Not sure if I should put this on stackoverflow or unix.stackexchange but I found some similar questions here, so here it goes.
I'm trying to create a script to be called by .bashrc that allows me to select one of two options based on a single keystroke. That wouldn't be hard normally but I want the two keys corresponding to the two options to be space and enter.
Here's what I got so far:
#!/bin/bash
SELECT=""
while [[ "$SELECT" != $'\x0a' && "$SELECT" != $'\x20' ]]; do
echo "Select session type:"
echo "Press <Enter> to do foo"
echo "Press <Space> to do bar"
read -s -N 1 SELECT
echo "Debug/$SELECT/${#SELECT}"
[[ "$SELECT" == $'\x0a' ]] && echo "enter" # do foo
[[ "$SELECT" == $'\x20' ]] && echo "space" # do bar
done
The following output is what I get if I press enter, space, backspace and x:
:~$ bin/sessionSelect.sh
Select session type:
Press <Enter> to start/resume a screen session
Press <Space> for a regular ssh session
Debug//0
Select session type:
Press <Enter> to start/resume a screen session
Press <Space> for a regular ssh session
Debug//0
Select session type:
Press <Enter> to start/resume a screen session
Press <Space> for a regular ssh session
Debug//1
Select session type:
Press <Enter> to start/resume a screen session
Press <Space> for a regular ssh session
Debug/x/1
So both enter and space result in an empty SELECT. No way to distinguish the two. I tried to add -d 'D' to the read options, but that didn't help. Maybe someone can point me in the right direction.
The bash version would be 4.2 btw.
Try setting the read delimiter to an empty string then check the builtin $REPLY variable:
read -d'' -s -n1
For some reason I couldn't get it to work specifying a variable.
#!/bin/bash
SELECT=""
# prevent parsing of the input line
IFS=''
while [[ "$SELECT" != $'\x0a' && "$SELECT" != $'\x20' ]]; do
echo "Select session type:"
echo "Press <Enter> to do foo"
echo "Press <Space> to do bar"
read -s -N 1 SELECT
echo "Debug/$SELECT/${#SELECT}"
[[ "$SELECT" == $'\x0a' ]] && echo "enter" # do foo
[[ "$SELECT" == $'\x20' ]] && echo "space" # do bar
done
There are a couple of things about read
that are relevant here:
- It reads a single line
- The line is split into fields as with word splitting
Since you're reading one character, it implies that entering Enter would result into an empty variable.
Moreover, by default rules for word splitting, entering Space would also result into an empty variable. The good news is that you could handle this part by setting IFS
.
Change your read
statement to:
IFS= read -s -n 1 SELECT
and expect a null string instead of $'\x0a'
when entering Enter.
来源:https://stackoverflow.com/questions/22842896/bash-scripting-read-single-keystroke-including-special-keys-enter-and-space