Have a directory with files. how to grep all files with a particular keyword and files should be created today.
tried this :
grep -r -n "pass" *.json
this gives the line number and only searches json file but how to add pipe to search for only files that are created today
Use the find command as the source of the list of files to search.
grep -n "pass" $(find . -daystart -ctime 0 -name '*.json')
-ctime says "creation date should have been 0 days ago", and -daystart says to base it on the beginning of the day for "days ago", not "24 hours ago" ago.
Related
Say I have a directory /home/ and within it I have 3 subdirectories /home/red/ /home/blue/ /home/green/
And each subdirectory contains a file each like
/home/red/file1 /home/blue/file2 /home/green/file3
Now I want to find how many times file1,file2, file3 contains the word "hello" within them.
For example,
/home/red/file1 - 23
/home/blue/file2 - 6
/home/green/file3 - 0
Now, going to the locations of file and running the grep command is actually very inefficient when this problem scales.
I have tried using this grep command from the /home/ directory
grep -rnw '/path/to/somewhere/' -e 'pattern'
But this is just giving the occurrences rather than the count.
Is there any command through which I can get what I am looking for?
If the search term occurs at maximum once per line, you can use grep's -c option to report the count instead of the matching lines. So, the command will be grep -rc 'search' (add other options as needed).
If there can be more than one occurrence per line, I'd recommend using ripgrep. Note that rg recursively searches by default, so you can use something like rg -co 'search' from within the home directory (add other options as needed). Add --hidden if you need to search hidden files as well. Add --include-zero if you want to show files even if they didn't have any match.
Instead of grep you can use this find | gnu-awk solution:
cd /home
find {red/file1,blue/file2,green/file3} -type f -exec awk '
{c += gsub(/pattern/, "&")} ENDFILE {print FILENAME, "-", c; c=0}' {} +
I was doing some adb shell stuff on windows and stuck at a point. Here's what I was doing..
I was printing all installed apps on my phone and getting their exact path.
zeroltetmo:/ # pm list packages -f
package:/system/app/FilterProvider/FilterProvider.apk=com.samsung.android.provider.filterprovider
package:/system/priv-app/CtsShimPrivPrebuilt/CtsShimPrivPrebuilt.apk=com.android.cts.priv.ctsshim
package:/system/app/YouTube/Youtube.apk=com.google.android.youtube
package:/system/app/vsimservice/vsimservice.apk=com.sec.vsimservice
package:/system/priv-app/WallpaperCropper/WallpaperCropper.apk=com.android.wallpapercropper
package:/system/framework/framework-res.apk=android
package:/system/framework/samsung-framework-res/samsung-framework-res.apk=com.samsung.android.framework.res
package:/data/app/com.whatsapp-1/base.apk=com.whatsapp
package:/data/app/ru.meefik.busybox-2/base.apk=ru.meefik.busybox
package:/data/app/com.google.android.play.games-1/base.apk=com.google.android.play.games
But,
I want this to print only system/app directory but only upto folder name instead of the full path. What i'm doing is piping this to grep and using this pattern to get the result.
zeroltetmo:/ # pm list packages -f | grep -o "system/app.*\/"
system/app/FilterProvider/
system/app/RootPA/
system/app/YouTube/
system/app/ClipboardSaveService/
system/app/TetheringAutomation/
system/app/GoogleExtShared/
system/app/WfdBroker/
system/app/vsimservice/
system/app/USBSettings/
system/app/EasyOneHand3/
But the problem is this / at the end of folder name that I'm stuck with.
You can filter the trailing slashes out with sed like that:
pm list packages -f | grep -o "system/app.*/" | sed 's,/$,,'
Explanation of the sed command:
s stands for substitution
, delimits command name from its arguments - it's easier to use something different / when we want to replace /
/$ - string to be replaced. In this case it means slash at the end of the line
The string to replace /$ with is empty because we want to remove it.
I have a bunch of files: some contain the word star, some contain the word start, some contain both.
I'd like to grep for files that contain the word star, but not the word start.
How can this be accomplished using only grep?
grep has some options for inverting the matches at the line or file level. You want the latter option, with the -L switch. The following will print the names of all the files in a folder that don't contain the text start:
grep -LF start *
-F tells grep that start is a literal string and not a regex. It's optional here, but might speed things up a tiny bit.
You can use the resulting list to search for files that contain star:
grep -lF star $(grep -LF start *)
-l prints only the names of files containing a match, not any line-by-line or match-by-match details. If this is not exactly what you want, man grep is your friend.
This uses an additional shell construct to run the inverted match, but it technically doesn't call any additional programs that aren't grep.
Update
Since you mention wanting to look through all the files starting with a given root folder, change -LF to -LFr. Replace * with your root folder if you don't want to change working directories.
-r tells grep to recurse into directories, and search every file it finds along the way.
With GNU grep for -w:
$ cat file
foo star bar
oof start rab
$ grep -w star *
foo star bar
or if you just want the names of the files containing star:
$ grep -lw star *
file
and to just find files to look in:
$ find . -maxdepth 1 -type f -exec grep -w 'star' {} \;
foo star bar
Im greping a bunch of files in a directory as below
grep -EIho 'abc|def' *|sort|uniq -c >>counts.csv
My output is
150 abc
130 def
What I need is Current date (-1) and the result of grep like below to be inserted to counts.csv
5/21/2018 150,130
grep..|sort|uniq -c
|awk -v d="$(date -d '1 day ago' +%D)" 'NR==1{printf "%s",d}{printf "%s",","$1;}END{print ""}'
will do it.
With your example data, it gives:
05/21/18,150,130
I have a CSV file (foo.csv) with 200,000 rows. I need to break it into four files (foo1.csv, foo2.csv... etc.) with 50,000 rows each.
I already tried simple ctrl-v/-c using gui text editors, but the my computer slows to a halt.
What unix command(s) could I use to accomplish this task?
I don't have a terminal handy to try it out, but it should be just split -d -l 50000 foo.csv.
Hopefully the naming isn't terribly important because with the -d option, the output files will be named foo.csv00 .. foo.csv03. You can add the -a 1 option so that the suffixes are 0-3, but there's no simple way to get the suffix to be injected into the middle of the filename.
you should use head and tail.
head -n 50000 myfile > part1.csv
head -n 100000 myfile | tail -n 50000 > part2.csv
head -n 150000 myfile | tail -n 50000 > part3.csv
etc ...
Else, but with no control on file names, you can use unix command split.
sed -n 2000,4000p somefile.txt
will print from lines 2000 to 4000 to stdout.
split -l50000 foo.csv
You can use sed
I wrote this little shell script for this topic very similar at yours.
This shell script + awk works fine for me:
#!/bin/bash
awk -v initial_line=$1 -v end_line=$2 '{
if (NR >= initial_line && NR <= end_line)
print $0
}' $3
Used with this sample file (file.txt):
one
two
three
four
five
six
The command (it will extract from second to fourth line in the file):
edu#debian5:~$./script.sh 2 4 file.txt
Output of this command:
two
three
four
Of course, you can improve it, for example by testing that all argument values are the expected :-)