Note that $ allows a trailing newline, but \z doesn't.
Arpad
Stut wrote:
Chris Boget wrote:
<?php
echo 'Is String: [' . ( is_string( 'a1b2c3' ) && preg_match(
'/[A-Za-z]+/', 'a1b2c3' )) . ']<br>';
echo 'Is Numeric: [' . ( is_numeric( 'a1b2c3' ) && preg_match(
'/[0-9]+/', 'a1b2c3' )) . ']<br>';
echo 'Is String: [' . ( is_string( 'abcdef' ) && preg_match(
'/[A-Za-z]+/', 'abcdef' )) . ']<br>';
echo 'Is Numeric: [' . ( is_numeric( '123456' ) && preg_match(
'/[0-9]+/', '123456' )) . ']<br>';
?>
Why is the first "Is String" check returning true/showing 1?
preg_match should fail because 'a1b2c3' contains numbers and, as
such, doesn't match the pattern...
It does match the pattern. The expression says "1 or more A-Za-z in
sequence". If you want to check against the whole string you need to
add the start and end markers...
preg_match( '/^[A-Za-z]+$/', 'a1b2c3' ))
Look at the manual page for preg_match and read up on the third
parameter. Use it to see what is matching the expression.
-Stut
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