创建一个使用cat并具有3个参数的简单bash脚本

Question

I have this bash command I am using to cat the output of two java programs that take 1 parameter each into a file:

cat file.ext | java program1 argument1 | java program2 argument2 > argument3

argument3 is a optional and if it is not provided it should just be file.srtd

How can I create a bash script for this?

Answer 1

You can specify a default value for a variable if it's not set with the syntax ${var:-default} .

java program1 "$1" < file.ext | java program2 "$2" > "${3:-file.srtd}"

You should also remember to quote the variables in case they contain special characters.

Answer 2

In bash, the variable $n is the n ^th argument. The variable $# is the number of arguments given by the user.

So you can do something like this:

#!/bin/bash

arg3=$3
if (( $# < 3 )) ; then
    arg3="file.srtd"
fi;

cat file.ext | java program1 $1 | java program2 $2 > $arg3

You can look at shell parameters and shell parameter expansion for more information.

Answer 3

You could do :

file="$3"
[ -z "$file" ] && file="file.srtd"
#then use `file` for the rest of the program.