Extract value of column from a line (variable)

Question

Okay, so I have a variable ($line) that is defined in the bash/shell script as

$line = "abc:123:def:345"

need to get this column2 = "123"

How do I extract the value of the 2nd column i.e. "123" and name it as a different variable which can be summed later on? I know you have to separate it based on the delimiter ':' but I don't know how to transfer to different variable whilst taking input from $line variable. I only ask this because for some weird reason my code reads the first line of text file BUT doesn't perform the awk on just the first line only so hence the sum is wrong.

FILE=$1
while read line
do
 awk -F: '{summation += $3;}END{print summation;}'
done < $FILE

-code via shell script

Thanks.

Léa Gris · Answer

Using plain POSIX features:
#!/usr/bin/env sh

line='abc:123:def:345'

IFS=:
# Split line on IFS into arguments
set -- $line

printf %s
 "$2"

Alternate method:
#!/usr/bin/env sh

line='abc:123:def:345'

# Strip out first column
a="${line#*:}"

# Strip out remaining columns except 1st
b="${a%%:*}"

printf %s
 "$b"

user4453924 · Answer

way in bash using expr

Line2=$(expr $line : "[^:]*:([^:]*)")

or if the fields are always 3 characters

Line2=${line:4:3}

Jotne · Answer

You can use awk to get second field:

line="abc:123:def:345"

awk -F: '{print $2}' <<< "$line"
123

glenn jackman · Answer

To assign a variable in the shell, no $ on the left-hand side, no spaces around the =, and < and > are not valid quote characters line="abc:123:def:345" In bash, you would do this: IFS=: read -ra fields <<< "$line" temporarily set IFS to a colon use the $line variable as input to the read command (a here-string) and read the values into the fields array. Bash arrays are indexed starting from zero, so to extract the 2nd field: echo "${fields[1]}" # => 123

Extract value of column from a line (variable)

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