using a Bash variable in place of a file as input

问题:

I have an executable that is used in a way such as the following:

executable -v -i inputFile.txt -o outputFile.eps

In order to be more efficient, I want to use a Bash variable in place of the input file. So, I want to do something like the following:

executable -v -i ["${inputData}"] -o outputFile.eps

Here, the square brackets represent some clever code.

Do you know of some trick that would allow me to pipe information into the described executable in this way?

Many thanks for your assistance

回答1:

You can use the construct

<(command)

to have bash create a fifo with commands output for you. So just try:

-i <(echo "$inputData")

回答2:

Echo is not safe to use for arbitrary input.

To correctly handle pathological cases like inputdata='\ntest' or inputdata='-e', you need

executable -v -i <(cat <<< "$inputData")

In zsh, the cat is not necessary

---

Edit: even this adds a trailing newline. To output the exact variable contents byte-by-byte, you need

executable -v -i <(printf "%s" "$inputData")

回答3:

Note: zsh only:

To get a filename containing the contents of ${variable}, use:

<(<<<${variable})

Note:

• <<<${variable} redirects STDIN to come from ${variable}
• <<<${variable} is equivalent to (but faster than) cat <<<${variable}

So for the OP's case:

executable -v -i <(<<<${inputData}) -o outputFile.eps

回答4:

Assuming that your executable file name is exec and defined variable name is x and you want to pass the value of x as input to the executable. Then do this:

$ echo $x > stdin && ./exec < stdin

this will do the job

回答5:

executable -v -i <<<"${inputData}" -o outputFile.eps

will do the trick in bash.