Grep string and put in csv file Unix

Question

I have gz file that have some data now i want to grep two diffrent pattern and put that data into a csv file .For the same i want to write a shell script how can we do this please help me in this.

Below are the two command with i want to grep the data line by line and then put into a csv file .

Commands :

zgrep "Time"  file.txt.gz
zgrep "requests" file.txt.gz

Please suggest how could i use these command in shell and get the data in a csv file

This is the output i am getting after doing :

zgrep -E 'Time|requests' file.txt.gz

 Time 27-Apr-2016 07:24:15 CDT,
 requests currently being processed, 1 
  Time 27-Apr-2016 07:24:15 CDT,
 requests currently being processed, 2 ,

I want the ouput like Time 27-Apr-2016 07:24:15 CDT | requests currently being processed, 1

Is file.txt.gz a single file or you need to recursively grep inside that directory? — Vishal R
– Vishal R, Commented Apr 28, 2016 at 8:01
@anubhava this is working but i want both the output as pipe seprated like TimeOutput | RequestOut\ — Developer
– Developer, Commented Apr 28, 2016 at 8:06
I suggest you show your expected output as there can be multiple matches of each word in the file. — anubhava
– anubhava, Commented Apr 28, 2016 at 8:10

anubhava · Accepted Answer · 2016-04-28 08:39:15Z

3

You can use awk with gzat:

gzcat file.txt.gz | awk '/Time/{p=$0} /requests/{print p, "|", $0}'

answered Apr 28, 2016 at 8:39

anubhava

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3 Comments

anubhava Over a year ago

Hmm in that case you will need to gunzip the file first using gunzip

J. Chomel Over a year ago

is it gzat or gzcat?

anubhava Over a year ago

@J.Chomel: gcat is gnu utility but on OSX we get gzcat

Andreas Louv · Accepted Answer · 2016-04-28 08:36:10Z

1

Use awk to format it output from grep:

zgrep -E ... | awk 'NR%2==0{print l, "|", $0}{l=$0}'
 Time 27-Apr-2016 07:24:15 CDT, |  requests currently being processed, 1 
  Time 27-Apr-2016 07:24:15 CDT, |  requests currently being processed, 2 ,

answered Apr 28, 2016 at 8:36

Andreas Louv

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Developer Over a year ago

where is the file name ??

Yuri Lachin · Accepted Answer · 2016-04-28 08:48:28Z

0

I assume, that 'Time ...' and ;requests ...' in grep output are on one line (not wrapped as in your example).

zgrep -E 'Time|requests' file.txt.gz | sed -r -e 's/\s*,\s*/ | /' > file.csv

To remove trailing comma as in 'being processed, 2 ,' use this variant

zgrep -E 'Time|requests' file.txt.gz | sed -r -e 's/\s*,\s*/ | /; s/\s*,\s*$//'

Alternative (pure sed without grep):

gzip -dc file.txt.gz | sed -r -e '/Time|requests/!d; s/\s*,\s*/ | /; s/\s*,\s*$//'

answered Apr 28, 2016 at 8:48

Yuri Lachin

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J. Chomel · Accepted Answer · 2016-04-28 09:27:45Z

0

You could also use paste to do so:

zgrep -E 'Time'     file.txt.gz >f1
zgrep -E 'requests' file.txt.gz >f2

paste f1 f2

answered Apr 28, 2016 at 9:27

J. Chomel

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