awk / sed to print only up to the underscore character

Question

How can I use awk or sed to print a string only up to the first underscore character?

Before:

host100_044 2
host101_045 2
host102_046 2

After:

host100
host101
host102

Answer 1

This can be done with cut:

cut -d _ -f 1

e.g.

$ echo host100_044 2 | cut -d _ -f 1
host100

Also with awk can be done with awk -F_ '{print $1}' (probably there is a cleaner way of doing that)

Answer 2

Another alternative for sed:

echo 'host100_044 2' | sed 's/^\(.*\)_.*$/\1/'

If you have these in a file, you could call it as follows;

cat fileName | sed 's/^\(.*\)_.*$/\1/'

Answer 3

$ echo "host100_044 2" | awk -F'_' '{print $1}'
host100

The -F'_' instructs awk to use underscores as field delimiters. The awk script itself just prints the first field.

Answer 4

echo host100_044 2 host101_045 2 host102_046 2| sed 's/_/ /g' | awk 'BEGIN { RS="host"} {printf("host%s ", $1)}'  | cut -d ' ' -f2-

Output:

host100 host101 host102

With newlines:

echo host100_044 2 host101_045 2 host102_046 2| sed 's/_/ /g' | awk 'BEGIN { RS="host"} $1 ~ /[0-9]/ {print "host"$1}'

Output:

host100
host101
host102

Answer 5

Using sed:

echo host100_044 2 | sed 's;_.*;;'

Using sed in place edit option,

sed -i.old 's;_.*;;' infile

Answer 6

This will print the output before _ :

sed 's/_.*//' -filename