Regular expression that matches all specified characters in any order at the start of the line

Question

I'm looking for a regexp that matches any line that contains 'B', 'R', 'A' and 'S' (in any order) at the start. It would match all the following lines, except the last two.

BRASIL
BSRAIL
BARSILERO
BRASILERA
BRASILEÑA
BRASILEÑO
BARBADOS
BOSNIA AND HERZEGOVINA

I tried the following:

^(B|R|A|S){4}.*$
^(?=.*B)(?=.*R)(?=.*A)(?=.*S).*$
^(?=.{4})(?=.*B)(?=.*R)(?=.*A)(?=.*S).*$
^(?=.*B)(?=.*R)(?=.*A)(?=.*S){4}.*$
^(?=.*B){1}(?=.*R){1}(?=.*A){1}(?=.*S){1}.*$

Answer 1

You can use

^([BRAS])(?!\1)([BRAS])(?!\1|\2)([BRAS])(?!\1|\2|\3)([BRAS])

See the regex demo. Details:

• ^ - start of string
• ([BRAS]) - Group 1: B, R, A or S
• (?!\1)([BRAS]) - Group 2: B, R, A or S but not as in Group 1
• (?!\1|\2)([BRAS]) - Group 3: B, R, A or S but not as in Group 1 and 2
• (?!\1|\2|\3)([BRAS]) - Group 4: B, R, A or S but not as in Group 1, 2 and 3.

Note that when using the pattern with MySQL v.8, you need to double the backslashes:

"^([BRAS])(?!\\1)([BRAS])(?!\\1|\\2)([BRAS])(?!\\1|\\2|\\3)([BRAS])"

Answer 2

There are only 24 permutations :)

^(ABRS|BARS|RABS|ARBS|BRAS|RBAS|RBSA|BRSA|SRBA|RSBA|BSRA|SBRA|SARB|ASRB|RSAB|SRAB|ARSB|RASB|BASR|ABSR|SBAR|BSAR|ASBR|SABR)

You can shorten it a bit by grouping pairs of two:

^((AB|BA)(RS|SR)|(AR|RA)(BS|SB)|(AS|SA)(BR|RB)|(BR|RB)(AS|SA)|(BS|SB)(AR|RA)|(RS|SR)(AB|BA))

Each double pair matches 4 inputs, e.g., (AB|BA)(RS|SR) can match:

ABRS
ABSR
BARS
BASR