PHP regex: each word must end with dot - php

Question

Can someone help me how to specific pattern for preg_match function?
Every word in string must end with dot
First character of string must be [a-zA-Z]
After each dot there can be a space
There can't be two spaces next to each other
Last character must be a dot (logicaly after word)
Examples:
"Ing" -> false
"Ing." -> true
".Ing." -> false
"Xx Yy." -> false
"XX. YY." -> true
"XX.YY." -> true
Can you help me please how to test the string? My pattern is
/^(([a-zA-Z]+)(?! ) \.)+\.$/
I know it's wrong, but i can't figure out it. Thanks

Answer 1

Check how this fits your needs.
/^(?:[A-Z]+\. ?)+$/i
^ matches start
(?: opens a non-capture group for repetition
[A-Z]+ with i flag matches one or more alphas (lower & upper)
\. ? matches a literal dot followed by an optional space
)+ all this once or more until $ end
Here's a demo at regex101
If you want to disallow space at the end, add negative lookbehind: /^(?:[A-Z]+\. ?)+$(?<! )/i

Answer 2

Try this:
$string = "Ing
Ing.
.Ing.
Xx Yy.
XX. YY.
XX.YY.";
if (preg_match('/^([A-Za-z]{1,}\.[ ]{0,})*/m', $string)) {
// Successful match
} else {
// Match attempt failed
}
Result:
The Regex in detail:
^ Assert position at the beginning of a line (at beginning of the string or after a line break character)
( Match the regular expression below and capture its match into backreference number 1
[A-Za-z] Match a single character present in the list below
A character in the range between “A” and “Z”
A character in the range between “a” and “z”
{1,} Between one and unlimited times, as many times as possible, giving back as needed (greedy)
\. Match the character “.” literally
[ ] Match the character “ ”
{0,} Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
)* Between zero and unlimited times, as many times as possible, giving back as needed (greedy)