CODEWITHSUNDEEP
bashvariablescommand-lineescaping
Dan
Dan

Reputation: 13

bash - escape variable SSH command without altering it

I am running a command in a bash script that uses the content of a file as an argument. Here a simplified example:

USER=$(<$fileA)
PASSWORD=$(<$fileB)
wget -u $USER -p $PASSWORD

However, the content of the file can be set to any value by the user. I assume I need to escape this, but I don't know how to do that, especially in a way that $PASSWORD will NOT be altered. Otherwise the password will obviously not work.

Thanks

Upvotes: 1

Views: 284

Answers (1)

vdavid
vdavid

Reputation: 2544

You probably simply need to add quotes around the variables.

wget -u "$USER" -p "$PASSWORD"

If you happen to capture a newline at the end of the variable, you may want to strip it with:

wget -u "${USER%$'\n'}" -p "${PASSWORD%$'\n'}"

The ${var} syntax is equivalent to $var but curly braces allows to manipulate variables with bash. By adding %$'\n' before the ending curly brace, the % means to remove the text $'\n', which means removing the trailing newline, if any.

In the worst case scenario, you may get newline and carriage returns (probably if the files were generated by a text editor in Windows), but you can also deal with it:

wget -u "${USER%%[$'\n'$'\r']*}" -p "${PASSWORD%%[$'\n'$'\r']*}"

This time around, the searched text is [$'\n'$'\r']* which is means "the first newline (\n) or carriage return (\r) and everything after" and the %% operator means "remove the longest sequence at the end".

Upvotes: 2

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