Hi All,
Thanks for all your help, but I have one more, final question--Please Help--

How to I display a varibale that was inputed in the form? That is;
name: John (this was entered in the form)

I thought that if I included the variable ($user_name), I also tried echo, but neither worked.
Any help would be great
Thank you very much,
John

You should be able to just do:


Make sure the variable is inside the quotes.

Hi dbodner,
Thanks for your reply, but I get a parse error with that.

Tried with echo with-out echo, many different ways
I have been trying to figure this out all night, I feel really stuck

Thanks so much for your help

John,

It would probably help if we could look at a full copy of your script. Could you post the PHP script in here:
http://pastebin.com/

and send us the link to it? Thanks

Hi pmeserve,
Thanks so much for your help, I really apreiate it very much
I posted the script on the site you gave, that is cool!

I also have the script here:
http://67.59.131.1/script.txt
Hopefully, you can help me with this.
Basically what I am trying to do is when the user fills the form, is to give a message about that.
Example:
your entered:user_name=John
or something like that

Thanks so much,

The thank you message is never ended with a closing ". So the echo (a new PHP call) will give you a parse error, as the assignment of $thankyou_message was never finished. Try changing it to:


Also, i before e, except after c

Hi dbodner,
Thanks so much for your help, that did not do the trick
When I used the echo, like in your code, it always displayed what was echoed, but not just after when the form was filled out. That is, I just want to display the name=whatever after the form was filled out. therefore I changed the echo code to this:

this made it so it only displayed after the form was filled out, but did not pass the variables
Thanks again for your help,
John

John, is that code above the lines like:
$name = stripslashes($_POST['user_name']);

?

If so that will cause issues. One thing I always do when writing a new PHP script is add these lines to the very top:


You'll want to comment them out once the script is done, but while you're writing code it will make it much easier to see errors and warnings, for example if you try to output an unset variable(which I suspect it what's happening). It's often a lot easier than trying to track down problems based on warning messages than based on the actual behavior of the problem

Hi pmeserve,
Thanks for your help. I used your code, and I see that I get this error:


I don't understand while it says that the variables are undenified. As when I use the form, it sends the data to the e-mail, but I can't display it on the page. Oh no, I am so lost on this

Can you re-post the updated code?

Hi dbodner,
Thanks for your help, I have been looking at this for 2 hours
I am going to buy a book on php.
Thanks for all your help,
here is the code,
Thanks,
John

John,

I rearranged the code a bit and I believe it now works as expected. Try this:

Hi pmeserve,
thank you so very much
I can't tell you how much it means to me that you took the time to figure out how to make this problem work.
The code you wrote works like a dream, thanks so much.

Please let me know if there is anything I can ever do for you, thanks for all your help
You can see the test page at:
http://67.59.131.1/NutritionATC/add_event.php

Thank you so much!
John