undefined variable error

smilley654

here is the line of code entered:
<form action="you_are.php">
Please enter your name:<br />
I am...
<?php print('<input type="text" name="person" value="' . $person .
'"size="15" />');
?>
<input type="submit" value="Go!" size="15" />
</form>

here is the error message:
Notice: Undefined variable: person in who_are_you.php on line 9

...

My question is what is wrong with using ". $person ."? This is based on the who_are_you.php code in "Mastering PHP 4.1" from Sybex Publishing. I'm using PHP 4.2.3 for Windows using Apache Server 2.0.40 for Windows.

Thanks for your time.

stuartc1

You can only use $person if the value of $person has already been set. Also you may have to use superglobals to access he value if the $person variable has come from a form (post or get) or a session. for example, if it from a form that uses post as the method you may need to use: $_POST['person'] to access it.

HTH🙂

cbailster

The reason this is happening is because of the error reporting settings. In php > 4.2.1 (i think) error reporting is set to E_ALL which means that thinks such as undefined variables produce error messages. To stop php doing this to you set error reporting to E_ALL & ~E_NOTICE (i.e. All errors apart from run-time notices)

stuartc1

Its not a good idea to supress error messages, as you will probably end up with a very buggy app that you think is sound but is really a bag of shite.

cbailster

True, but all you're doing by using ~E_NOTICE is suppressing run-time notifications, such as undefined variables, and in some cases the design of the script to use unitialised variables is intentional. Failing that, if you do want to keep all errors but not to see them on the pages themseves (e.g. just log them to event viewer), you can just turn display_errors to Off

smilley654

You were right on the mark! I plugged in a name before the php statement to see what it would do, i.e.:
<?php $person="newperson" ?>
<?php print('<input type="text" name="person" value="' . $person . '"size="15" />'); ?>

and, the results:
Please enter your name:
I am... [newperson ] [GO button]

However, I believe the book stated it wrong. For example, as an excerpt, it says:
"If you are familiar with other programming languages, you noticed that $person was never declared or initialized. Strongly typed languages require a variable to be declared and its type to be defined before the variable can be used to store data. This is not the case with PHP, as you will soon learn..."

(Notice my underlines!)

I now understand based on other reviews I've read of books on PHP and other related material is that some books have some things either confused or taken out. This may be one of the cases.

But, isn't there a way to not initialize a variable in PHP?

jmcneese

try this:

i am <? if ($isset($person) and !empty($person)) { echo $person; } ?>.

smilley654

ok, I just discovered that it doesn't print the name or any name I enter into the input form from what_you_are.php. For example, here is the other code for "you_are.php". Here's what it's supposed to print after the what_you_are.php code:
<?php
print('Well, hello ' . $person . ', nice to meet you!');
print('<br />');
print('<a href="who_are_you.php?person=' . urlencode($person) .'">
Back to Who are You?</a>');
?>

However, it doesn't print the variable in $person, because it didn't pass the variable to the you_are.php file. It comes up with undefined variable msg again.

I also just tried making the variable global and it got rid of the error message in you_are.php, (i.e. undefined variable error msg), but I still don't see the contents of $person printed on output.

smilley654

here's what I entered:
I am...
<?php global $person;
print('<input type="text" name="person" value="' . $person .
'"size="15" />');
if (isset($person) and !empty($person)) { echo $person; }
?>

But, nothing outputed on screen. I know it's something I'm missing. I wonder, is there something supposed to showup in phpinfo() in order for variables to work?

cbailster

This is because by default in php V4.2.3 register globals is off. This means that if you have a variable in a form called person instead of it being registered in your script as $person it will be placed in the $GET array of $POST array depending on whether the information from the form was passed in get or post data.

In your case, the information passed in the form will be in $GET['person']. Note however, that you cannot use the isSet function on the $GET/$_POST array to test if any information was passed.

(see http://www.php.net/manual/en/reserved.variables.php for more information on $GET, $POST etc.)

smilley654

All I had to do was set register_globals on in php.ini!! Excited
Anyways, is there anything to read up on all changes since PHP 4.1? I don't want to find out something else I would have several posts on(not that I'm not enjoying learning all of this). So, anything docs concerning new things added would be great.

Again, thanks all for your time.