I've looked around StackExchange sites but I haven't found anything that's quite what I'm looking for. Here are two use cases of grep:
Printing items before/after a match
Print a certain match
I'm trying to parse a log file, and I want to return the last error in the log which is, predictably, at the end of the file. However, sometimes the errors are multiple lines. The answers for 'how to grep the last match' all involve either tail or head, and only work with a single line.
In my case, I want to simply return everything in the file, starting with the last match. Typically, this won't be any more than 10-15 lines maximum, so a grep -A 15 does the trick there. But, I still need to only get the last one of these, so that alone doesn't produce the right output.
The naive approach is to use a two-part match, to first get what the last match is and then everything after that. This won't work for me, because I can't guarantee that the last match is unique.
Is it possible to do this with grep somehow, or would there be better tools for this?
There is a way to get sed to do this but I can't remember.
If you are open to using a combination of commands here is something that might work:
# Get the line number of teh last match
LNO=$( grep -n 'the error' the_file | tail -1 | cut -d":" -f1 )
# Now use sed to print all lines from that point:
sed -n "$LNO,\$p" the_file
I think there's an exact duplicate somewhere, but I found only these close ones:
How to get lines from the last match to the end of file?
grep last match and it's following lines
Here's one way to do it:
$ cat ip.txt
foo123
error 1
xyz
error 2
99999
88888
$ tac ip.txt | sed '/error/q' | tac
error 2
99999
88888
Related
I've search the web for this but couldn't find the solution I wanted.
I used grep to search for lines having the word one and that works fine.
I want grep to also count the matched lines as well. But I only get the number only.
I tried this but no cigar.
grep -cn 'one' foo
2
Option c is to count and option n is to show lines. But it still shows the count only.
Thus far, this is the only way to do it. Running grep twice :(
grep 'one' foo ; grep 'one' foo | wc -l
one two three
seven one blah
2
Is it possible for grep to show both the matches and the count without repeating the command again?
BTW, I know awk can do this as I've seen in my research. But I want to know if grep can do this. Thanks in advance :)
I have a need to perform multiple grep matches as part of the same grep command. When I run them individually, they work fine. But not when together. I hope someone could either show me a solution or perhaps can help me find a work-around. Here is sample stream:
(string start..) RollUp:"V" Enzyme:"ENZA ENZB ENZD ENZE" (..string end)
In the first command I am needing to isolate all RollUp substrings.Value is always A or V:
grep -o "RollUp:\"[AV]\""
In the second command I am needing to isolate all combinations of Enzyme values (1-20 total, spaces in between, don't know values names). This command works:
grep -oE 'Enzyme:[[:space:]]*"[^"]+"'
However, I need to match both patterns as part of same stream. When I try:
grep -oE "RollUp:\"[AV]\""\|Enzyme:[[:space:]]*"[^"]+""
, nothing is returned. I would be grateful for any ideas for getting this double grep pattern match to work. Thank you!
regex someting[^"]+ : this means string something followed by anything till next " is seen. Here + sign means , at least one or more match.
grep -oE 'RollUp:"[^"]+|Enzyme:[[:space:]]*"[^"]+"' file
I have literally been at this for 5 hours, I have busybox on my device, and I unfortunately do not have -X in grep to make my life easier.
edit;
I have two list both of them have mac addresses, essentially I am just wanting to achieve offline mac address lookup so I don't have to keep looking it up online
list.txt has vendor mac prefix of course this isn't the complete list but just for an example
00:13:46
00:15:E9
00:17:9A
00:19:5B
00:1B:11
00:1C:F0
scan will have list of different mac addresses unknown to which vendor they go to. Which will be full length mac addresses. when ever there is a match I want the line in scan to be output.
Pretty much it does that, but it outputs everything from the scan file, and then it will output matching one at the end, and causing duplicate. I tried sort -u, but it has no effect its as if there is two different output from two different methods, the reason why I say that is because it will instantly output scan file that has everything in it, and couple seconds later it will output the matching one.
From searching I came across this
#!/bin/bash
while read line; do
grep -F 'list' 'scan'
done < list.txt
which displays the duplicate result when/if found, the output is pretty much echoing my scan file then displaying the matched pattern, this creating duplicate
This is frustrating me that I have not found a solution after click on all the links in google up to page 9.
Please someone help me.
I don't know if the Busybox sed supports this out of the box, but it should be easy to do in Awk or Perl instead then.
Create a sed script to print lines from file2 which are covered by a prefix in file1 by transforming each line in file1 into a sed command to print a match for that regular expression:
sed 's%.*%/&/p%' file1 | sed -n -f - file2
The same in Awk:
awk 'NR==FNR { a[++i]="^" $0; next }
{ for (j=1; j<=i; ++j) if ($0 ~ a[j]) print }' file1 file2
Ok guys I did a nested for loop (probably very in efficient) but I got it working printing the matching mac addresses using this
#!/usr/bin/bash
for scanlist in `cat scan | cut -d: -f1,2,3`
do
for listt in `cat list`
do
if [[ $scanlist == $listt ]]; then
grep $scanlist scan
fi
done
done
if anyone can make this more elegant but it works for me for now. I think the problem I had was one list contained just 00:11:22 while my other list contained 00:11:22:33:44:55 that is why I cut it on my scanlist to make same length as my other list. So this only output the matches instead of doing duplicate output.
grep (GNU grep) 2.14
Hello,
I have a log file that I want to filter on a selected word. However, it tends to filter on many for example.
tail -f gateway-* | grep "P_SIP:N_iptB1T1"
This will also find words like this:
"P_SIP:N_iptB1T10"
"P_SIP:N_iptB1T11"
"P_SIP:N_iptB1T12"
etc
However, I don't want to display anything after the 1. grep is picking up 11, 12, 13, etc.
Many thanks for any suggestions,
You can restrict the word to end at 1:
tail -f gateway-* | grep "P_SIP:N_iptB1T1\>"
This will work assuming that you have a matching case which is only "P_SIP:N_iptB1T1".
But if you want to extract from P_SIP:N_iptB1T1x, and display only once, then you need to restrict to show only first match.
grep -o "P_SIP:N_iptB1T1"
-o, --only-matching show only the part of a line matching PATTERN
More info
At least two approaches can be tried:
grep -w pattern matches for full words. Seems to work for this case too, even though the pattern has punctuation.
grep pattern -m 1 to restrict the output to first match. (Also doable with grep xxx | head -1)
If the lines contains the quotes as in your example, just use the -E option in grep and match the closing quote with \". For example:
grep -E "P_SIP:N_iptB1T1\"" file
If these quotes aren't in the text file, and there's blank spaces or endlines after the word, you can match these too:
# The word is followed by one or more blanks
grep -E "P_SIP:N_iptB1T1\s+" file
# Match lines ending with the interesting word
grep -E "P_SIP:N_iptB1T1$" file
I need to find some matching conditions from a file and recursively find the next conditions in previously matched files , i have something like this
input.txt
123
22
33
The files where you need to find above terms in following files, the challenge is if 123 is found in say 10 files , the 22 should be searched in these 10 files only and so on...
Example of files are like f1,f2,f3,f4.....f1200
so it is like i need to grep -w "123" f* | grep -w "123" | .....
its not possible to list them manually so any easier way?
You can solve this using awk script, i ve encountered a similar problem and this will work fine
awk '{ if(!NR){printf("grep -w %d f*|",$1)} else {printf("grep -w %d f*",$1)} }' input.txt | sh
What it Does?
it reads input.txt line by line
until it is at last record , it prints grep -w %d | (note there is a
pipe here)
which is then sent to shell for execution and results are piped back
to back
and when you reach the end the pipe is avoided
Perhaps taking a meta-programming viewpoint would help. Have grep output a series of grep commands. Or write a little PERL program. Maybe Ruby, if the mood suits.
You can use grep -lw to write the list of file names that matched (note that it will stop after finding the first match).
You capture the list of file names and use that for the next iteration in a loop.