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.
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 variable name that appears in multiple locations of a text file. This variable will always start with the same string but not always end with the same characters. For example, it can be var_name or var_name_TEXT.
I'm looking for a way to extract the first occurrence in the text file of this string starting with var_name and ending with , (but I don't want the comma in the output).
Example1: var_name, some_other_var, another_one, ....
Output: var_name
Example2: var_name_TEXT, some_other_var, another_one, ...
Output: var_name_TEXT
grep -oPm1 '\bvar_name[^, ]*(?=,)' file | head -1
match and output only variables starting with var_name and ending with comma, do not include comma in the output, quit after the first line of match and pick the first match on that line (if there are more than one)
ps. you have to include space in the regex as well.
I suggest with GNU grep:
grep -o '\bvar_name[^,]*' file | head -n 1
All you need is (GNU awk):
$ awk 'match($0,/\<var_name[^,]*/,a){print a[0]; exit}' file
var_name_TEXT
To print the field only (i.e., var_name or var_name_TEXT only; not the line containing it) you could use awk:
awk -F, '{for (i=1;i<=NF;i++) if ($i~/^var_name/) print $i}' file
If you actually have spaces before or after the commas (as you show in your example) you can change to awk field separator:
awk -F"[, ]+" '{for (i=1;i<=NF;i++) if ($i~/^var_name/) print $i}' file
You can also use GNU grep with a word boundary assertion:
grep -o '\bvar_name[^,]*' file
Or GNU awk:
awk '/\<var_name/' file
If you want only one considered, add exit to awk or -m 1 to grep to exit after the first match.
How to use "grep" shell command to show specific word from a line starting with a specific word.
Ex:
I want to print a string "myFTPpath/folderName/" from the line starting with searchStr in the below mentioned line.
searchStr:somestring:myFTPpath/folderName/:somestring
Something like this with awk:
awk -F: '/^searchStr/{print $3}' File
From all the lines starting with searchStr, print the 3rd field (field seperator set as :)
Sample:
AMD$ cat File
someStr:somestring:myFTPpath/folderName/:somestring
someStr:somestring:myFTPpath/folderName/:somestring
searchStr:somestring:myFTPpath/folderName/:somestring
someStr:somestring:myFTPpath/folderName/:somestring
AMD$ awk -F: '/^searchStr/{print $3}' File
myFTPpath/folderName/
Remember that grep isn't the only tool that can usefully do searches.
In this particular case, where the lines are naturally broken into fields, awk is probably the best solution, as #A.M.D's answer suggests.
For more general case edits, however, remember sed's -n option, which suppresses printing out a line after edits:
sed -n 's/searchStr:[^:]*:\([^:]*\):.*/\1/p' input-file
The -n suppresses automatic printing of the line, and the trailing /p flag explicitly prints out lines on which there is a substitution.
This matching pattern is fiddly – use awk in this fielded case – but don't forget sed -n.
You could get the desired output with grep itself but you need to enable -P and -o parameters.
$ echo 'searchStr:somestring:myFTPpath/folderName/:somestring' | grep -oP '^searchStr:[^:]*:\K[^:]*'
myFTPpath/folderName/
\K discards the characters which are matched previously from printing at the final leaving only the characters which are matched by the pattern exists next to \K. Here we used \K instead of a variable length positive lookbehind assertion.
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