Need to ignore grep if the line starts with ; or # for a specific string in a file. file.ini contains below line
output_partition_key=FILE_CREATED_DATE
doing a grep as below returns values FILE_CREATED_DATE
grep -w "output_partition_key" file.ini | cut -d= -f2
but if say the line starts with ; or # then it should not grep anything
;output_partition_key=FILE_CREATED_DATE
I tried solutions from other posts but its not working.Can anyone tell me how to achieve the expected result
It seems like what you really want is to find lines that start with output_partition_key=. The simplest way to do that is:
grep ^output_partition_key= file.ini | cut -d= -f2
(where ^ means "the start of a line").
Related
I am trying to find which words happened in logfiles plus show the logfilename for anything that matches following pattern:
'BA10\|BA20\|BA21\|BA30\|BA31\|BA00'
so if file dummylogfile.log contains BA10002 I would like to get a result such as:
dummylogfile.log:BA10002
it is totally fine if the logfile shows up twice for duplicate matches.
the closest I got is:
for f in $(find . -name '*.err' -exec grep -l 'BA10\|BA20\|BA21\|BA30\|BA31\|BA00' {} \+);do printf $f;printf ':';grep -o 'BA10\|BA20\|BA21\|BA30\|BA31\|BA00' $f;done
but this gives things like:
./register-05-14-11-53-59_24154.err:BA10
BA10
./register_mdw_files_2020-05-14-11-54-32_24429.err:BA10
BA10
./process_tables.2020-05-18-11-18-09_11428.err:BA30
./status_load_2020-05-18-11-35-31_9185.err:BA30
so,
1) there are empty lines with only the second match and
2) the full match (e.g., BA10004) is not shown.
thanks for the help
There are a couple of options you can pass to grep:
-H: This will report the filename and the match
-o: only show the match, not the full line
-w: The match must represent a full word (string build from [A-Za-z0-9_])
If we look at your regex, you use BA01, this will match only BA01 which can appear anywhere in the text, also mid word. If you want the regex to match a full word, it should read BA01[[:alnum:]_]* which adds any sequence of word-constituent characters (equivalent to [A-Za-z0-9_]). You can test this with
$ echo "foo BA01234 barBA012" | grep -Ho "BA01"
(standard input):BA01
(standard input):BA01
$ echo "foo BA01234 barBA012" | grep -How "BA01"
$ echo "foo BA01234 barBA012" | grep -How "BA01[[:alnum:]_]*"
(standard input):BA01234
So your grep should look like
grep -How "\('BA10\|BA20\|BA21\|BA30\|BA31\|BA00'\)[[:alnum:]_]*" *.err
From your example it seems that all files are in one directory. So the following works right away:
grep -l 'BA10\|BA20\|BA21\|BA30\|BA31\|BA00' *.err
If the files are in different directories:
find . -name '*.err' -print | xargs -I {} grep 'BA10\|BA20\|BA21\|BA30\|BA31\|BA00' {} /dev/null
Explanation: the addition of /dev/null to the filename {} forces grep to report the matching filename
I am new to linux and I am experimenting with basic terminal commands. I found out that I can list all users using compgen -u but what if I only want to display the bottom line outputs ?
Ok lets say the output of compgen -u goes like this:
extra
extra
extra
extra
extra
extra
extra
extra
extra
John
William
Kate
Harold
I can only use grep to find a single text (ex. compgen -u | grep John). But what if I want to use grep to display John as well as all the remaining entries after it ?
sed or awk solution would be easier, but if you can only use grep, then the option --after-context (or -A) might do:
grep -A 5 John file
The drawback is that you need to know the number of lines to display after the matching (or use an arbitrary big number for the rest of the file).
compgen -u | grep -A$(compgen -u| wc -l) John
Explanation:
From man grep
-A NUM, --after-context=NUM
Print NUM lines of trailing context after matching lines. Places a line containing a group separator (described under --group-separator) between
contiguous groups of matches.
grep -A -- print number of rows after pattern
$() -- Execute unix command
compgen -u| wc -l --> Get total number of rows of output of command.
You can use the following one-liner :
n=$( compgen -u | grep -n John | head -1 | cut -d ":" -f 1 ) && compgen -u | tail -n +$n
This finds out the line number for first occurrence of John, and prints everything starting that line.
So I have a bash script which is below, and whenever I try to execute it, it says only one type of list may be specified, any clue as to whats wrong with this code? I interpret this as we use grep to find the word (-word) of the first character ($1) typed by the user in the file femalenames.txt and then using the cut command we print the 2nd field of characters 16-20? sort of confused there.
#!/bin/bash
grep -w $1 femalenames.txt | cut -f2 -c16-20
Just pipe it ( | ) again & it'll work.
Command guide here
#!/bin/bash
grep -w $1 femalenames.txt | cut -f2 | cut -c16-20
I have a bunch of strings that I have to fetch the 'port_num' from -
"76 : client=new; tags=circ, LINK; port_num=switch01; far_port=Gi1/0"
The word might be in a different place in the string and it might be a different length, but it always says 'port_num=' before it and ';' after it...
I only want this bit- 'switch01'
Currently I use-
| grep -Eo 'port_num=.+' | cut -d"=" -f2 | cut -d";" -f1'
But there has got to be a better way
You can try grep -oP '(?<=port_num=).+(?=;)', if you run this:
echo "76 : client=new; tags=circ, LINK; port_num=switch01; far_port=Gi1/0" \
| grep -oP '(?<=port_num=).+(?=;)'
result will be:
switch01
Updated answer: grep -oP '(?<=port_num=)[^;]+(?=;)'
This is what I would use:
... | grep -E 'port_num=.+' | sed 's/^.*port_num=\([^;]*\).*$/\1/'
This works with or without the -o on grep, and the availability of -P will depend on the version of grep you have. (e.g., my grep does not have it). I'm not saying the other answers that rely on -P aren't any good -- they look fine to me. But grep -P will be less portable.
IMHO, piping grep with sed allows each utility to do what it specializes in -- grep is for selecting lines, sed is for modifying lines.
This can be done in a simple sed command:
s="76 : client=new; tags=circ, LINK; port_num=switch01; far_port=Gi1/0"
sed 's/.*port_num=\([^;]*\);.*/\1/' <<< "$s"
switch01
... | grep -Po 'port_num.+(?=;)'
This uses grep's Perl Compatible Regular Expression (PCRE) syntax. The (?=;) is a look-ahead assertion which looks for a match with ";" but doesn't include it in the matched output.
This produces:
port_num=switch01
As #Vladimir Kovpak noted, if you want to exclude the "port_num=" string from this output, add a look-behind assertion:
... | grep -Po '(?<=port_num).+(?=;)'
I have a file that possibly contains bad formatting (in this case, the occurrence of the pattern \\backslash). I would like to use grep to return only the line numbers where this occurs (as in, the match was here, go to line # x and fix it).
However, there doesn't seem to be a way to print the line number (grep -n) and not the match or line itself.
I can use another regex to extract the line numbers, but I want to make sure grep cannot do it by itself. grep -no comes closest, I think, but still displays the match.
try:
grep -n "text to find" file.ext | cut -f1 -d:
If you're open to using AWK:
awk '/textstring/ {print FNR}' textfile
In this case, FNR is the line number. AWK is a great tool when you're looking at grep|cut, or any time you're looking to take grep output and manipulate it.
All of these answers require grep to generate the entire matching lines, then pipe it to another program. If your lines are very long, it might be more efficient to use just sed to output the line numbers:
sed -n '/pattern/=' filename
Bash version
lineno=$(grep -n "pattern" filename)
lineno=${lineno%%:*}
I recommend the answers with sed and awk for just getting the line number, rather than using grep to get the entire matching line and then removing that from the output with cut or another tool. For completeness, you can also use Perl:
perl -nE 'say $. if /pattern/' filename
or Ruby:
ruby -ne 'puts $. if /pattern/' filename
using only grep:
grep -n "text to find" file.ext | grep -Po '^[^:]+'
You're going to want the second field after the colon, not the first.
grep -n "text to find" file.txt | cut -f2 -d:
To count the number of lines matched the pattern:
grep -n "Pattern" in_file.ext | wc -l
To extract matched pattern
sed -n '/pattern/p' file.est
To display line numbers on which pattern was matched
grep -n "pattern" file.ext | cut -f1 -d: