拆分字符串（例如使用bash）但跳过部分字符串

Question

How can I split with bash (awk, sed, whatever) the following string:

in:

a,b,[c, d],e

output:

a
b
[c, d]
e

try 1)

$IFS=',' read -a tokens <<< "a,b,[c, d], e"; echo ${tokens[@]}
a b [c d] e

try 2)

$ IFS=',' 
$ line="a,b,[c, d], e"
$ eval x=($line)
$ echo ${x[1]}
b
$ echo ${x[0]}
a
$ echo ${x[2]}
[c  d]

But not ','!

Answer 1

This is just a specific instance of the general CSV problem of identifying commas inside quotes differently from those outside of quotes in order to replace either one with some other character (eg ; ). The idiomatic awk solution to that (besides using FPAT in GNU awk) is:

Replace inside the quotes:

$ echo 'a,b,"c, d",e' | awk 'BEGIN{FS=OFS="\""} {for (i=2;i<=NF;i+=2) gsub(/,/,";",$i)}1'
a,b,"c; d",e

Replace outside the quotes:

$ echo 'a,b,"c, d",e' | awk 'BEGIN{FS=OFS="\""} {for (i=1;i<=NF;i+=2) gsub(/,/,";",$i)}1'
a;b;"c, d";e

In your case the delimiters are [...] instead of "..." and the replacement character is a newline instead of a semi-colon but it's essentially the same problem:

Replace outside the "quotes" (square brackets):

$ echo 'a,b,[c, d],e' | awk 'BEGIN{FS="[][]"; OFS=""} {for (i=1;i<=NF;i+=2) gsub(/,/,"\n",$i)}1'
a
b
c, d
e

Note that the square brackets are gone because I set OFS to a blank char since there is no 1 single FS character to use. You can get them back with this if you actually do need them:

$ echo 'a,b,[c, d],e' | awk 'BEGIN{FS="[][]"; OFS=""} {for (i=1;i<=NF;i++) if (i%2) gsub(/,/,"\n",$i); else $i="["$i"]"}1'
a
b
[c, d]
e

but chances are you don't since their purpose was to group text that contained commas and now that's handled by the newlines being the field separators instead of commas.

Answer 2

You can for example use this grep:

grep -Po '([a-z]|\[[a-z], [a-z]\])'
           ^^^^^ ^^^^^^^^^^^^^^^^ 

See:

$ echo "a,b,[c, d],e" | grep -Po '([a-z]|\[[a-z], [a-z]\])'
a
b
[c, d]
e

That is, use grep to print only (hence the -o , to match only), either blocks of [az] letter or [ + [az], [az] + ] .

Or you can also make the opening [ and closing , [az]] block optional:

$ echo "a,b,[c, d],e" | grep -Po '(\[)?[a-z](, [a-z]\])?'
a
b
[c, d]
e

Answer 3

Match everything that starts with [ and ends with ] : \\[[^][]*\\] . Then match anything that's not a comma: [^,]\\+ :

echo 'a,b,[c, d],e' | grep -o -e '\[[^][]*\]' -e '[^,]\+'

Output:

a
b
[c, d]
e

Answer 4

echo "a,b,[c, d],e" | grep -o '\\[.*\\]\\|[^,]*'

Output:

a
b
[c, d]
e