CODEWITHSUNDEEP
bash
Woodsman
Woodsman

Reputation: 1179

Why separate declaration from export of variables in bash scripts?

I have a bash script like

#!/bin/bash
DATABASE_NAME=my_database
export DATABASE_NAME
run_some_other_command

They first declare the variable and set it to a value, then in a separate line, they export it. Personally, I like to just do it in one line like:

export DATABASE_NAME=my_database

Is there some style rule against this? I've seen the declaration and export be separated by others, but never knew why.

If it helps, this script runs on Linux Mint, but could run on other Linux's or even a Mac.

Upvotes: 3

Views: 715

Answers (1)

Léa Gris
Léa Gris

Reputation: 19545

Because export is a command; when assignment comes from a command or sub-shell output, its own return code will override/mask the return code of the assigning command:

# the return-code of the getDBName function call
# is masked by the export command/statement
export DATABASE_NAME=$(getDBName)

# Separate export
export DATABASE_NAME

# Allow capture and use of the getDBName return code
if ! DATABASE_NAME=$(getDBName)
then printf 'Failed getDBName with RC=%d\n' "$?" >&2
fi

Upvotes: 5

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