Bash script variable with a space

RonLat

7/8/24, 1:26 PM

I try to write a bash script for copying data from one folder to another.

Source="/My Images/"
Target=/Transfer/Media/
rsync -avu --progress $Source $Target

The problem is, that the script can not find the folder "/My Images/", because there is a space between those two words.

How do I tell the variable Source to accept this folder name?

1 + 1

bash

scripts

Raffa

7/8/24, 1:30 PM

Double quote `"$Source"` ... And as a good measure `"$Target"` as well as you should get into the habit of quoting parameters in your scripts ... i.e. `rsync -avu --progress "$Source" "$Target"`

Score:3

Ubuntu

Raffa

7/8/24, 1:47 PM

How do I tell the variable Source to accept this folder name?

It's not assigning that parameter:

Source="/My Images/"

but rather its expansion:

rsync -avu --progress $Source $Target

As it appears you already know that quoting the path/string "/My Images/" is necessary to handle the space in it.

You will also need to quote the parameter/variable containing it when you use it in the command to prevent word splitting by the shell like so:

rsync -avu --progress "$Source" "$Target"

You should get into the habit of double quoting all parameters in your scripts.

I assume you are using rsync locally on the same machine, but if you are using it to synchronize files to a remote machine using ssh for example, then you might need to double quote/escape ... See for example: Using scp to copy a remote file containing spaces requires double space escaping. Why?

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