Re: Split command problem

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I guess you are trying to create an array by the name 'name' and assign two 
elements to it by calling name($fname, $lname). Am I correct?

I think it does not work that way. Try list($fname, $lname) = ......
Then the variable $fname and $lname will contain the first and last names.

list($fname, $lname) = split('[/\s+/]', "$_POST[username]");

Cheers,

Prathap


---------- Original Message -----------
From: Russ <russbucket@xxxxxxxxxxxxxx>
To: PHP General <php-general@xxxxxxxxxxxxx>
Sent: Sun, 10 Apr 2005 18:27:34 -0700
Subject:  Split command problem

> I have been trying to get the following code working. I keep getting 
> an error on line nine. It looks simular to the example in the PHP 
> online manual. If I substitute a print command for line nine I get 
> the correct information from $_POST[username]. I must be missing or 
> have an extra a quote but I cannot figure out where.
> 
> <?php
> //check for required fields from the form
> if ((!$_POST[username]) || (!$_POST[password])) {
> 	header("Location: memberlogin15.7.php");
> 	exit;
> } 
> if(preg_match("/[A-Z]/", substr("$_POST[username]", 0, 1)))
> {
> name($fname, $lname) = split('[/\s+/]', "$_POST[username]");
> //echo "First Name: $fname; Last Name: $lname\n;
> echo "first letter is uppercase";
> }
> else
> {
> echo "first letter is not uppercase";
> }
> 
> Any help will be appreciated.
> -- 
> Russ
> 
> -- 
> PHP General Mailing List (http://www.php.net/)
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------- End of Original Message -------

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