Bash shell read error: 0: Resource temporarily unavailable

bashshell

When writing a bash script. Sometimes you are running a command which opens up another program such as npm, composer.. etc. But at the same time you need to use read in order to prompt the user.

Inevitable you hit this kind of error:

read: read error: 0: Resource temporarily unavailable

After doing some research there seems to be a solution by piping the STDIN of those programs which manipulate the STDIN of your bash script to /dev/null.

Something like:

npm install </dev/null

Other research has shown it has something to do with the fact that the STDIN is being set to some sort of blocking/noblocking status and it isn't being reset after the program finishes.

The question is there some sort of fool proof, elegant way of reading user prompted input without being affected by those programs that manipulate the STDIN and not having to hunt which programs need to have their STDIN redirected to /dev/null. You may even need to use the STDIN of those programs!

Best Answer

Usually it is important to know what input the invoked program expects and from where, so it is not a problem to redirect stdin from /dev/null for those that shouldn't be getting any.

Still, it is possible to do it for the shell itself and all invoked programs. Simply move stdin to another file descriptor and open /dev/null in its place. Like this:

exec 3<&0 0</dev/null

The above duplicates stdin file descriptor (0) under file descriptor 3 and then opens /dev/null to replace it.

After this any invoked command attempting to read stdin will be reading from /dev/null. Programs that should read original stdin should have redirection from file descriptor 3. Like this:

read -r var 0<&3

The < redirection operator assumes destination file descriptor 0, if it is omitted, so the above two commands could be written as such:

exec 3<&0 </dev/null
read -r var <&3

Related Solutions

Bash – How to check if a directory exists in a Bash shell script

To check if a directory exists in a shell script, you can use the following:

if [ -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY exists.
fi

Or to check if a directory doesn't exist:

if [ ! -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY doesn't exist.
fi

However, as Jon Ericson points out, subsequent commands may not work as intended if you do not take into account that a symbolic link to a directory will also pass this check. E.g. running this:

ln -s "$ACTUAL_DIR" "$SYMLINK"
if [ -d "$SYMLINK" ]; then 
  rmdir "$SYMLINK" 
fi

Will produce the error message:

rmdir: failed to remove `symlink': Not a directory

So symbolic links may have to be treated differently, if subsequent commands expect directories:

if [ -d "$LINK_OR_DIR" ]; then 
  if [ -L "$LINK_OR_DIR" ]; then
    # It is a symlink!
    # Symbolic link specific commands go here.
    rm "$LINK_OR_DIR"
  else
    # It's a directory!
    # Directory command goes here.
    rmdir "$LINK_OR_DIR"
  fi
fi

Take particular note of the double-quotes used to wrap the variables. The reason for this is explained by 8jean in another answer.

If the variables contain spaces or other unusual characters it will probably cause the script to fail.

Bash – Assigning default values to shell variables with a single command in bash

Very close to what you posted, actually. You can use something called Bash parameter expansion to accomplish this.

To get the assigned value, or default if it's missing:

FOO="${VARIABLE:-default}"  # If variable not set or null, use default.
# If VARIABLE was unset or null, it still is after this (no assignment done).

Or to assign default to VARIABLE at the same time:

FOO="${VARIABLE:=default}"  # If variable not set or null, set it to default.

Best Answer

Related Solutions

Bash – How to check if a directory exists in a Bash shell script

Bash – Assigning default values to shell variables with a single command in bash

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