html form for mysql using php

[tracy]

I'm redoing it now (as I type) and uploading it to a different location.  Might be a while.  Thanks, though. 

[tracy]

what is the first line of code referring to include secure.php?

I don't have that file...is that the problem?

[jsladek]

no not a problem the "#" in the front of it tells the php processor to ignore that line

[tracy]

ok...thanks.

[tracy]

It looks a lot like yours now, no pics obviously.

Had to make some changes, but I figured it out...thanks again.

Pretty cool.

That was where I was going with this...sort of...but enough to get me moving more forward with the coding part.

Thanks a lot.  Really.

[jsladek]

No Problem, should get you started.  What did you have to change to get it to work? Was it something in the script?  The pictures are easy you can just copy them from the links I supplied or even straigt off the site I showed you.

-John

[tracy]

Login kind of stuff mainly...I think...a few syntax here and there...

My goal was to learn the 'how-to' of the code to make it work, not copy it.  This helped a lot.

I have done a lot of other research too, but this seemed to be geared to more of what I wanted to do...

not exactly, but enough to solve some of the issues to take the next step.  Thanks again.  Sounds like that

program you wrote is pretty cool...and working!  Congratulations.

[tracy]

I keep getting this error when I try to use the script provided (or similar) in testtable.php:

Parse error: syntax error, unexpected T_VARIABLE (says it's on line 4)

Here's the first part of the code...

line one (where the php question mark is...)

# include ("secure.php");
$pgflow = $_GET['pgflow'];     <<<===this would be line four!

//conditional statements included below determine program flow
switch ($pgflow) {

case 'addrecord':
new_record();
break;
case 'insertrecord':
insert_record();
break;
case 'deleterecord':
delete_record();
break;
case 'editrecord':
edit_record();
break;
case 'showchanges':
show_changes();
break;
case 'verified':
update_database();
break;
default:
show_data();
}


etc.....

any ideas? 

[jsladek]

That Code if failing for you?  I thought it was working earlier?

What happens if you comment out the line?

There could be something a head of that line causing the problem.

Did you cut and paste this code? Maybe some strange formating got in there. Try looking at it in note pad and see if anything strange there.


Let me know what you find.

-John

[tracy]

I tried to use the principles of PHP to make some modifications...didn't work.

I don't really want to just cut and paste.  What is the signifigance of #  ?

I'm not familiar with the "#" in PHP.  I think it is similar to // right?

Also, there is nothing above line 4 to cause an issue.  Is PHP reading the entire code and then coming back to line four for that particular line's command and then announcing the issue? 

Thanks.

[jsladek]

Yes you are correct # is like // just an alternative way to comment things, I picked it using some programming language, PERL maybe, I forget...

Why not post your code and let's take a look @ it.

-John