通过Bash脚本捕获文本组

Question

A file that looks like the following is created from an API call.

{
  "success" : true,
  "messages" : [ "traces loaded successfully" ],
  "traces" : [ "CRXJ-ZCKP-3XVD-4J36", "8C31-QMHZ-XVF0-ZV4Q", "40P7-MT6L-2YFP-5Q6Q", "JE7J-J4WQ-7GGL-2U8N", "10PW-AZ8M-FPTK-XWYL", "9YGF-5HOP-8GTY-9AF4", "SNXZ-GW6O-BTJT-3XUX" ]
}

How can I capture each traces separately? I will need to iterate through them in a subsequent step.

The following doesn't return anything unless I remove the parentheses. Then it just gives the entire "traces" line.

grep "([A-Z0-9]{4}-[A-Z0-9]{4}-[A-Z0-9]{4}-[A-Z0-9]{4})"

The answer must be able to run on all Unix machines without loading libraries.

Answer 1

You can use:

grep -oE '([A-Z0-9]+-?){4}' file

---

Options:

-o, --only-matching       show only the part of a line matching PATTERN
-E, --extended-regexp     PATTERN is an extended regular expression (ERE)

---

Output:

CRXJ-ZCKP-3XVD-4J36
8C31-QMHZ-XVF0-ZV4Q
40P7-MT6L-2YFP-5Q6Q
JE7J-J4WQ-7GGL-2U8N
10PW-AZ8M-FPTK-XWYL
9YGF-5HOP-8GTY-9AF4
SNXZ-GW6O-BTJT-3XUX

---

Regex Demo and Explanation

Answer 2

Use grep -oE :

grep -oE '[A-Z0-9]{4}-[A-Z0-9]{4}-[A-Z0-9]{4}-[A-Z0-9]{4}' file

Output:

CRXJ-ZCKP-3XVD-4J36
8C31-QMHZ-XVF0-ZV4Q
40P7-MT6L-2YFP-5Q6Q
JE7J-J4WQ-7GGL-2U8N
10PW-AZ8M-FPTK-XWYL
9YGF-5HOP-8GTY-9AF4
SNXZ-GW6O-BTJT-3XUX

Answer 3

You could also stay in bash:

#!/usr/bin/env bash

regex='("(([[:alnum:]]{4}-?){4})"(, )?)+'

while read -r line
do
    if [[ "$line" =~ $regex ]]
    then
        for trace in ${BASH_REMATCH[0]//[\",]/}
        do
            echo "do stuff with $trace"
        done
    fi
done<input_file