I am trying to search the file dep/playlist for 'ohn'. Then, I would like to take this result and apply it to a new grep command that searches the file employee list end then echoes the results on screen. The code below isn't behaving as expected.
grep ohn dep/playlist > search
grep $(cat search) employeelist > newlist
cat newlist
Thanks,
Tim
You need to tell grep to obtain the patterns from a file, use the -f option:
-f FILE, --file=FILE
Obtain patterns from FILE, one per line. The empty file
contains zero patterns, and therefore matches nothing. (-f is
specified by POSIX.)
So the command would look like:
grep -f search employeelist > newlist
Using process substitution you could obviate the need of a temporary file. So the two grep commands could be written into one as:
grep -f <(grep ohn dep/playlist) employeelist > newlist
xargs:
grep ohn dep/playlist | xargs -I name grep name employeelist
This searches dep/playlist for 'ohn', and then upon finding a result, that result is then used in grep X employeelist, where X is the result from the first grep.
Related
Tried the following command, it lists all the lines including file names
which are not matching the given pattern.
grep -nrv "^type.* = .*"
"But what we need is list of file names in a folder with content
which does not have even a single occurrence of above pattern."
Your help will be really appreciated.
You need the -L option:
grep -rL '^type.* = .*' directory_name
From the GNU grep manual:
-L, - -files-without-match
Suppress normal output; instead print the name of each input file from which no output would normally have been printed. The scanning will stop on the first match.
I have a file with partial matches to lines in another file. In order to do this I was looking to generate a while loop with read and substituting a variable for each line of partial matches into a grep command to search a database files with a partial match but for some reason, I am not getting an output (an empty outputfile.txt).
Here is my current script
while read -r line; do
grep $line /path/to/databasefile >> /path/to/folder/outputfile.txt
done < "/partial_matches.txt"
the database has multiple lines with a sequence name then DNA sequence after:
>transcript_ab
AGTCAGTCATGTC
>transcript_ac
AGTCAGTCATGTC
>transctipt_ad
AGTCAGTCATGTC
and the partial matching search file has lines of text:
ab
ac
and I'm looking for a return of:
>transcript_ab
>transcript_ac
any help would be appreciated. Thanks.
If you are using GNU grep, then its -f option is what you are looking for:
grep -f /partial_matches.txt /path/to/databasefile
(if you don't have any pattern in partial_matches.txt but only strings, then use grep -F instead of grep)
you can use a for loop instead:
for i in $(cat partial_matches.txt); do
grep $i /path/to/databasefile >> /path/to/folder/outputfile.txt
done
Also, check if you have a typo:
"/partial_matches.txt" -> "./partial_matches.txt"
so its pretty hard to describe for me what I want to do, but I'll try it:
(Because of some private information I changed the names)
I want to "diff" a command output with a text file created from me.
The command output looks like:
'Blabla1' '12.34.56.78' (24 objects + dependencies), STATUS: 'RUNNING'
'Blabla3' '12.34.56.89' (89 objects + dependencies), STATUS: 'RUNNING'
And the txtfile:
Blabla1
Blabla2
If it finds Blabla1 anywhere in the command output its fine. But you see, he will not find Blabla2 anywhere in the command output and this difference I want as an output.
I hope you understand what I mean and you could possible help me.
Greetings,
Can
UPDATE::::
#hek2mgl
So my command is:
./factory.sh listapplications | grep -i running
This command shows this:
'ftp' '1' (7 objects + dependencies), STATUS: 'RUNNING' - 'XSD Da
'abc' '5.1.0' (14 objects + dependencies), STATUS: 'RUNNING' - '2017-10-13: Fix fuer Bug 2150'
'name' '1.0.2' (5 objects + dependencies), STATUS: 'RUNNING'
And I want to compare that output with my textfile:
ftp
abc
name
missing
alsomissing
So if I compare this 2 now it should check if he finds the words from my textfile ANYWHERE in the command output. If it does find it anywhere -> not output.
And as you see he'll not find "missing" and "alsomissing". I want this two as an output at the end.
What you might be interested in is grep in combination with 'process substitution'. If your file with patterns is file.txt and your command to execute is cmd then you can use
grep -o -F -f file.txt <(cmd) | grep -v -F -f - file.txt
This will output the patterns is file.txt which are not matched in the output of cmd.
In case of the Blabla example, the above line will output
Blabla2
How it works is the following. The first part will search for all patterns listed in file.txt in the output of cmd and will only output the matched parts. This means that
% grep -o -F -f file.txt <(cmd)
Blabla1
This output is now piped to another command that will try to find all lines in file.txt which do not match any of the patterns comming from the pipe (-f -)
% grep -o -F -f file.txt <(cmd) | grep -v -F -f - file.txt
Blabla2
So ... this seems to do it, using bash process substitution:
$ cat file1
'Blabla1' '12.34.56.78' (24 objects + dependencies), STATUS: 'RUNNING'
'Blabla3' '12.34.56.89' (89 objects + dependencies), STATUS: 'RUNNING'
$ cat file2
Blabla1
Blabla2
$ grep -vFf <(awk '{gsub(/[^[:alnum:]]/,"",$1);print $1}' file1) file2
Blabla2
The awk script takes the first field, strips non-alphanumeric characters from it (i.e. the single quotes) and outputs just that first field. The grep option -f uses the "virtual" file created by the aforementioned process substitution as a list of fixed strings to search for within the input file (file2), and the -v reverses the search, showing you only what was not found.
If the regex in the gsub() is too greedy, you might replace it with something like $1=substr($1,2,length($1)-2).
You could alternately do this in (POSIX) awk alone, without relying on bash process substitution:
$ awk 'NR==FNR{a[substr($1,2,length($1)-2)];next} $1 in a{next} 1' file1 file2
Blabla2
This reads the stripped first field of file1 into the keys of an array, then for each line of file2 checks for the existence of that key in the array, skipping lines that match and printing any left over. (The 1 at the end of the script is short-hand for "print this line".)
You can also use awk only:
awk '
# Store patterns of text.file in an array (p)atterns.
# Initialize their count of occurrence with 0
NR==FNR{
p[$0]=0
next
}
# Replace the quotes around BlaBla... in cmd output.
# Increase the count of occurrence of the pattern
{
gsub("'\''", "")
p[$1]++
}
# At the end of the input print those patterns which
# did not appear in cmd output, meaning their count of
# occurrence is zero.
END{
for(i in p){
if(p[i]==0){
print i
}
}
}' text.file cmd.txt
PS: Alternatively you use process substitution instead of storing the command output in a file. Replace cmd.txt by <(cmd) then.
I want to print out the name of a file if and only if it does not contain the string foo. However, if file contains foo and I run this
grep -lv 'foo' file
file is outputted. Why does this happen and what can I do to work around it?
-v means to match any line that doesn't match the pattern. So -lv means to list any file that contains any line that doesn't match the pattern. That's not the same as a file where none of the lines match the pattern.
Use the -L option to list all files that don't have any match for the pattern.
grep -L 'foo' file
-L, --files-without-match
Suppress normal output; instead print the name of each input file from which no output would normally have been printed. The
scanning will stop on the first match.
Don't use -v with this. -L already inverts which files are listed, and -v inverts the way lines are matched, so -Lv is the same as -l.
grep -lv means: list every file that contains a line that does not match.
You're looking for grep -L.
grep -L 'foo' file
you should remove -v option, because it means it will omit the text which is given in any line. So you should try:
grep -L 'foo' Input_file
Also if you want to know all the files which are having string(foo) in a directory or etc then you could try following too.
grep -L 'foo' /path/to/files/*.txt
An example of grep -L where it will print all the .txt file names which have string foo in them.
EDIT: If in case you are interested in awk solution, you could try following too.
awk 'val{close(val)} FNR==1{val=FILENAME} /Var/{;print FILENAME;nextfile}' *.txt
When using the Grep command to find a search string in a set of files, how do I dump the results to a text file?
Also is there a switch for the Grep command that provides cleaner results for better readability, such as a line feed between each entry or a way to justify file names and search results?
For instance, a away to change...
./file/path: first result
./another/file/path: second result
./a/third/file/path/here: third result
to
./file/path: first result
./another/file/path: second result
./a/third/file/path/here: third result
grep -n "YOUR SEARCH STRING" * > output-file
The -n will print the line number and the > will redirect grep-results to the output-file.
If you want to "clean" the results you can filter them using pipe | for example:
grep -n "test" * | grep -v "mytest" > output-file
will match all the lines that have the string "test" except the lines that match the string "mytest" (that's the switch -v) - and will redirect the result to an output file.
A few good grep-tips can be found in this post
Redirection of program output is performed by the shell.
grep ... > output.txt
grep has no mechanism for adding blank lines between each match, but does provide options such as context around the matched line and colorization of the match itself. See the grep(1) man page for details, specifically the -C and --color options.
To add a blank line between lines of text in grep output to make it easier to read, pipe (|) it through sed:
grep text-to-search-for file-to-grep | sed G