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Trouble using strlen method and another newbie question.

Answer 1

+5 A:

PHP variable names are case-sensitive. Instead of $Firstname you should be referencing $FirstName:

$firstNameCount=strlen($FirstName);

As far as styling practices are concerned there is no language-wide standard. The best approach for formatting is to write code that is clean and readable. For your snippet above I would do it like this:

<?php
  $FirstName = "Sergio";
  $LastName = "Tapia";
  $firstNameCount = strlen($FirstName);
?>
<html>
  <head>
  </head>
  <body>
    <h1>It works!</h1>
    <?php echo "Why hello there Mr. " . $FirstName . " " . $LastName . "So, your name appears to have " . $firstNameCount . " letters." ?>
  </body>
</html>

The echo function can be used in either fashion (echo('') or echo ''), the only recommendation I would make is to pick one and stick with it.

Cryo 2010-02-13 03:48:06

Your call to strlen is still using the wrong-cased version in your example code. (The 'n' isn't capitalized.)

clee 2010-02-16 00:16:28

@clee Thanks for pointing that out. That example was for code styling, the bug fix was above that but I've updated my answer just in case.

Cryo 2010-02-16 05:32:18

Answer 2

+1 A:

Variable names are case sensitive. Notice how you spell $FirstName and $Firstname.

Vincent Ramdhanie 2010-02-13 03:48:52

Answer 3

+1 A:

$Firstname <> $FirstName

codingguy3000 2010-02-13 03:49:20

In PHP, that would be '!='. As this '<>' does not equal this '!=' in PHP.

Mark Tomlin 2010-02-13 04:22:35

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Trouble using strlen method and another newbie question.

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