How to make preg_match that if there are characters in the line except letters and space then return false
by pattern
^[A-zА-ЯёЁ ] well, or something like that
Visman
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dfhsfhgfj dfhsfhgfj
1,113 5 gold signs 22 silver marks 51 bronze marks
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1 answer
preg_match('/^[a-zа-яё ]+$/ui', 'ab c А бв '); 
romeo romeo
4,855 11 silver marks 28 bronze marks
- and not so? ^ means the beginning of $ the end seems to be like, but here it is the negation of preg_match ('/ [^ a-za-yo] + / i', 'ab c A bv'); - dfhsfhgfj 3:04 pm
- @dfhsfhgfj> ^ means the beginning of $ the end seems to be all right, and therefore preg_match returns false if everything except letters. $ subject = 'ab c A bv'; / * If false * / if (! Preg_match ('/ ^ [a-za-ya] + $ / ui', $ subject)) {echo 'The value must contain only alphabetic characters and a space'; } Do not forget to use the u modifier, since you can probably use Cyrillic in the string. - romeo
- thanks for the detailed answer - dfhsfhgfj pm
- Will this code work correctly? preg_match ('/ [^ a-za-ya] + / i', 'ab c A bv'); I did not understand, is it an analogue of what you wrote or not? - dfhsfhgfj
- @dfhsfhgfj Yes, it will, but the value will be the opposite. preg_match ('/ ^ [^ a-za-ya] + $ / iu', '# * _ 122'); will print true if the string does not contain alphabetic characters and a space. Then the above condition will look like this: / * If true / if (preg_match ('/ ^ [^ a-za-ya] + $ / ui', $ subject)) {echo 'The value must contain only alphabetic characters and a space' ; } The characters of the beginning and end of the search string (** ^ ... $ * ) in the pattern are required. If this is not done, then the array (3 argument of the preg_match function) will get part of the string that satisfies this condition, i.e. will be returned ... - - romeo
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