$JBOSS_HOME=/home/user/app -----line from file.
I want to grep only /home/user/app as the output from the file.
I tried but it not working
With GNU grep:
grep -Po -m 1 '\$JBOSS_HOME=\K(.+)' file
outputs
/home/user/app
-m 1 - stops at the first matching line
-P - use Perl compatible regular expression
-o - display only the matched part
\K reset the regular expression match to start after =
Alternatively you can try:
grep 'JBOSS_HOME' $YOUR_FILE | cut -d'=' -f2
Related
I am just simply trying to grab the commit ID, but not quite sure what I'm missing:
➜ ~ curl https://github.com/microsoft/vscode/releases -s | grep -oE 'microsoft/vscode/commit/(.*?)/hovercard'
microsoft/vscode/commit/ccbaa2d27e38e5afa3e5c21c1c7bef4657064247/hovercard
The only thing I need back from this is ccbaa2d27e38e5afa3e5c21c1c7bef4657064247.
This works just fine on regex101.com and in ruby/python. What am I missing?
If supported, you can use grep -oP
echo "microsoft/vscode/commit/ccbaa2d27e38e5afa3e5c21c1c7bef4657064247/hovercard" | grep -oP "microsoft/vscode/commit/\K.*?(?=/hovercard)"
Output
ccbaa2d27e38e5afa3e5c21c1c7bef4657064247
Another option is to use sed with a capture group
echo "microsoft/vscode/commit/ccbaa2d27e38e5afa3e5c21c1c7bef4657064247/hovercard" | sed -E 's/microsoft\/vscode\/commit\/([^\/]+)\/hovercard/\1/'
Output
ccbaa2d27e38e5afa3e5c21c1c7bef4657064247
The point is that grep does not support extracting capturing group submatches. If you install pcregrep you could do that with
curl https://github.com/microsoft/vscode/releases -s | \
pcregrep -o1 'microsoft/vscode/commit/(.*?)/hovercard' | head -1
The | head -1 part is to fetch the first occurrence only.
I would suggest using awk here:
awk 'match($0,/microsoft\/vscode\/commit\/[^\/]*\/hovercard/){print substr($0,RSTART+24,RLENGTH-34);exit}'
The regex will match a line containing
microsoft\/vscode\/commit\/ - microsoft/vscode/commit/ fixed string
[^\/]* - zero or more chars other than /
\/hovercard - a /hovercard string.
The substr($0,RSTART+24,RLENGTH-34) will print the part of the line starting at the RSTART+24 (24 is the length of microsoft/vscode/commit/) index and the RLENGTH is the length of microsoft/vscode/commit/ + the length of the /hovercard.
The exit command will fetch you the first occurrence. Remove it if you need all occurrences.
You can use sed:
curl -s https://github.com/microsoft/vscode/releases |
sed -En 's=.*microsoft/vscode/commit/([^/]+)/hovercard.*=\1=p' |
head -n 1
head -n 1 is to print the first match (there are 10)grep -o will print (only) everything that matches, including microsoft/ etc.
Your task can not be achieved with Mac's grep. grep -o prints all matching text (compared to default behaviour of printing matching lines), including microsoft/ etc. A grep which implemented perl regex (like GNU grep on Linux) could make use of look ahead/behind (grep -Po '(?<=microsoft/vscode/commit/)[^/]+(?=/hovercard)'). But it's just not available on Mac's grep.
On MacOS you don't have gnu utilities available by default. You can just pipe your output to a simple awk like this:
curl https://github.com/microsoft/vscode/releases -s |
grep -oE 'microsoft/vscode/commit/[^/]+/hovercard' |
awk -F/ '{print $(NF-1)}'
ccbaa2d27e38e5afa3e5c21c1c7bef4657064247
3a6960b964327f0e3882ce18fcebd07ed191b316
f4af3cbf5a99787542e2a30fe1fd37cd644cc31f
b3318bc0524af3d74034b8bb8a64df0ccf35549a
6cba118ac49a1b88332f312a8f67186f7f3c1643
c13f1abb110fc756f9b3a6f16670df9cd9d4cf63
ee8c7def80afc00dd6e593ef12f37756d8f504ea
7f6ab5485bbc008386c4386d08766667e155244e
83bd43bc519d15e50c4272c6cf5c1479df196a4d
e7d7e9a9348e6a8cc8c03f877d39cb72e5dfb1ff
I want grep a word in my file, but only the part I gave to grep.
Example : in my file, i've "hell\nhell:o", i want grep hell but not hello.
How can I do that?
Give this a try:
grep -E "\bhell(\s|$)" file
add o if you just want the matched word:
kent$ echo "hell\nhell:o"|grep -oE "\bhell(\s|$)"
hell
or:
grep -oP '\bhell(?=\s|$)'
Using the -o option of grep :
grep -o hell your_file
From grep manual page :
-o, --only-matching
Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
Your question (and comments) aren't overly clear but you could try either:
"negative lookahead"
using the "end of line anchor"
Negative Lookahead
The following regex will match any hell that isn't followed by a : (adapt as suitable):
hell(?!:)
grep doesn't support this, so you'll need to use perl or something else:
echo -e "hell\nhell:o" \
| perl -ne 'print if /hell(?!:)/'
End of Line Anchor
The following will work with grep, matching only where the hell touches the end of the line ($):
hell$
echo -e "hell\nhell:o" \
| grep 'hell$'
Try to use -oP
echo "hell hello" | grep -oP '\bhell\b'
Given the following content:
title="Bar=1; Fizz=2; Foo_Bar=3;"
I'd like to match the first occurrence of Bar value which is 1. Also I don't want to rely on soundings of the word (like double quote in the front), because the pattern could be in the middle of the line.
Here is my attempt:
$ grep -o -m1 'Bar=[ ./0-9a-zA-Z_-]\+' input.txt
Bar=1
Bar=3
I've used -m/--max-count which suppose to stop reading the file after num matches, but it didn't work. Why this option doesn't work as expected?
I could mix with head -n1, but I wondering if it is possible to achieve that with grep?
grep is line-oriented, so it apparently counts matches in terms of lines when using -m[1]
- even if multiple matches are found on the line (and are output individually with -o).
While I wouldn't know to solve the problem with grep alone (except with GNU grep's -P option - see anubhava's helpful answer), awk can do it (in a portable manner):
$ awk -F'Bar=|;' '{ print $2 }' <<<"Bar=1; Fizz=2; Foo_Bar=3;"
1
Use print "Bar=" $2, if the field name should be included.
Also note that the <<< method of providing input via stdin (a so-called here-string) is specific to Bash, Ksh, Zsh; if POSIX compliance is a must, use echo "..." | grep ... instead.
[1] Options -m and -o are not part of the grep POSIX spec., but both GNU and BSD/OSX grep support them and have chosen to implement the line-based logic.
This is consistent with the standard -c option, which counts "selected lines", i.e., the number of matching lines:
grep -o -c 'Bar=[ ./0-9a-zA-Z_-]\+' <<<"Bar=1; Fizz=2; Foo_Bar=3;" yields 1.
Using perl based regex flavor in gnu grep you can use:
grep -oP '^(.(?!Bar=\d+))*Bar=\d+' <<< "Bar=1; Fizz=2; Foo_Bar=3;"
Bar=1
(.(?!Bar=\d+))* will match 0 or more of any characters that don't have Bar=\d+ pattern thus making sure we match first Bar=\d+
If intent is to just print the value after = then use:
grep -oP '^(.(?!Bar=\d+))*Bar=\K\d+' <<< "Bar=1; Fizz=2; Foo_Bar=3;"
1
You can use grep -P (assuming you are on gnu grep) and positive look ahead ((?=.*Bar)) to achieve that in grep:
echo "Bar=1; Fizz=2; Foo_Bar=3;" | grep -oP -m 1 'Bar=[ ./0-9a-zA-Z_-]+(?=.*Bar)'
First use a grep to make the line start with Bar, and then get the Bar at the start of the line:
grep -o "Bar=.*" input.txt | grep -o -m1 "^Bar=[ ./0-9a-zA-Z_-]\+"
When you have a large file, you can optimize with
grep -o -m1 "Bar=.*" input.txt | grep -o -m1 "^Bar=[ ./0-9a-zA-Z_-]\+"
I have a file as show below
1.2.3.4.ask
sanma.nam.sam
c.d.b.test
I want to remove the last field from each line, the delimiter is . and the number of fields are not constant.
Can anybody help me with an awk or sed to find out the solution. I can't use perl here.
Both these sed and awk solutions work independent of the number of fields.
Using sed:
$ sed -r 's/(.*)\..*/\1/' file
1.2.3.4
sanma.nam
c.d.b
Note: -r is the flag for extended regexp, it could be -E so check with man sed. If your version of sed doesn't have a flag for this then just escape the brackets:
sed 's/\(.*\)\..*/\1/' file
1.2.3.4
sanma.nam
c.d.b
The sed solution is doing a greedy match up to the last . and capturing everything before it, it replaces the whole line with only the matched part (n-1 fields). Use the -i option if you want the changes to be stored back to the files.
Using awk:
$ awk 'BEGIN{FS=OFS="."}{NF--; print}' file
1.2.3.4
sanma.nam
c.d.b
The awk solution just simply prints n-1 fields, to store the changes back to the file use redirection:
$ awk 'BEGIN{FS=OFS="."}{NF--; print}' file > tmp && mv tmp file
Reverse, cut, reverse back.
rev file | cut -d. -f2- | rev >newfile
Or, replace from last dot to end with nothing:
sed 's/\.[^.]*$//' file >newfile
The regex [^.] matches one character which is not dot (or newline). You need to exclude the dot because the repetition operator * is "greedy"; it will select the leftmost, longest possible match.
With cut on the reversed string
cat youFile | rev |cut -d "." -f 2- | rev
If you want to keep the "." use below:
awk '{gsub(/[^\.]*$/,"");print}' your_file
The command below in OSX checks whether an account is disabled (or not).
I'd like to grep the string "isDisabled=X" to create a report of disabled users, but am not sure how to do this since the output is on three lines, and I'm interested in the first 12 characters of line three:
bash-3.2# pwpolicy -u jdoe -getpolicy
Getting policy for jdoe /LDAPv3/127.0.0.1
isDisabled=0 isAdminUser=1 newPasswordRequired=0 usingHistory=0 canModifyPasswordforSelf=1 usingExpirationDate=0 usingHardExpirationDate=0 requiresAlpha=0 requiresNumeric=0 expirationDateGMT=12/31/69 hardExpireDateGMT=12/31/69 maxMinutesUntilChangePassword=0 maxMinutesUntilDisabled=0 maxMinutesOfNonUse=0 maxFailedLoginAttempts=0 minChars=0 maxChars=0 passwordCannotBeName=0 validAfter=01/01/70 requiresMixedCase=0 requiresSymbol=0 notGuessablePattern=0 isSessionKeyAgent=0 isComputerAccount=0 adminClass=0 adminNoChangePasswords=0 adminNoSetPolicies=0 adminNoCreate=0 adminNoDelete=0 adminNoClearState=0 adminNoPromoteAdmins=0
Your ideas/suggestions are most appreciated! Ultimately this will be part of a Bash script. Thanks.
This is how you would use grep to match "isDisabled=X":
grep -o "isDisabled=."
Explanation:
grep: invoke the grep command
-o: Use the --only-matching option for grep (From grep manual: "Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line."
"isDisabled=.": This is the search pattern you give to grep. The . is part of the regular expression, it means "match any character except for newline".
Usage:
This is how you would use it as part of your script:
pwpolicy -u jdoe -getpolicy | grep -oE "isDisabled=."
This is how you can save the result to a variable:
status=$(pwpolicy -u jdoe -getpolicy | grep -oE "isDisabled=.")
If your command was run some time prior, and the results from the command was saved to a file called "results.txt", you use it as input to grep as follows:
grep -o "isDisabled=." results.txt
You can use sed as
cat results.txt | sed -n 's/.*isDisabled=\(.\).*/\1/p'
This will print the value of isDisbaled.