I am new to regex and I have been going round and round on this problem.

PHP: Check alphabetic characters from any latin-based language? gives the brilliant regex to check for any characters in the Latin script, which is part of what I need.

^\p{Latin} $

and provides a working example at https://regex101.com/r/I5b2mC/1

If I use the regex in PHP by using

echo preg_match('/^\p{Latin} $/', $testString);

and $testString contains only Latin letters, the output will be 1. If there is any non-Latin letters, the output will be 0. Brilliant.

To add numbers in I tried ^\p{Latin} [[:alnum:]]*$ but that allows any characters in the Latin script OR non-Latin letters and numbers (letters without accents — grave, acute, cedilla, umlaut etc.) as it is the equivalent to [a-zA-Z0-9].

If you add any numbers with characters in the Latin script, echo preg_match('/^\p{Latin} [[:alnum:]]*$/', $testString); returns a 0. All numbers return a 0 too. This can be confirmed by editing the expression in https://regex101.com/r/I5b2mC/1

How do I edit the expression in echo preg_match('/^\p{Latin} $/', $testString); to output a 1 if there are any characters in the Latin script, any numbers and/or spaces in $testString? For example, I wish for a 1 to be output if $testString is Café ßüs 459.

CodePudding user response：

There are at least two things to change:

• Add u flag to support chars other than ASCII (/^\p{Latin} $/ => /^[\p{Latin}] $/u)
• Create a character class for letters, digits and whitespace patterns (/^\p{Latin} $/u => ^[\p{Latin}] $/u)
• Then add the digit and whitespace patterns. If you need to support any Unicode digits, add \d. If you need to support only ASCII digits, add 0-9.

Thus, you can use

preg_match('/^[\p{Latin}\s0-9] $/u', $testString) // ASCII only digits
preg_match('/^[\p{Latin}\s\d] $/u', $testString)  // Any digits

Also, \s with u flag will match any Unicode whitespace chars.