I need to get a number of version from file. My version file looks like this:
#define MINOR_VERSION_NUMBER 1
I try to use sed command:
VERSION_MINOR=`sed -i -e 'MINOR_VERSION_NUMBER\s+\([0-9]+\).*/\1/p' $WORKSPACE/project/common/version.h`
but I get error:
sed: -e expression #1, char 2: extra characters after command
The "address" that selects matching lines needs to be enclosed in /.../ (or \X...X for any X).
sed -ne '/MINOR_VERSION_NUMBER/{ s/.*\([0-9]\).*/\1/;p }'
Don't use -i, it changes the file in place and doesn't output anything.
The more common way would be to use awk to find the line and extract the wanted column:
awk '(/MINOR_VERSION_NUMBER/){print$3}'
using grep
grep MINOR_VERSION_NUMBER | grep -o '[0-9]*$'
Demo :
$echo "#define MINOR_VERSION_NUMBER 1" | grep -o '[0-9]*$'
1
$echo "#define MINOR_VERSION_NUMBER 1123" | grep -o '[0-9]*$'
1123
$
Here is a correction of your attempt. Change your line:
VERSION_MINOR=`sed -i -e 'MINOR_VERSION_NUMBER\s+\([0-9]+\).*/\1/p' $WORKSPACE/project/common/version.h`
into:
VERSION_MINOR=`sed -n -e '/^#define\s\+MINOR_VERSION_NUMBER\s\+\([0-9]\+\).*/ s//\1/p' $WORKSPACE/project/common/version.h`
This can be made more readable with GNU sed's -r option:
VERSION_MINOR=`sed -n -r -e '/^#define\s+MINOR_VERSION_NUMBER\s+([0-9]+).*/ s//\1/p' $WORKSPACE/project/common/version.h`
As stated by choroba, awk would be more suited than sed for this kind of processing (see his answer).
However, here is another solution using bash's read builtin, together with GNU grep:
read x x VERSION_MINOR x < <(grep -F -w -m1 MINOR_VERSION_NUMBER $WORKSPACE/project/common/version.h)
VERSION_MINOR=$(echo "#define MINOR_VERSION_NUMBER 1" | tr -s ' ' | cut -d' ' -f3)
Related
I'm trying to fetch the first line in a log file which contain a date.
Here is an example of the log file :
SOME
LOG
2021-1-1 21:50:19.0|LOG|DESC1
2021-1-4 21:50:19.0|LOG|DESC2
2021-1-5 21:50:19.0|LOG|DESC3
2021-1-5 21:50:19.0|LOG|DESC4
In this context I need to get the following line:
2021-1-1 21:50:19.0|LOG|DESC1
An other log file example :
SOME
LOG
21-1-3 21:50:19.0|LOG|DESC1
21-1-3 21:50:19.0|LOG|DESC2
21-1-4 21:50:19.0|LOG|DESC3
21-1-5 21:50:19.0|LOG|DESC4
I need to fetch :
21-1-3 21:50:19.0|LOG|DESC1
At the moment I tried the following command :
cat /path/to/file | grep "$(date +"%Y-%m-%d")" | tail -1
cat /path/to/file | grep "$(date +"%-Y-%-m-%-d")" | tail -1
cat /path/to/file | grep -E "[0-9]+-[0-9]+-[0-9]" | tail -1
In case you are ok with awk, could you please try following. This will find the matched regex first line and exit from program, which will be faster since its NOT reading whole Input_file.
awk '
/^[0-9]{2}([0-9]{2})?-[0-9]{1,2}-[0-9]{1,2} [0-9]{2}:[0-9]{2}:[0-9]{2}\.[0-9]+/{
print
exit
}' Input_file
Using sed, being not too concerned about exactly how many digits are present:
sed -En '/^[0-9]+-[0-9]+-[0-9]+ [0-9]+:[0-9]+:[0-9]+[.][0-9]+[|]/ {p; q}' file
$ grep -m1 '^[0-9]' file1
2021-1-1 21:50:19.0|LOG|DESC1
$ grep -m1 '^[0-9]' file2
21-1-3 21:50:19.0|LOG|DESC1
If that's not all you need then edit your question to provide more truly representative sample input/output.
A simple grep with -m 1 (to exit after finding first match):
grep -m1 -E '^([0-9]+-){2}[0-9]+ ([0-9]{2}:){2}[0-9]+\.[0-9]+' file1
2021-1-1 21:50:19.0|LOG|DESC1
grep -m1 -E '^([0-9]+-){2}[0-9]+ ([0-9]{2}:){2}[0-9]+\.[0-9]+' file2
21-1-3 21:50:19.0|LOG|DESC1
This sed works with either GNU or POSIX sed:
sed -nE '/^[[:digit:]]{2,4}-[[:digit:]]{1,2}-[[:digit:]]{1,2}/{p;q;}' file
But awk, with the same BRE, is probably better:
awk '/^[[:digit:]]{2,4}-[[:digit:]]{1,2}-[[:digit:]]{1,2}/{print; exit}' file
I'm trying to get a pattern over multiple lines. I would like to ensure the line I'm looking for ends in \r\n and that there is specific text that comes after it at some point. The two problems I've had are I often get unmatched parenthesis in groupings or I get a positive match when there is none. Here are two simple examples.
echo -e -n "ab\r\ncd" | grep -U -c -z -E $'(\r\n)+.*TEST'
grep: Unmatched ( or \(
What exactly is unmatched there? I don't get it.
echo -e -n "ab\r\ncd" | grep -U -c -z -E $'\r\n.*TEST'
1
There is no TEST in the string, so why does this return a count of 1 for matches?
I'm using grep (GNU grep) 2.16 on Ubuntu 14. Thanks
Instead of -E you can use -P for PCRE support in gnu grep to use advanced regex like this:
echo -ne "ab\r\ncd" | ggrep -UczP '\r\n.*TEST'
0
echo -ne "ab\r\ncd" | ggrep -UczP '\r\n.*cd'
1
grep -E matches only in single line input.
I am using sed with grep command to replace a string. Old string is in 8 files at home location and I want to replace all of these with new string. I am using this:
#! /bin/bash
read oldstring
read newstring
sed -i -e 's/'Soldstring'/'$newstring'/' grep "$oldstring" /home/*
Now this command works but I am getting an warning:
sed: can't read grep: No such file or directory
sed: can't read oldstring: No such file or directory
Any ideas?
You probably wanted
sed -i -e "s|Soldstring|$newstring|" $(grep -l "$oldstring" /home/*)
However that form is unsafe. Better use xargs:
grep -l "$oldstring" /home/* | xargs sed -i -e "s|Soldstring|$newstring|"
And another if possible is to store on arrays:
readarray -t files < <(exec grep -l "$oldstring" /home/*)
sed -i -e "s|Soldstring|$newstring|" "${files[#]}"
You are not executing grep, you are giving it as a parameter to sed.
are you missing backticks?
sed -i -e 's/'Soldstring'/'$newstring'/' `grep "$oldstring" /home/*`
sed -i -e "s/$oldstring/$newstring/g" `grep -l "$oldstring" /home/*`
Just in order to clearly point out the various typos in your code:
#! /bin/bash
# ^
# extra space here (not really an error I think -- but unusual)
read oldstring
read newstring
sed -i -e 's/'Soldstring'/'$newstring'/' grep "$oldstring" /home/*
# ^ ^ ^
# `S` instead of `$` here | |
# here and there
# missing backticks (`)
As a side note, I suggest backticks above, but, since you are using bash, the syntax $(grep ....) is probably better than the classic Bourne Shell syntax `grep ....`. Finally, as suggested by konsolebox, "command nesting" might be unsafe, for example, in this case, if some file names contain spaces.
I have a string of text in a file that I am parsing out, I almost got it but not sure what I am missing
basic expression I am using is
cat cred.txt | grep -m 1 -o '&CD=[^&]*'
I am getting a results of
&CD=u8AA-RaF-97gc_SdZ0J74gc_SdZ0J196gc_SdZ0J211
I do not want the &CD= part in the resulting string, how would I do that.
The string I am parsing from is:
webpage.asp?UserName=username&CD=u8AA-RaF-97gc_SdZ0J74gc_SdZ0J196gc_SdZ0J211&Country=USA
If your grep knows Perl regex:
grep -m 1 -oP '(?<=&CD=)[^&]*' cred.txt
If not:
sed '1s/.*&CD=\([^&]*\).*/\1/' cred.txt
Many ways to skin this cat.
Extend your pipe:
grep -o 'CD=[^&]*' cred.txt | cut -d= -f2
Or do a replacement in sed:
sed -r 's/.*[&?]CD=([^&]*).*/\1/' cred.txt
Or get really fancy and parse the actual QUERY_STRING in awk:
awk -F'?' '{ split($2, a, "&"); for(i in a){split(a[i], kv, "="); out[kv[1]]=kv[2];} print out["CD"];}'
file.txt contains:
##w##
##wew##
using mac 10.6, bash shell, the command:
cat file.txt | grep [[:alpha:]]* -o
outputs nothing. I'm trying to extract the text inside the hash signs. What am i doing wrong?
(Note that it is better practice in this instance to pass the filename as an argument to grep instead of piping the output of cat to grep: grep PATTERN file instead of cat file | grep PATTERN.)
What shell are you using to execute this command? I suspect that your problem is that the shell is interpreting the asterisk as a wildcard and trying to glob files.
Try quoting your pattern, e.g. grep '[[:alpha:]]*' -o file.txt.
I've noticed that this works fine with the version of grep that's on my Linux machine, but the grep on my Mac requires the command grep -E '[[:alpha:]]+' -o file.txt.
sed 's/#//g' file.txt
/SCRIPTS [31]> cat file.txt
##w##
##wew##
/SCRIPTS [32]> sed 's/#//g' file.txt
w
wew
if you have bash >3.1
while read -r line
do
case "$line" in
*"#"* )
if [[ $line =~ "^#+(.*)##+$" ]];then
echo ${BASH_REMATCH[1]}
fi
esac
done <"file"