Split a string in bash based on delimiter

Question

I have a file log_file which has contents such as

CCO O-MR1 Sync:No:3:No:346:Yes
CCO P Sync:No:1:No:106:Yes
CCO P Checkout:Yes:1:No:10:No
CCO O-MR1 Checkout(2.2):Yes:1:No:10:No

I am trying to obtain the 4 fields based on ":" delimiter The script that I have is

#!/bin/bash
log_file=$1

for i in `cat $log_file` ; do
        echo $i
        field_a=`echo $i | awk -F '[:]' '{print $1}'`
        echo $field_a
        field_b=`echo $i | awk -F '[:]' '{print $2}'`
        echo $lfield_b
        ...
done

but the value that this code gives for field_a is wrong, it splits the line based on " " delimiter. echo $i also prints wrong value.

What else can I use to correct this?

See DontReadLinesWithFor. And as an aside, see also BashPitfalls #14 -- echo $foo should be echo "$foo". shellcheck.net is also a useful and relevant resource. — Charles Duffy
– Charles Duffy, Commented Aug 23, 2018 at 16:22

2 revs, 2 users 92% · Accepted Answer · 2018-08-23 16:40:30Z

5

This is covered in detail in BashFAQ #1. To summarize, use a while read loop with IFS set to contain (only) the characters that should be used to split fields.

while IFS=: read -r field_a field_b other_fields; do
  echo "field_a is $field_a"
  echo "field_b is $field_b"
  echo "Remaining fields are $other_fields"
done <"$log_file"

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