[Resolved] can't get script to switch active DBs

jacksonpt

I've got some code that is supposed to update one of two databases depending on the value of a specified variable. I put in a couple of echo statements to make sure the proper code is being run depending on the value of the specified variable. As near as I can tell (from the echo statements), the code is being run correctly, but for some reason, all data is being fed into the "tour" DB.

Any thoughts? Here's a snippet of code...

  // variable declaration
  $VisitType = NULL;
       if ($VisitDate == 'March 27' OR
           $VisitDate == 'April 17' OR
           $VisitDate == 'April 18' OR
           $VisitDate == 'April 24' OR
           $VisitDate == 'April 25') {
	          $VisitType = 'VisitDay';
	       } else {
	          $VisitType = 'Tour';
	       }

//   *snip*

  // connect to DB
  $db = mysql_connect("localhost", "root");
  if ($VisitType == 'Tour') {
     mysql_select_db("tour",$db);
	 echo "Database set to tour<br><br>";
  } elseif ($VisitType == 'VisitDay') {
     mysql_select_db("visitdays",$db);
	 echo "Database set to visitdays<br><br>";
  }



//   SQL Query 
$sql = "INSERT INTO registrations ..."

//    *snip*

pete_bisby

I've just ran your code and it runs fine.

Are you sure that the $VisitDate variable is being passed correctly, or is spelled correctly on the previous screen - PHP is case sensitive.

jacksonpt

Originally posted by pete_bisby
Are you sure that the $VisitDate variable is being passed correctly, or is spelled correctly on the previous screen - PHP is case sensitive.

Yes.

Here's a quick rundown of what the script does (in order)...

Extract data from a form
Use that data to set variables, including $VisitType and $VisitDate
Select appropriate DB based on $VisitType
Update DB with data extracted from the form
Display a confirmation message

In the confirmation message, I use both $VisitDate and $VisitType without any problems. However, I'll add in couple of printf statements to display the values being passed as $VisitDate and $VisitType at the time the database is supposedly selected.

Here is why I am confused...
If you look at the code above, the code that sets the DB is absolute (no variables involved). The only variable is in the if statement which controls which mysql_select_db is run. I put the echo statements in after each mysql_select_db call to make sure the code was running the correct branch of the if structure, which it is. So, because of the echo statements, I know that the code is following the correct branch of the if structure. The db names used in the mysql_select_db calls match the actual db names. I even commented out the if structure and tested each mysql_select_db call independantly - they work just fine. But as soon as I add back in the if, the switch never happens.

mtimdog

You could make the database name a variable.....chosen by an if statement. And then only have one select_db statement.

jacksonpt

Originally posted by mtimdog
You could make the database name a variable.....chosen by an if statement. And then only have one select_db statement.

No change.

That does basically the same thing I was doing with my original code, right? It's not fundamentally different?

jacksonpt

OK, I've tried a few different things, put in a few more error checks, and still the same deal.

The script makes the correct mysql_select_db call, but the data goes into the "tours" DB regardless.

jacksonpt

OK, I must have an error somewhere else in my code. I just went back and commented out all the code related to the auto-db selection. I coded in (no variables, just straight code) for the script to use the visitdays db - that's the db I haven't been able to get data into. I ran the script, and data is still going to the tour db... so something is wrong somewhere else.

Thanks for the help, but I have to dig though my code and see what's going on. I'll post back when I know more.

jacksonpt

OK, I got it working. I have no idea what the problem was, but I ended up recreating the target table in both databases, as well as a couple chunks of php code.

There must have been an error somewhere that I was just overlooking. I've been staring at this code for 2 days now, so it wouldn't surprise me if I missed something obvious - my guess would be in the table definition(s).