regex - 使用 '=~' 运算符的 Bash 正则表达式匹配意外失败

Question

这是一个小单元脚本，用于由 =~ 调用的良好的旧 bash 正则表达式匹配

#!/bin/bash

# From "man bash"
# An additional binary operator, =~, is available, with the same
# precedence as == and !=. When it is used, the string to the right of
# the operator is considered an extended regular expression  and  matched
# accordingly (as  in regex(3)).  The return value is 0 if the string
# matches the pattern, and 1 otherwise.  If the regular expression
# is syntactically incorrect, the conditional expression's return value
# is 2.

# The above should say regex(7) of course

match() {
   local REGEX=$1
   local VAL=$2
   [[ $VAL =~ $REGEX  ]]
   RES=$?
   case $RES in
      0) echo "Match of '$VAL' against '$REGEX': MATCH" >&2 ;;
      1) echo "Match of '$VAL' against '$REGEX': NOMATCH" >&2 ;;
      2) echo "Error in regex expression '$REGEX'" >&2 ;;
      *) echo "Unknown returnvalue $RES" >&2 ;;
   esac
   echo $RES
}

v() {
   SHALL=$1
   IS=$2
   if [ "$SHALL" -eq "$IS" ]; then echo "OK"; else echo "NOT OK"; fi
}

unit_test() {
   v 0 "$(match A                A  )"
   v 0 "$(match A.               AB )"
   v 0 "$(match A[:digit:]?      A  )"
   v 0 "$(match A[:digit:]       A6 )"
   v 0 "$(match \"A[:digit:]*\"  A6 )"  # enclosing in quotes needed otherwise fileglob happens
   v 0 "$(match A[:digit:]+      A6 )"
   v 0 "$(match A                BA )"
   v 1 "$(match ^A               BA )"
   v 0 "$(match ^A               Ab )"
   v 0 "$(match 'A$'             BA )"
   v 1 "$(match 'A$'             Ab )"
}

unit_test

看起来很简单，但运行它会产生:

Match of 'A' against 'A': MATCH
OK
Match of 'AB' against 'A.': MATCH
OK
Match of 'A' against 'A[:digit:]?': MATCH
OK
Match of 'A6' against 'A[:digit:]': NOMATCH
NOT OK
Match of 'A6' against 'A[:digit:]*': MATCH
OK
Match of 'A6' against 'A[:digit:]+': NOMATCH
NOT OK
Match of 'BA' against 'A': MATCH
OK
Match of 'BA' against '^A': NOMATCH
OK
Match of 'Ab' against '^A': MATCH
OK
Match of 'BA' against 'A$': MATCH
OK
Match of 'Ab' against 'A$': NOMATCH
OK

人们期待

Match of 'A6' against 'A[:digit:]'

和

Match of 'A6' against 'A[:digit:]+'

成功。

我做错了什么？

Answer 1

请记住将字符类括在方括号 [] 中，以将它们匹配为字符列表，即 [[:digit:]]

string="A6"
[[ $string =~ A[[:digit:]] ]]
echo $?
0

在 Bracket-Expressions 上查看更多信息.