regex - 为什么空格分隔而不是 FPAT

Question

我正在尝试拆分由 ',' 分隔的字符串，除非 ',' 位于用方括号括起来的子字符串中。修改这里的其他解决方案和文档中的示例我尝试了这个测试:

awk -v FPAT='([^,]+)|(\([^\))+\))' '{
    for (i=1; i<=NF; i++) {
        printf("%s\n", $i)
    }                     
}' <<< 'one two (1one), three four (3three, 4four), five six, seven eight, nine ten eleven (8ten)'
one
two
(1one),
three
four
(3three,
4four),
five
six,
seven
eight,
nine
ten
eleven
(8ten)

FPAT 没有像我预期的那样覆盖默认分隔符。很明显我错过了一些东西。

我想要的输出是:

one two (1one),
three four (3three, 4four),
five six,
seven eight, 
nine ten eleven (8ten)

Answer 1

使用 gnu grep:

s='one two (1one), three four (3three, 4four), five six, seven eight, nine ten eleven (8ten)'
grep -oP '\s*\K([^,(]*\([^)]*\))*[^,]*(,|$)' <<< "$s"

one two (1one),
three four (3three, 4four),
five six,
seven eight,
nine ten eleven (8ten)

如果你没有gnu grep那么你可以使用

grep -oE '([^,(]*\([^)]*\))*[^,]*(,\s*|$)' <<< "$s"

这将在逗号后留下尾随空格。

对于 regex explanation see this demo .