ZSH: Read command fails within bash function “read:1: -p: no coprocess”

bash-scriptingzsh

Edit:

Seems to work within bash. It appears the problem is related to zsh. If there is a better site to post this issue on let me know.

I am writing a simple script that creates a series of directories. I want the user to give a confirmation before I do so. I'm using the following as a basis, but cannot seem to get it to run within a bash function. If I put it outside of a function it works fine. Here is an isolated example:

read.sh

#!/bin/bash
test() {
  read -p "Here be dragons. Continue?" -n 1 -r
  if [[ $REPLY =~ ^[Yy]$ ]]
  then
    echo "You asked for it..."
  fi
}

code from this SO post.

Sourcing the file and/or test results in the following error: read:1: -p: no coprocess. Same output when I place it in my .bashrc

Edit::

@hennes

  1. I want the function to be in a config file, so I can call it from whatever directory (ideally my .bashrc or .zshrc)
  2. I've corrected the formatting of the first commented line. Problem still exists in zsh
  3. Bash version is 3.2, but you've helped me figure out that the problem is with zsh and not bash.

Best Answer

The –p option doesn’t mean the same thing to bash’s read built-in command and zsh’s read built-in command.  In zsh’s read command, –p means –– guess –– “Input is read from the coprocess.”  I suggest that you display your prompt with echo or printf.

You may also need to replace –n 1 with –k or –k 1.

The zsh equivalent of bash's read -p prompt is

read "?Here be dragons. Continue?"

Anything after a ? in the first argument is used as the prompt string.

And of course you can specify a variable name to read into (and this may be better style):

read "brave?Here be dragons. Continue?"
if [[ "$brave" =~ ^[Yy]$ ]]
then
    ...
fi

(Quoting shell variables is generally a good idea, too.)

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