GREP to show files WITH text and WITHOUT text - linux

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

Related

grep: Invalid regular expression

I have a text file which looks like this:
haha1,haha2,haha3,haha4
test1,test2,test3,test4,[offline],test5
letter1,letter2,letter3,letter4
output1,output2,[offline],output3,output4
check1,[core],check2
num1,num2,num3,num4
I need to exclude all those lines that have "[ ]" and output them to another file without all those lines that have "[ ]".
I'm currently using this command:
grep ",[" loaded.txt | wc -l > newloaded.txt
But it's giving me an error:
grep: Invalid regular expression
Use grep -F to treat the search pattern as a fixed string. You could also replace wc -l with grep -c.
grep -cF ",[" loaded.txt > newloaded.txt
If you're curious, [ is a special character. If you don't use -F then you'll need to escape it with a backslash.
grep -c ",\[" loaded.txt > newloaded.txt
By the way, I'm not sure why you're using wc -l anyways...? From your problem description, it sounds like grep -v might be more appropriate. -v inverts grep's normal output, printing lines that don't match.
grep -vF ",[" loaded.txt > newloaded.txt
An alternative method to Grep
It's unclear if you want to remove lines that might contain either bracket [], or only the ones where the brackets specifically surround characters. Regardless of which method you intend to use, sed can easily remove lines that fit a definitive pattern:
To delete only lines that contained both brackets surrounding characters [...]:
sed '/\[.*\]/d' loaded.txt > newloaded.txt
Another approach might be to remove any line that contained either bracket:
sed '/\[/d;/\]/d' loaded.txt > newloaded.txt
(eg. lines containing either [ or ] would be deleted)
Your grep command doesn't seem to be excluding anything. Also, why are you using wc? I thought you want the lines, not their count.
So if you just want the lines, as you say, that don't have [], then this should work:
grep -v "\[" loaded.txt > new.txt
You can also use awk for this:
awk -F\[ 'NF==1' file > newfile
cat newfile
haha1,haha2,haha3,haha4
letter1,letter2,letter3,letter4
num1,num2,num3,num4
Or this:
awk '!/\[/' file

How do I grep in a list of files targeted by a previous grep?

I am using grep to get a list of files that I want to use for another grep search (and not simply piping it).
For example I got as an output:
file1.h:XXX: linecontent
file2.h:XXX: linecontent
file3.h:XXX: linecontent
file4.h:XXX: linecontent
and I want to grep only file1.h, file2.h ...
I'm assuming you want to search for files that contain two different patterns. If so this is what you want:
grep 'your pattern 2' `grep -l 'your pattern 1' *`
The contents of the back quotes will be executed first and the output substituted into the command line. Use of the -l flag will restrict the output of the grep command to just the file names.
If there are a very large number of files that match against your pattern 1 this could fail. The solution for that is to use xargs
grep -l 'your pattern 1' * | xargs grep 'your argument 2'
Assuming what you want is the names of files that contain 'lineofcontent', you could use:
grep -l 'lineofcontent' file*.h

grep -o and display part of filenames using ls

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]+'

Grep Search all files in directory for string1 AND string2

How can I make use of grep in cygwin to find all files that contain BOTH words.
This is what I use to search all files in a directory recursively for one word:
grep -r "db-connect.php" .
How can I extend the above to look for files that contain both "db-connect.php" AND "version".
I tried this: grep -r "db-connect.php\|version" . but this is an OR i.e. it gets file that contain one or the other.
Thanks all for any help
grep -r db-connect.php . | grep version
If you want to grep for several strings in a file which have different lines, use the following command:
grep -rl expr1 | xargs grep -l expr2 | xargs grep -l expr3
This will give you a list of files that contain expr1, expr2, and expr3.
Note that if any of the file names in the directory contains spaces, these files will produce errors. This can be fixed by adding -0 I think to grep and xargs.
grep "db-connect.php" * | cut -d: -f1 | xargs grep "version"
I didn't try it in recursive mode but it should be the same.
To and together multiple searches, use multiple lookahead assertions, one per thing looked for apart from the last one:
instead of writing
grep -P A * | grep B
you write
grep -P '(?=.*A)B' *
grep -Pr '(?=.*db-connect\.php)version' .
Don’t write
grep -P 'A.*B|B.*A' *
because that fails on overlaps, whereas the (?=…)(?=…) technique does not.
You can also add in NOT operators as well. To search for lines that don’t match X, you normally of course use -v on the command line. But you can’t do that if it is part of a larger pattern. When it is, you add (?=(?!X).)*$) to the pattern to exclude anything with X in it.
So imagine you want to match lines with all three of A, B, and then either of C or D, but which don’t have X or Y in them. All you need is this:
grep -P '(?=^.*A)(?=^.*B)(?=^(?:(?!X).)*$)(?=^(?:(?!Y).)*$)C|D' *
In some shells and in some settings. you’ll have to escape the ! if it’s your history-substitution character.
There, isn’t that pretty cool?
In my cygwin the given answers didn't work, but the following did:
grep -l firststring `grep -r -l secondstring . `
Do you mean "string1" and "string2" on the same line?
grep 'string1.*string2'
On the same line but in indeterminate order?
grep '(string1.*string2)|(string2.*string1)'
Or both strings must appear in the file anywhere?
grep -e string1 -e string2
The uses PCRE (Perl-Compatible Regular Expressions) with multiline matching and returns the filenames of files that contain both strings (AND rather than OR).
grep -Plr '(?m)db-connect\.php(.*\n)*version|version(.*\n)*db-connect\.php' .
Why to stick to only grep:
perl -lne 'print if(/db-connect.php/&/version/)' *

Pipe output to use as the search specification for grep on Linux

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.

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