将 awk 拆分数组返回到 bash 变量

Question

I have a requirement to split a string on a multi-character delimiter and return the values into an array in Bash for further processing

IFS can take a single character delimiter.

a="2;AAAAA;BBBBB;1111_MultiCharDel_2;CCCC;DDDDDD;22222_MultiCharDel_2;EEEE;FFFFFFF;22222" 
awk'{split($0,ArrayDeltaMulDep,"_MultiCharDel_")}' <<< $a

The input string can have several substrings separated by the MultiCharDel delimiter.

How can i access this array ArrayDeltaMulDep fur further processing in Bash?

Answer 1

Your example string, a , does not contain newlines. If that is true in general, then:

a="2;AAAAA;BBBBB;1111_MultiCharDel_2;CCCC;DDDDDD;22222" 
readarray -t b <<< "${a//MultiCharDel/$'\n'}"

We can verify that this split the string properly using declare -p to show the value of b :

$ declare -p b
declare -a b=([0]="2;AAAAA;BBBBB;1111_" [1]="_2;CCCC;DDDDDD;22222")

How it works:

1. readarray -tb

   This reads lines from stdin and puts then in a bash array b .

2. <<< "${a//MultiCharDel/$'\\n'}"

   ${a//MultiCharDel/$'\\n'} uses pattern substitution to replace MultiCharDel with a newline character. <<< provides the result as stdin to the command readarray .

Hat tip: Chepner

More general solution

A bash string will never contain a null character (hex 00). Using GNU sed:

b=()
while read -d '' -r line
do
   b+=("$line")
done < <(sed 's/MultiCharDel/\x00/g; s/$/\x00/' <<<"$a")

This again creates an array with the desired splitting:

$ declare -p b
declare -a b=([0]="2;AAAAA;BBBBB;1111_" [1]="_2;CCCC;DDDDDD;22222")