Split string using 2 different delimiters in Bash

Question

I am trying to split a string in BASH based on 2 delimiters - Space and the \. This is the string:-

var_result="Pass results_ADV__001__FUNC__IND\ADV__001__FUNC__IND_08_06_14_10_04_34.tslog"

I want it to split in 3 parts as follows:-

part_1=Pass
part_2=results_ADV__001__FUNC__IND
part_3=ADV__001__FUNC__IND_08_06_14_10_04_34.tslog

I have tried using IFS and it splits the first one well. But the second split some how removes the "\" and sticks the entire part and I get the split as :-

test_res= Pass
log_file_info=results_ADV__001__FUNC__INDADV__001__FUNC__IND_08_06_14_10_04_34.tslog

The IFS I used is as follows:-

echo "$var_result"
IFS=' ' read -a array_1 <<< "$var_result"
echo "test_res=${array_1[0]}, log_file_info=${array_1[1]}"

Thanks in advance.

You need to pass `-r` to `read` to tell it not to evaluate escapes. — Etan Reisner, Aug 06 '14 at 14:53
@EtanReisner you were right in both the cases. I missed the use of -r. Thank you. :) — geekyjazzy, Aug 06 '14 at 15:15

score 11 · Accepted Answer · answered Aug 06 '14 at 14:56

11

I think you need this:

IFS=' |\\' read -ra array_1 <<< "$var_result"

answered Aug 06 '14 at 14:56

Mark Setchell

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Can someone please give answer for https://stackoverflow.com/questions/46556202/split-string-using-r-n-in-bash ? – Jitesh Sojitra Oct 04 '17 at 02:43

score 0 · Answer 2 · answered Aug 06 '14 at 14:58

This should do it

#!/bin/bash
var_result="Pass results_ADV__001__FUNC__IND\ADV__001__FUNC__IND_08_06_14_10_04_34.tslog"
field1=$(echo "$var_result" | awk -F ' ' '{print $(NF-1)}');
field2=$(echo "$var_result" | awk -F  \\ '{print $(NF-1)}');
field1=$(echo "$field1" | awk -F ' ' '{print $(NF-1)}');
field3=$(echo "$var_result" | awk -F  \\ '{print $2}');
echo $field1;
echo $field2;
echo $field3;

Split string using 2 different delimiters in Bash

2 Answers2