How to cut the string from bash shell

Question

From linux Bash can you help me on how to show only the

Monitoringmetric=1.5Count

instead of

Monitoringmetric=1.5Count;1;2

You can use bash for this: `s='Monitoringmetric=1.5Count;1;2'; echo "${s%%;*}"` — anubhava, Oct 04 '16 at 17:01
Thanks Jean and anubhva, both suggestions works like charm can I ask one more on how can I only cut and show only = 1.5 — Apps Tester, Oct 04 '16 at 17:10
Please don't turn the comment section into a Q-and-A session. If you have an answer, post it as an answer. @AppsTester If you have a follow-up question, post a new question. — chepner, Oct 04 '16 at 17:34
FInally showed it but seems to be a long command, please advise if there is a way I can shorten this $ echo "Monitoringmetric=1.5Count;1;2" | cut -f2 |sed -e 's/Monitoringmetric=//g' | cut -f1 -d ";" | sed -e 's/Count//g' — Apps Tester, Oct 04 '16 at 17:50

score 0 · Answer 1 · answered Oct 04 '16 at 17:51

0

FInally showed it but seems to be a long command, please advise if there is a way I can shorten this

$ echo "Monitoringmetric=1.5Count;1;2" | cut -f2 |sed -e 's/Monitoringmetric=//g' | cut -f1 -d ";" | sed -e 's/Count//g'

answered Oct 04 '16 at 17:51

Apps Tester

Why are you doing the **cut -f2** that doesn't seem to accomplish anything? What are you trying to extrace from the input? Just **1.5** ? – blackpen Oct 04 '16 at 18:38

score 0 · Accepted Answer · answered Oct 04 '16 at 18:45

0

Since you seem to be interested only capturing the number after the Monitoringmetric=, you can do:

echo "Monitoringmetric=1.5Count;1;2" | sed -n -e 's/^Monitoringmetric=\([0-9.]*\).*/\1/p'

answered Oct 04 '16 at 18:45

blackpen

score 0 · Answer 3 · answered Oct 04 '16 at 19:25

0

If your string is in a variable, you could also use ${a%%;*}, for example:

a="Monitoringmetric=1.5Count;1;2"
echo ${a%%;*}

answered Oct 04 '16 at 19:25

Ljm Dullaart

3 Answers3