parent script name in sourced child

Question

I am having two scripts: parent and child.

>>cat fake_parent.sh

#!/bin/sh my_child(){ echo "I am parent " ./fake_child.sh }

>>cat fake_child.sh

#!/bin/sh echo "I am child" echo "I want to know my parent ...i.e fake_parent.sh here"

I am sourcing fake_parent.sh and calling my_child function; which should print "fake_parent.sh" at the end of the echo command. I have tried $BASH_SOURCE and $0 but both are giving me "fake_child.sh"

Please help.

Please take a look at this question: stackoverflow.com/questions/20572934/… IMHO this should be what you need. — ahuemmer
– ahuemmer, Commented Jul 14, 2020 at 10:19
Does this answer your question? Get the name of the caller script in bash script — Cristian Ramon-Cortes
– Cristian Ramon-Cortes, Commented Jul 14, 2020 at 10:39
@nikitaB : Why are you printing $BASH_SOURCE in fake_parent.sh and not in fake_child.sh? — user1934428
– user1934428, Commented Jul 14, 2020 at 10:42
@ahuemmer PARENT_COMMAND=$(ps -o comm= $PPID) is printing "bash" and PARENT_COMMAND=$(ps -o args= $PPID) is printing "-bash" — nikitaB
– nikitaB, Commented Jul 14, 2020 at 10:54

UtLox · Accepted Answer · 2020-07-14 12:47:48Z

You can try this:

# fake_parent.sh my_child(){ echo "I am parent " export PARENT="$(basename $0)" ./fake_child.sh } # fake_child.sh #!/bin/sh echo "I am child" echo "I want to know my parent ... $PARENT"

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