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Grep from match to the end of file

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# 1  
Old 12-05-2016
Grep from match to the end of file
Hi

i have this script :

Code:
#!/bin/bash
DATE=$(date '+%Y-%m-%d %H:%M' -d  "1 hour ago")

PASS=$(grep -A99999999999  '$DATE'  /var/log/asterisk/full| grep -i 'Wrong password')

it a script that ment to go over log file a hour back from now until
the end of file.

right now im using -A999 - do you have more prudactive way ?
when i searched all lead to use awk but as im looking for 1 hour backwords its becoming complicated.

ideas ?
Last edited by Scrutinizer; 12-05-2016 at 11:59 AM.. Reason: icode tags => code tags
# 2  
Old 12-05-2016
If you can get the line number from the file in some way, you can display the file from that point on with (amongst other things) a sed

If you were to do something like this, would it help?:-
Code:
last_line=$(grep -n "Your search text here" input_file | tail -1 | cut -f 1 -d ":")

((last_line=$last_line-1))               # To include current line in the next step

sed -n "$last_line,\$"p input_file



I hope that this is useful,
Robin
# 3  
Old 12-05-2016
Any string search to match $DATE will work only if the exact minute is found in the log file. E.g.
Code:
sed -n "/$DATE/,$ p" file

If that fails, you'd need e.g. awk to do a "larger than" comparison. Try (untested)
Code:
awk -v DT="$DATE" '($1 " " $2) >= DT' file

- given the time stamp is in $1 and $2, separated by a space.
# 4  
Old 12-06-2016
thanks for your answer!

rbatte1:

Code:
last_line=$(grep -n "Your search text here" input_file | tail -1 | cut -f 1 -d ":")

((last_line=$last_line-1))               # To include current line in the next step

sed -n "$last_line,\$"p input_file

what it did is to give me the line in the log which it have found my word:
Code:
last_line=$(grep -n "Wrong password" /var/log/asterisk/full-20161205 | tail -1 | cut -f 1 -d ":")

gave me : 37171

then i did
Code:
((last_line=$last_line-1))

whichgave me this :
37171-1 as a varible

and when i ran
Code:
sed -n "$last_line,\$"p  /var/log/asterisk/full-20161205

i got :
Code:
sed: -e expression #1, char 6: unknown command: `-'

------------------------------------------------------------------------------

RudiC:

thanks ! it did the trick :

Code:
awk -v DT="$DATE" '($1 " " $2) >= DT'   /var/log/asterisk/full-20161205 | grep "password"

finds from hour back to the end of file! Smilie
Moderator's Comments:
Mod Comment Please use CODE tags (not ICODE tags) when displaying full line and multi-line sample input, sample output, and code segments.
Last edited by Don Cragun; 12-06-2016 at 02:54 AM.. Reason: Change most ICODE tags to CODE tags.
# 5  
Old 12-06-2016
Why use grep? Try
Code:
awk -v DT="$DATE" '/password/ && ($1 " " $2) >= DT' /var/log/asterisk/full-20161205

# 6  
Old 12-06-2016
Quote:
Originally Posted by RudiC
Why use grep? Try
Code:
awk -v DT="$DATE" '/password/ && ($1 " " $2) >= DT' /var/log/asterisk/full-20161205

thanks!
another thing that was missing for me was "[" befor the DATE variable

Code:
awk -v DT="[$DATE" '/password/ && ($1 " " $2) >= DT' /var/log/asterisk/full-20161205


THANKS
Last edited by rbatte1; 12-06-2016 at 06:24 AM.. Reason: Changed ICODE tags to CODE tags
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