How to make a shell script take an argument passed in the command line?

Question

I have a shell script (count_reviews.sh) that contains the following:

#!/bin/bash
grep -c "Author" "#1"

I have tried using "$1" as I found that this takes the first argument in the command line but I still get the error mentioned below. I have also used chmod +x to make my file executable.

The script counts the number of times "Author" appears in the file. I am required to be able to make the command line take an input of "% ./count_reviews.sh hotel_72572.dat" where hotel_72572.dat is an example file name. The number of times author appears will then be printed out underneath. When I do this however, I am getting an error -bash: fg: %: no such job. What is causing this and how do I fix it? My count_reviews.sh file is in the same directory as all of my hotel data files if that matters.

The `%` is a common default prompt used in the c-shell (csh), I'm wondering if that's where it came from (others, like bash, use `$` for ordinary users, `#` for root). — cdarke, Feb 15 '16 at 22:58

score 1 · Accepted Answer · answered Feb 15 '16 at 20:25

1

You should indeed use $1, but when you try to run your script just use ./your_script.sh your_argument don't add the % in the beginning.

answered Feb 15 '16 at 20:25

haltode

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But my specification requires it to be there? What function does it hold? – John Smith Feb 15 '16 at 20:41
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The `%` is used in job control : https://en.wikipedia.org/wiki/Job_control_%28Unix%29#Commands. That's why you get the `fg: no current job` error. – haltode Feb 15 '16 at 20:46
and how could I alter this to make it so that it checks for "Author" in all files in a directory? – John Smith Feb 15 '16 at 20:53
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What do you mean by that ? Do you want grep to recursively check every file in a given directory ? If so, just add the argument `-r` to the grep command. – haltode Feb 15 '16 at 20:55
In place of the -c I assume? – John Smith Feb 15 '16 at 21:08
No you can use both of them, like this `grep -cr "Author" $1`. – haltode Feb 16 '16 at 06:32

How to make a shell script take an argument passed in the command line?

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