How can I check in bash if a shell is running in interactive mode?

Solution 1:

According to man bash:

PS1 is set and $- includes i if bash is interactive, allowing a shell script or a startup file to test this state.

So you can use:

if [[ $- == *i* ]]
then
    do_interactive_stuff
fi

Also:

When an interactive shell that is not a login shell is started, bash reads and executes commands from /etc/bash.bashrc and ~/.bashrc, if these files exist.

So ~/.bashrc is only sourced for interactive shells. Sometimes, people source it from ~/.bash_profile or ~/.profile which is incorrect since it interferes with the expected behavior. If you want to simplify maintenance of code that is common, you should use a separate file to contain the common code and source it independently from both rc files.

It's best if there's no output to stdout from login rc files such as ~/.bash_profile or ~/.profile since it can interfere with the proper operation of rsync for example.

In any case, it's still a good idea to test for interactivity since incorrect configuration may exist.

Solution 2:

the test tool can check for this (from the man page):

 -t FD          True if FD is opened on a terminal.

So you can use for example:

 if [ -t 0 ] ; then
    echo stdin is a terminal
   .....
 fi

or

if [ -t 1 ] ; then
    echo stdout is a terminal
 fi

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Solution 3:

Use:

if tty -s; then echo interactive; fi

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Solution 4:

I typically look at the output of the program tty.

If you're on a tty, it will tell you which tty you're on. If you're not in interactive mode, it will typically tell you something like "not a tty".

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Solution 5:

This is how red hat does it... Guessing it's the fastest way...

if [ "$-#*i" == "$-" ]; then

It means get the bash parameters, do a substring removal, shortest possible, match everything between the beginning and i. Then check if it's the same as the original bash parameters.

Check you did your job by connecting to the user using sftp, it will fail if non interactive sessions have output