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I am beginner in unix programming and a way to automate my work
I want to run a list a grep commands and get the output of all the grep command in a in a single delimited file .

i am using the following bash script. But it's not working .

Mockup sh file:

!/bin/sh
grep -l abcd123
grep -l abcd124
grep -l abcd125

and while running i used the following command 

$ ./Mockup.sh > output.txt

Is it the right command?

How can I get both the grep command and output in the output file?

how can i delimit the output after each command and result?

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4
  • Things: you need #!/bin/sh to make it work. grep needs at least two arguments: the search string and the file where to look for. grep -l abcd123 needs more info. Commented Nov 5, 2014 at 12:54
  • You need to show us what your input and desired output look like. Commented Nov 5, 2014 at 12:55
  • when i give the following input inside a script' grep -l abcd123, grep -l abcd124' i want to get a output some thing like ' grep -l abcd123 <some kind of delimiter> sample1.txt , grep -l abcd124 <some kind of delimiter> sample2.txt Commented Nov 6, 2014 at 6:50
  • Please help me on this Commented Nov 6, 2014 at 6:53

4 Answers 4

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How can I get both the grep command and output in the output file

You can use bash -v (verbose) to print each command before execution on stderr and it's output will be as usual be available on stdout:

bash -v ./Mockup.sh > output.txt 2>&1

cat output.txt

Working Demo

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5 Comments

@Downvoter: Question was How can I get both the grep command and output in the output file?
Disclaimer: the downvoter is someone else! That said, if I were forced to find a defect in your answer I'd say that Mockup.sh, as posted in the original question, would benefit from some small correction. Ciao
i tried but am getting only the output in that file. The input grep command is not available in the file.
That's not true, if you use bash -v it does print each and every command on stderr. Make sure you copy/paste command as is using > output.txt 2>&1
@Kaleem: I have included a working demo in my answer.
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A suitable shell script could be

#!/bin/sh
grep -l 'abcd123\|abcd124\|abcd125' "$@"

provided that the filenames you pass on the invocation of the script are "well behaved", that is no whitespace in them. (Edit Using the "$@" expansion takes care of generic whitespace in the filenames, tx to triplee for his/her comment)

This kind of invocation (with alternative matching strings, as per the \| syntax) has the added advantage that you have exactly one occurrence of a filename in your final list, because grep -l prints once the filename as soon as it finds the first occurrence of one of the three strings in a file.

Addendum about "$@"

% ff () { for i in "$@" ; do printf "[%s]\n" "$i" ; done ; }
% # NB "a s d" below is indeed "a SPACE s TAB d"
% ff "a s d" "    ert  " '345
345'
[a s    d]
[    ert  ]
[345
345]
%
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1 Comment

+1 but you should be using "$@" instead of $*. Both the different variable name and the double quotes are significant. That way, you no longer have any constraint on what file names you can use with this script.
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cat myscript.sh 
########################
#!/bin/bash
echo "Trying to find the file contenting the below string, relace your string with below string"
grep "string" /path/to/folder/* -R -l
########################

save above file and run it as below 

sh myscript.sh > output.txt

once the command prmpt get return you can check the output.txt for require output.

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0

Another approach, less efficient, that tries to address the OP question

How can I get both the grep command and output in the output file?

% cat Mockup
#!/bin/sh
grep -o -e string1 -e string2 -e string3 "$@" 2> /dev/null | sort -t: -k2  | uniq

Output: (mocked up as well)

% sh Mockup file{01..99}
file01:string1
file17:string1
file44:string1
file33:string2
file44:string2
file48:string2
%

looking at the output from POV of a consumer, one foresees problems with search strings and/or file names containing colons... oh well, that's another Q maybe

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