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)
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 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".
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 file and i want to sort it according to a word and to remove the special characters.
The grep command is used to search for the characters
-b Display the block number at the beginning of each line.
-c Display the number of matched lines.
-h Display the matched lines, but do not display the filenames.
-i Ignore case sensitivity.
-l Display the filenames, but do not display the matched lines.
-n Display the matched lines and their line numbers.
-s Silent mode.
-v Display all lines that do NOT match.
-w Match whole word
but
How to use the grep command to do the file sort and remove the special character and number.
grep searches inside all the files to find matching text. It doesn't really sort and it doesn't really chop and change output. What you want is probably to use the sort command
sort <filename>
and the output sent to either the awk command or the sed command, which are common tools for manipulating text.
sort <filename> | sed 's/REPLACE/NEW_TEXT/g'
something like above I'd imagine.
The following command would do it.
sort FILE | tr -d 'LIST OF SPECIAL CHARS' > NEW_FILE
according to the manual for grep,
-l, --files-with-matches
Suppress normal output; instead print the name of each input
file from which output would normally have been printed. The
scanning will stop on the first match.
grep -l, this seems fine in that when a match is found, the file name containing the match is echoed.
However when i do a grep -ln, grep echoes every line of the occurrence.
Does grep -l really mean to stop when the first occurrence of the match is found and stop scanning, while grep -ln will ignore the -l flag?
Those options are incompatible. Use grep -Hnm 1 if you want to display the line number of the first match (and only the first match) in each file.
-H, --with-filename
Print the filename for each match.
-n, --line-number
Prefix each line of output with the line number within its input file.
-m NUM, --max-count=NUM
Stop reading a file after NUM matching lines.