extract the value from the log using shell script

Question

Hi I am new to shell script. I am trying to extract the specific value from the log.

When I filter the data by using specific keyword. It looks like these.

cat hive-server2.log | grep user

The output of the data is

2018-01-18T16:20:39,464 WARN  [67272380-f3e9-40da-8e8e-a209c05eb4fe HiveServer2-Handler-Pool: Thread-37([])]: util.CurrentUserGroupInformation (CurrentUserGroupInformation.java:getGroupNameFromUser(52)) - user a8197zz (auth:PROXY) via hive (auth:SIMPLE) has no primary groupName, setting groupName to be a8197zz.

In the above data I want to extract the specific value for the user like this.

a8197zz

I tried like this.

awk 'BEGIN{ print "User" }
 /\<user\>/{ u=$10 }
 //{ print u }' OFS=',' hive-server2.log

It prints blank lines only. Any help will be appreciated.

Cyrus · Accepted Answer · 2018-01-20 07:53:22Z

2

With GNU grep:

grep -Po '[^ ]*(?=\.$)' file

or

grep -Po 'user \K[^ ]*' file

With awk:

awk -F "[. ]" '{print $(NF-1)}' file

or

awk -F "user " '{split($2,array," "); print array[1]}' file

or search for field with string user and print next field:

awk '{for(i=1; i<=NF; i++) if ($i=="user") print $(i+1) }' file

Output:

a8197zz

edited Jan 20, 2018 at 7:53

answered Jan 20, 2018 at 7:29

Cyrus

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Kalanidhi · Accepted Answer · 2018-01-20 07:40:54Z

1

Try this sed command also

sed 's/.*user \([^ ]\+\).*/\1/' fileName

Output:

a8197zz

answered Jan 20, 2018 at 7:40

Kalanidhi

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2 Comments

daniu Over a year ago

I'd suggest sed -n '/user/s/.*user \([^ ]\+\).*/\1/p' fileName to avoid getting spammed with the lines not containing "user".

Kalanidhi Over a year ago

@daniu, Thanks for your suggestion

RavinderSingh13 · Accepted Answer · 2018-01-20 07:35:39Z

0

Following awk may help you in same.

awk '{sub(/\./,"",$NF);print $NF}'  Input_file

Output will be as follows.

a8197zz

answered Jan 20, 2018 at 7:35

RavinderSingh13

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