Unable to do a word count of a file through ssh in unix shell scripting

Question

I need to go another server and perform a word count. Based on the count variable I will perform a if else logic. However i am unable to do a word count and further unable to compare the variable value in if condition.

Error:

wc: cannot open the file v.txt

Script:

#!/bin/bash
ssh u1@s1 "cd ~/path1/ | fgrep-f abc.csv xyz.csv > par.csv | a=$(wc -l par.csv)| if ["$a" == "0"];
then echo "success"
fi"

Answer 1

First, although the wc program is named for 'word count', wc -l actually counts lines not words. I assume that is what you want even though it isn't what you said.

A shell pipline one | two | three runs things in parallel with (only) their stdout and stdin connected; thus your command runs one subshell that changes directory to ~/path1 and immediately exits with no effect on anything else, and at the same time tries to run fgrep-f (see below) in a different subshell which has not changed the directory and thus probably can't find any file, and in a third subshell does the assignment a= (see below) which also immediately exits so it cannot be used for anything.

You want to do things sequentially:

ssh u@h 'cd path1; fgrep -f abc.csv xyz.csv >par.csv; a=$(wc -l par.csv); if [ "$a" == "0" ] ...'
# you _might_ want to use && instead of ; so that if one command fails
# the subsequent ones aren't attempted (and possibly go further wrong)

Note several other important changes I made:

• the command you give ssh to send the remote must be in singlequotes ' not doublequotes " if it contains any dollar as yours does (or backtick); with " the $(wc ...) is done in the local shell before sending the command to the remote

• you don't need ~/ in ~/path1 because ssh (or really sshd) always starts in your home directory

• there is no common command or program fgrep-f; I assume you meant the program fgrep with the flag -f, which must be separated by a space. Also fgrep although traditional is not standard (POSIX); grep -F is preferred

• you must have a space after [ and before ]

However, this won't do what you probably want. The value of $a will be something like 0 par.csv or 1 par.csv or 999 par.csv; it will never equal 0 so your "success" branch will never happen. In addition there's no need to do these in separate commands: if your actual goal is to check that there are no occurrences in xyz.csv of the (exact/non-regexp) strings in abc.csv both in path1, you can just do

ssh u@h 'if ! grep -qFf path1/abc.csv path1/xyz.csv; then echo success; fi'
# _this_ case would work with " instead of ' but easier to be consistent

grep (always) sets its exit status to indicate whether it found anything or not; flag -q tells it not to output any matches. So grep -q ... just sets the status to true if it matched and false otherwise; using ! inverts this so that if grep does not match anything, the then clause is executed.

If you want the line count for something else as well, you can do it with a pipe

'a=$( fgrep -Ff path1/abc.csv path1/xyz.csv | wc -l ); if [ $a == 0 ] ...'

Not only does this avoid the temp file, when the input to wc is stdin (here the pipe) and not a named file, it outputs only the number and no filename -- 999 rather than 999 par.csv -- thus making the comparison work right.