Sorry, there was an error...
if (!empty($_POST['surname']))
$surname=strtoupper( $_POST['surname'] );
else
echo "Empty name";
Luis Moreira wrote:
Try reading the variable before you use it :-)
For best protection, see first if something went through...
if (!empty($_POST['surname']))
$surname=strtoupper($surname);
else
echo "Empty name";
Luis
Martyn Kinder wrote:
Hi,
I am not sure if this is the correct group for this problem, so
apologies if
I have got this totally wrong.
I have just installed PHP5 on a Windows XP machine with IIS (PWS).
phpinfo() seems OK.
However, when I try the following test script:
name.html looks like:
<html>
<body>
<center>
<FORM METHOD="POST" ACTION="name2.php">
<table>
<tr>
<th width="15%" nowrap><b>First Name</b></th>
<td width="35%"><input type=text name = "name"></td>
<th width="15%" nowrap><b>Last Name</b></th>
<td width="35%"><input type=text name = "surname"></td>
</tr>
</table>
<table>
<INPUT TYPE="SUBMIT" VALUE="Submit" NAME="Submit">
<INPUT TYPE="RESET" VALUE="Reset" NAME="Reset">
</table>
</Form>
</center>
</body>
</html>
and
name2.php looks like:
<html>
<body>
<?php
$surname=strtoupper($surname);
?>
<?php echo("<h3>Hello"), $name, (" "), $surname, ("</h3>");?>
</body>
</html>
When I run the name.html script:
I get
Notice: Undefined variable: surname in c:\Inetpub\wwwroot\name2.php
on line
5
Hello
Notice: Undefined variable: name in c:\Inetpub\wwwroot\name2.php on
line 9
I think this works OK on PHP4 and Apache.
I have looked through the FAQ's etc. But I'm not sure if this is an
idiot
fault or a configuration fault.
Is there a requirement to declare variables in PHP5 now?
Any suggestions please!
Thanks
Martyn
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