Establishing Variable For Use Later in Code

ejwf

Hello
Can anyone tell me whether this is correct coding procedure and whether it will work...thanks

I want to perform a query that takes several variables from different places in the database.

I understand that I have to define the variables first and have been using the following code to do so...

$query = "SELECT SectionRef FROM tblUser WHERE CLogIn = '$username'";
$result = mysql_query($query)
or die ("no info");
$row = mysql_fetch_array($result);
$SectionRef = $row["SectionRef"];

Here I have the $username information from a login form on the first page and I can use that information to get the SectionRef variable from the User table....

However can I then run a second query in the same format using the $SectionRef to get another variable e.g.

$query = "SELECT MachineId FROM tblMachine WHERE SectionRef = $SectionRef";
$result1 = mysql_query($query)
or die ("no Machine info");
$row1 = mysql_fetch_array($result1);
$MachineRef = $row1["MachineId"];

And then can I use the Machine Ref as part of a bigger query, that uses $username, $MachineRef and results from a form on the previous page to INSERT info into the database?

Is it correct coding to run subsequent queries in this way to gather information that will be recognised later in the script?

ALSO (!) can anyone see why I keep getting a parse error on this script, it shows an error on the first line of the second query :-

$query = "SELECT MachineId FROM tblMachine WHERE SectionRef = $SectionRef";

thanks to you all in advance..........

laserlight

Is it correct coding to run subsequent queries in this way to gather information that will be recognised later in the script?

Yes.

$query = "SELECT MachineId FROM tblMachine WHERE SectionRef = $SectionRef";

Looks more like a potential SQL problem.
You should enclose names in backticks, values in single quotes, e.g.

$query = "SELECT `MachineId` FROM `tblMachine` WHERE `SectionRef` = '$SectionRef'";

Aside from that I dont see any obvious errors that would cause PHP to be unable to parse the code.

ejwf

thank you, I have added the backticks and the parse error has gone......

However, the first query isn't working and I am getting the error message 'no info' that I put in the or die code.

I am thinking that the reason it isn't working is because it can't find the information for $username.

$username is part of the login form on page 1, it is used on page 2 as part of another form, and now I am running it in another code here on page 3.

I thought it would bring $username as I have got it set up as a variable in the session code
code on page 1 =

		 $_SESSION['username'] = $row[1];
		 $_SESSION['user_id'] = $row[0];

and then on every subsequent page I have put

session_start ();

at the very top of the page

I thought that this meant that $username would be taken across to every page as a valid / readable variable - is this right?

Sorry to keep asking questions but just as I think I am getting somewhere with php, I keep finding new problems!!!

thanks

laserlight

instead of printing "no info" print mysql_error(), i.e.
die(mysql_error());

ejwf

I put the die(mysql_error()); error message in the code and it gave me the information that no database was selected

I have added a line into the code to call up the database connection and it now seems to work OK.

In future I will use this error message - it seems to give you all the answers !!