I am attempting to search multiple files within a directory, using multiple egrep pipes I am organize the data I need, but it matches every single line. I am only need it to match one line, and continue to the next file in the directory.
Ex:
egrep -i "stringname" * | egrep -i "anotherstringname"
Is there another way? Any recommendations, I am new to Unix.
You probably want something like:
egrep -m 1 "string1|string2|string3" *.txt
(where I assume your files have names matching *.txt). For instance, on my own computer if I type:
egrep -m 1 "def|append" *.py
I will search at most 1 line in each python file matching either "def" or "append".
Related
i am trying to get the count of files which has matching keywords in directory. Code i used is:
grep -r -i --include=\*.sas 'keyword'
Can any one help me to, how to get the count of the files which contains the keyword.
Thanks
You will need to do two things. The first is to suppress normal output from grep and print only the file name with -l. The second is to pipe the output through to wc -l to get the count of the lines, hence the count of the files.
grep -ril "keyword" --include="*.sas" * | wc -l
I have a directory full of files with names such as:
file_name_is_001
file_name_001
file_name_is_002
file_name_002
file_name_is_003
file_name_003
I want to copy only the files that don't contain 'is'. I'm not sure how to do this. I have tried to search for it, but can't seem to google the right phrase to find the results.
Details depend on operating system, shell, etc.
For a unix system a quite verbose but easy to understand approach could look like this (please mind that I didn't test it):
mkdir some_temporary_directory
mv *_is_* some_temporary_directory
cp * where_ever_you_want_to_copy_it
mv some_temporary_directory/* .
rmdir some_temporary_directory
You can do this using bash. First, here's a command to get you a list of files that don't contain the text _is_:
ls | grep -v "_is_"
This takes the output of ls and matches all values with DO NOT contain _is_ using grep -v.
In order to then copy these files, we need to turn the lines output by grep into arguments of cp. We can do this using xargs:
ls | grep -v "_is_" | xargs -J % cp % new_folder
From the xargs man page, it is a tool to "build and execute command lines from standard input".
How can I use ls (or other commands) and grep together to search from specific files for a certain word inside that file?
Example I have a file - 201503003_315_file.txt and I have other files in my dir.
I only want to search files that have a file name that contains _315_ and inside that file, search for the word "SAMPLE".
Hope this is clear and thanks in advance for any help.
You can do:
ls * _315_* | xargs grep "SAMPLE"
The first part: ls * _315_* will list only files that have 315 as part of the file name, this list of files is piped to grep which will scan each one of them and look for "SAMPLE"
UPDATE
A bit easier (and actually safer) approach was mentioned by David in the comments bellow:
grep "SAMPLE" *_315_*.txt
The reason why it's safer is that ls doesn't handle well special characters.
Another option, as mentioned by Charles Duffy in the comments below:
printf '%s\0' *_315_* | xargs -0 grep
Change to that directory (using cd dir) and try:
grep SAMPLE *_315_*
If you really MUST use ls AND grep try this:
ls *_315_* | xargs grep SAMPLE
The first example, however, requires less typing...
This question already has answers here:
How to pass command output as multiple arguments to another command
(5 answers)
Read expression for grep from standard input
(1 answer)
Closed last month.
I am looking for insight as to how pipes can be used to pass standard output as the arguments for other commands.
For example, consider this case:
ls | grep Hello
The structure of grep follows the pattern: grep SearchTerm PathOfFileToBeSearched. In the case I have illustrated, the word Hello is taken as the SearchTerm and the result of ls is used as the file to be searched. But what if I want to switch it around? What if I want the standard output of ls to be the SearchTerm, with the argument following grep being PathOfFileToBeSearched? In a general sense, I want to have control over which argument the pipe fills with the standard output of the previous command. Is this possible, or does it depend on how the script for the command (e.g., grep) was written?
Thank you so much for your help!
grep itself will be built such that if you've not specified a file name, it will open stdin (and thus get the output of ls). There's no real generic mechanism here - merely convention.
If you want the output of ls to be the search term, you can do this via the shell. Make use of a subshell and substitution thus:
$ grep $(ls) filename.txt
In this scenario ls is run in a subshell, and its stdout is captured and inserted in the command line as an argument for grep. Note that if the ls output contains spaces, this will cause confusion for grep.
There are basically two options for this: shell command substitution and xargs. Brian Agnew has just written about the former. xargs is a utility which takes its stdin and turns it into arguments of a command to execute. So you could run
ls | xargs -n1 -J % grep -- % PathOfFileToBeSearched
and it would, for each file output by ls, run grep -e filename PathOfFileToBeSearched to grep for the filename output by ls within the other file you specify. This is an unusual xargs invocation; usually it's used to add one or more arguments at the end of a command, while here it should add exactly one argument in a specific place, so I've used -n and -J arguments to arrange that. The more common usage would be something like
ls | xargs grep -- term
to search all of the files output by ls for term. Although of course if you just want files in the current directory, you can this more simply without a pipeline:
grep -- term *
and likewise in your reversed arrangement,
for filename in *; do
grep -- "$#" PathOfFileToBeSearched
done
There's one important xargs caveat: whitespace characters in the filenames generated by ls won't be handled too well. To do that, provided you have GNU utilities, you can use find instead.
find . -mindepth 1 -maxdepth 1 -print0 | xargs -0 -n1 -J % grep -- % PathOfFileToBeSearched
to use NUL characters to separate filenames instead of whitespace
i have a directory called testDir and it contains 1000 file, some of them contains telephone numbers and some of them doesn't, the telephone number format is "12-3456789"
how to get the number of files that contains telephone numbers ?
EDIT: i am not familiar with unix, so i couldn't answer the question.
A simple solution could be:
grep -lE "[0-9]{2}-[0-9]{7}" * | wc -l
EDIT:
grep seeks for pattern in files.
-E activates regular expressions (you could use egrep instead)
-l filters grep results, only the file name will be printed
wc counts
-l lines will be count (-w counts words, but it could provide incorrect results in case of spaces in filenames)