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ctype_alpha - how to add space

Posted on 2009-05-19
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Last Modified: 2012-05-07
j.e ( if ctype_alpha => true if there are white spaces inside)
function _surname($v) {
	if(!empty($v)) {
		if(!ctype_alpha($v)) $e .= '» Please use only alphabet(no digits) characters in surname (+ no spaces)'."\r";
		if(!lenght($v,2)) $e .= '» Surname must be above 2 letters lenght.'."\r";
	} return $e;
}

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Question by:AndyPSV
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4 Comments
 
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Expert Comment

by:Richard Quadling
ID: 24431357
I don't think you can alter ctype.

But a regex will do ...

if (0 == preg_match('`^^[a-z](?!.*  .*)[a-z ]+[a-z]$`i', $v)) $e .= '» Please use only alphabet(no digits) characters in surname (+ spaces)'."\r";

The regex is saying must start and end with a letter and may contain spaces and letters but not 2 spaces next to each other.


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Author Comment

by:AndyPSV
ID: 24438297
The regex is saying must start and end with a letter and may contain spaces and letters but not 2 spaces next to each other.

How to modify this to trim the empty 2 spaces to one?
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Richard Quadling earned 500 total points
ID: 24439036
Something like this (and sorry for the typo of the double ^. Cut and paste error).

if (0 == preg_match('`^(?! )[a-z ]+(?<! )$`i', preg_replace('` +`', ' ', trim($v, ' ')), $a_Matches))

and then use $a_Matches[0] as the new cleaned value of $v


Below is a usage test.

The output is ...

' '                                     : Failed to pass test.
'  '                                    : Failed to pass test.
'       starts with spaces'             : Passed test and was returned as 'starts with spaces'.
'ends with spaces         '             : Passed test and was returned as 'ends with spaces'.
'      wrapped with spaces          '   : Passed test and was returned as 'wrapped with spaces'.
'one space'                             : Passed test and was returned as 'one space'.
'nospace'                               : Passed test and was returned as 'nospace'.
'lots    of    spaces'                  : Passed test and was returned as 'lots of spaces'.


<?php
$a_Tests = array
	(
	' ',
	'  ',
	'       starts with spaces',
	'ends with spaces         ',
	'      wrapped with spaces          ',
	'one space',
	'nospace',
	'lots    of    spaces'
	);
 
foreach($a_Tests as $s_Test)
	{
	echo str_pad("'$s_Test' ", 40) . ": ";
	if (0 == preg_match('`^(?! )[a-z ]+(?<! )$`i', preg_replace('` +`', ' ', trim($s_Test, ' ')), $a_Matches))
		{
		echo "Failed to pass test.", PHP_EOL;
		}
	else
		{
		$s_Test = $a_Matches[0];
		echo "Passed test and was returned as '$s_Test'.", PHP_EOL;
		}
	}

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Expert Comment

by:Richard Quadling
ID: 24448505
Slight update. The regex can be simplified. See comments in the code below.
<?php
$a_Tests = array
        (
        ' ',
        '  ',
        '       starts with spaces',
        'ends with spaces         ',
        '      wrapped with spaces          ',
        'one space',
        'nospace',
        'lots    of    spaces'
        );
 
foreach($a_Tests as $s_Test)
        {
        echo str_pad("'$s_Test' ", 40) . ": ";
	// As we are trimming the string, no leading or trailing spaces.
	// As we replace 2+ spaces with 1 only single spaces.
	// So, the regex can be a lot simpler. Must match at least 1 letter or space
	// (though 1 space will never actually get to the regex test).
	// If you want to say there has to be at least 10 letters, use a regex of ...
	// '`^[a-z ]{10,}$`i'
	// The {10,} means at least 10 lettes.
	// If you want between 10 and 40 letters ...
	// '`^[a-z ]{10,40}$`i'
        if (0 == preg_match('`^[a-z ]+$`i', preg_replace('` +`', ' ', trim($s_Test, ' ')), $a_Matches))
                {
                echo "Failed to pass test.", PHP_EOL;
                }
        else
                {
                $s_Test = $a_Matches[0];
                echo "Passed test and was returned as '$s_Test'.", PHP_EOL;
                }
        }

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