how to find -l or --log-file string in a file and store absolute path passed with that as a variable?

Question

how to find -l or --log-file string in a file and store absolute path passed with that as a variable ?

Sample file 1:-

$ cat server.log
server_options='-l /tmp/server_log'

Sample file 2:-

$ cat server.log
server_options='--log-file=/tmp/server1_log'

Requirement

if -l or --log-file is found in server_options variable then find the absolute path of the file and store it in a variable

Note:- 1. Don't consider lines starting with "#" 2. if "-l" option is specified between the -l and absolute path there will be a space and if "--log-file" is specified between the "--log-file" and absolute path there will be a equal to sign "="

for example

server_options='-l /tmp/server_log'
server_options='--log-file=/tmp/server1_log'

Desired output of the sample 1 file stored in a variable "LOG_FILE"

$ echo $LOG_FILE
/tmp/server_log

Desired output of the sample 2 file stored in a variable "LOG_FILE"

$ echo $LOG_FILE
/tmp/server1_log

Tried:-

$ grep "^[^#;]" server.log | awk '/\-\l|/{ print $2 }'
/tmp/server_log'

Need more reliable way to achieve this

Update the question that there could be more flags following `-l /tmp/server_log` in that variable — Inian, Nov 01 '17 at 12:15
With GNU grep: `LOG_FILE=$(grep -Po '(-l|--log-file) \K.*(?=.)' file)` — Cyrus, Nov 01 '17 at 12:19
See: [The Stack Overflow Regular Expressions FAQ](http://stackoverflow.com/a/22944075/3776858) — Cyrus, Nov 01 '17 at 12:19
@Cyrus: You should post it as answer, and add `^[^#].*` before, as OP wants to ignore lines starting with `#` — Inian, Nov 01 '17 at 12:21
@Cyrus how to ignore lines that starts with "#" with the one-liner you posted — asokan, Nov 01 '17 at 12:45

score 1 · Answer 1 · answered Nov 01 '17 at 12:15

Other than awk, there is also a cut command which can have similar result. For example:

grep "-l|--log-file" server.log | cut -d'\'' -f1 | cut -d' ' -f1

cut -d'\'' -f1 = cut by ' character, choose 2nd string

cut -d' ' -f1 = cut by "space" character, choose 2nd string

asokan · Answer 2 · 2017-11-01T12:55:57.023

0

grep "^[^#;]" server.log |grep -Po '(-l|--log-file) \K.*(?=.)'

edited Nov 01 '17 at 12:55

answered Nov 01 '17 at 12:37

asokan

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You could just combine it as `grep -Po '^[^#;].*(-l|--log-file) \K.*(?=.)' /etc/ddn/ime/ime-server.conf` – Inian Nov 01 '17 at 12:48
if "-l" option is specified between the -l and absolute path there will be a space and if "--log-file" is specified between the "--log-file" and absolute path there will be a equal to sign "=" – asokan Nov 02 '17 at 07:41
server_options='-l /tmp/server_log' server_options='--log-file=/tmp/server1_log' – asokan Nov 02 '17 at 07:43
and this one liner doesn't work with "--log-file" option – asokan Nov 02 '17 at 07:43

score 0 · Answer 3 · answered Nov 01 '17 at 15:29

0

this is a bit long shot, but understandable

egrep '\-l|\-\-log\-file' server.log | grep -v ^# | gawk -F"'" '{print $2}' | cut -d" " -f2

answered Nov 01 '17 at 15:29

Potato

3,320
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how to find -l or --log-file string in a file and store absolute path passed with that as a variable?

3 Answers3