I have a file containing text in separate lines.
text1
text2
text3
textN
I have a directory with many files. I want to grep for each line in the of this specific directory. What is an easy way to do this?
There is no need to loop, you can do use grep with the -f option to get patterns from a file:
grep -f pattern_file files*
From man grep:
-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.)
Test
$ cat a1
hello
how are you?
$ cat a2
bye
hello
$ cat pattern
hello
bye
$ grep -f pattern a*
a1:hello
a2:bye
a2:hello
You can use standard bash loop for this as well :
for i in text*; do grep "pattern" $i; done
or even better option without loop :
grep "pattern" text*
If you press tab after the * then shell will expand it to the files that satisfy the condition.
Related
I am trying to search for files with specific text but excluding a certain text and showing only the files.
Here is my code:
grep -v "TEXT1" *.* | grep -ils "ABC2"
However, it returns:
(standard input)
Please suggest. Thanks a lot.
The output should only show the filenames.
Here's one way to do it, assuming you want to match these terms anywhere in the file.
grep -LZ 'TEXT1' *.* | xargs -0 grep -li 'ABC2'
-L will match files not containing the given search term
use -LiZ if you want to match TEXT1 irrespective of case
The -Z option is needed to separate filenames with NUL character and xargs -0 will then separate out filenames based on NUL character
If you want to check these two conditions on same line instead of anywhere in the file:
grep -lP '^(?!.*TEXT1).*(?i:ABC2)' *.*
-P enables PCRE, which I assume you have since linux is tagged
(?!regexp) is a negative lookahead construct, so ^(?!.*TEXT1) will ensure the line doesn't have TEXT1
(?i:ABC2) will match ABC2 case insensitively
Use grep -liP '^(?!.*TEXT1).*ABC2' if you want to match both terms irrespective of case
(standard input)
This error is due to use of grep -l in a pipeline as your second grep command is reading input from stdin not from a file and -l option is printing (standard input) instead of the filename.
You can use this alternate solution in a single awk command:
awk '/ABC2/ && !/TEXT1/ {print FILENAME; nextfile}' *.* 2>/dev/null
I have the following situation:
source.txt
ID1:email1#domain1.com
ID2:email2#domain2.com
ID3:email3#domain3.com
...
IDs are numeric strings, e.g. 1234, 23412, 897... (one or more digits).
exclude.txt
emailX#domainX.com
emailY#domainY.com
emailZ#domainZ.com
...
i.e. only emails, no IDs.
I want to remove all lines from source.txt which contain emails listed in exclude.txt, preserving the ID:email pairs for the lines which are not removed.
How can I do that with linux command line tools (or simple bash script if needed)?
You can do it easily with awk:
awk -F":" 'NR==FNR{a[$1];next}(!($2 in a))' exclude.txt source.txt
Alternative with grep:
grep -v -F -f exclude.txt source.txt
Use grep with care, since grep does a regex matching. You might need to add also -w option to grep (word matching)
I have a directory which has many directories inside it with the pattern of their name as :
YYYYDDMM_HHMISS
Example: 20140102_120202
I want to extract only the YYYYDDMM part.
I tried ls -l|awk '{print $9}'|grep -o ^[0-9]* and got the answer.
However i have following questions:
Why doesnt this return any results: ls -l|awk '{print $9}'|grep -o [0-9]* . Infact it should have returned all the directories.
Strangely just including '^' before [0-9] works fine :
ls -l|awk '{print $9}'|grep -o ^[0-9]*
Any other(simpler) way to achieve the result?
Why doesnt this return any results: ls -l|awk '{print $9}'|grep -o [0-9]*
If there are files in your current directory that start with [0-9], then the shell will expand them before calling grep. For example, if I have two files a1, a2 and a3 and run this:
ls | grep a*
After the filenames are expanded, the shell will run this:
ls | grep a1 a2 a3
The result of which is that it will print the lines in a2 and a3 that match the text "a1". It will also ignore whatever is coming from stdin, because when you specify filenames for grep (2nd argument and beyond), it will ignore stdin.
Next, consider this:
ls | grep ^a*
Here, ^ has no special meaning to the shell, so it uses it verbatim. Since I don't have filenames starting with ^a, it will use ^a* as the pattern. If I did have filenames like ^asomething or ^another, then again, ^a* would be expanded to those filenames and grep would do something I didn't really intend.
This is why you have to quote search patterns, to prevent the shell from expanding them. The same goes for patterns in find /path -name 'pattern'.
As for a simpler way for what you want, I think this should do it:
ls | sed -ne 's/_.*//p'
To show only the YYDDMM part of the directory names:
for i in ./*; do echo $(basename "${i%%_*}"); done
Not sure what you want to do with it once you've got it though...
You must avoid parsing ls output.
Simple is to use this printf:
printf "%s\n" [0-9]*_[0-9]*|egrep -o '^[0-9]+'
I want to write a bash script that sorts the input by rules in different files. The first rule is to write all chars or strings in file1. The second rule is to write all numbers in file2. The third rule is to write all alphanumerical strings in file3. All specials chars must be ignored. Because I am not familiar with bash I don t know how to realize this.
Could someone help me?
Thanks,
Haniball
Thanks for the answers,
I wrote this script,
#!/bin/bash
inp=0 echo "Which filename for strings?"
read strg
touch $strg
echo "Which filename for nums?"
read nums
touch $nums
echo "Which filename for alphanumerics?"
read alphanums
touch $alphanums
while [ "$inp" != "quit" ]
do
echo "Input: "
read inp
echo $inp | grep -o '\<[a-zA-Z]+>' > $strg
echo $inp | grep -o '\<[0-9]>' > $nums
echo $inp | grep -o -E '\<[0-9]{2,}>' > $nums
done
After I ran it, it only writes string in the stringfile.
Greetings, Haniball
Sure can help. See here:
How To Ask Questions The Smart Way
Help Vampires: A Spotter’s Guide
cool site about the bash is here: http://wiki.bash-hackers.org/doku.php
for sorting try man sort
for pattern matching try man grep
other useful tools: man sed man awk man strings man tee
And it is always correct tag your homework as "homework" ;)
You can try something like:
<input_file strings -1 -a | tee chars_and_strings.txt |\
grep "^[A-Za-z0-9][A-Za-z0-9]*$" | tee alphanum.txt |\
grep "^[0-9][0-9]*$" > numonly.txt
The above is only for USA - no international (read unicode) chars, where things coming a little bit more complicated.
grep is sufficient (your question is a bit vague. If I got something wrong, let me know...)
Using the following input file:
this is a string containing words,
single digits as in 1 and 2 as well
as whole numbers 42 1066
all chars or strings
$ grep -o '\<[a-zA-Z]\+\>' sorting_input
this
is
a
string
containing
words
single
digits
as
in
and
as
well
all single digit numbers
$ grep -o '\<[0-9]\>' sorting_input
1
2
all multiple digit numbers
$ grep -o -E '\<[0-9]{2,}\>' sorting_input
42
1066
Redirect the output to a file, i.e. grep ... > file1
Bash really isn't the best language for this kind of task. While possible, ild highly recommend the use of perl, python, or tcl for this.
That said, you can write all of stdin from input to a temporary file with shell redirection. Then, use a command like grep to output matches to another file. It might look something like this.
#!/bin/bash
cat > temp
grep pattern1 > file1
grep pattern2 > file2
grep pattern3 > file3
rm -f temp
Then run it like this:
cat file_to_process | ./script.sh
I'll leave the specifics of the pattern matching to you.
How do I pipe the output of grep as the search pattern for another grep?
As an example:
grep <Search_term> <file1> | xargs grep <file2>
I want the output of the first grep as the search term for the second grep. The above command is treating the output of the first grep as the file name for the second grep. I tried using the -e option for the second grep, but it does not work either.
You need to use xargs's -i switch:
grep ... | xargs -ifoo grep foo file_in_which_to_search
This takes the option after -i (foo in this case) and replaces every occurrence of it in the command with the output of the first grep.
This is the same as:
grep `grep ...` file_in_which_to_search
Try
grep ... | fgrep -f - file1 file2 ...
If using Bash then you can use backticks:
> grep -e "`grep ... ...`" files
the -e flag and the double quotes are there to ensure that any output from the initial grep that starts with a hyphen isn't then interpreted as an option to the second grep.
Note that the double quoting trick (which also ensures that the output from grep is treated as a single parameter) only works with Bash. It doesn't appear to work with (t)csh.
Note also that backticks are the standard way to get the output from one program into the parameter list of another. Not all programs have a convenient way to read parameters from stdin the way that (f)grep does.
I wanted to search for text in files (using grep) that had a certain pattern in their file names (found using find) in the current directory. I used the following command:
grep -i "pattern1" $(find . -name "pattern2")
Here pattern2 is the pattern in the file names and pattern1 is the pattern searched for
within files matching pattern2.
edit: Not strictly piping but still related and quite useful...
This is what I use to search for a file from a listing:
ls -la | grep 'file-in-which-to-search'
Okay breaking the rules as this isn't an answer, just a note that I can't get any of these solutions to work.
% fgrep -f test file
works fine.
% cat test | fgrep -f - file
fgrep: -: No such file or directory
fails.
% cat test | xargs -ifoo grep foo file
xargs: illegal option -- i
usage: xargs [-0opt] [-E eofstr] [-I replstr [-R replacements]] [-J replstr]
[-L number] [-n number [-x]] [-P maxprocs] [-s size]
[utility [argument ...]]
fails. Note that a capital I is necessary. If i use that all is good.
% grep "`cat test`" file
kinda works in that it returns a line for the terms that match but it also returns a line grep: line 3 in test: No such file or directory for each file that doesn't find a match.
Am I missing something or is this just differences in my Darwin distribution or bash shell?
I tried this way , and it works great.
[opuser#vjmachine abc]$ cat a
not problem
all
problem
first
not to get
read problem
read not problem
[opuser#vjmachine abc]$ cat b
not problem xxy
problem abcd
read problem werwer
read not problem 98989
123 not problem 345
345 problem tyu
[opuser#vjmachine abc]$ grep -e "`grep problem a`" b --col
not problem xxy
problem abcd
read problem werwer
read not problem 98989
123 not problem 345
345 problem tyu
[opuser#vjmachine abc]$
You should grep in such a way, to extract filenames only, see the parameter -l (the lowercase L):
grep -l someSearch * | xargs grep otherSearch
Because on the simple grep, the output is much more info than file names only. For instance when you do
grep someSearch *
You will pipe to xargs info like this
filename1: blablabla someSearch blablabla something else
filename2: bla someSearch bla otherSearch
...
Piping any of above line makes nonsense to pass to xargs.
But when you do grep -l someSearch *, your output will look like this:
filename1
filename2
Such an output can be passed now to xargs
I have found the following command to work using $() with my first command inside the parenthesis to have the shell execute it first.
grep $(dig +short) file
I use this to look through files for an IP address when I am given a host name.