Extract CSV column containing blank space to Bash array

Question

I would like to export a CSV bookmarks file as

Physics,physics.stackexchange.com Stack Overflow,stackoverflow.com Unix & Linux,unix.stackexchange.com 

into two Bash arrays. Each array would contain one column of the file. The first array would be

"Physics" "Stack Overflow" "Unix & Linux" 

I have troubles to deals with the blank spaces. I tried to use cut like so

declare names=($(cut -d ',' -f1 ~/bookmarks.csv)) 

which generates an array wherein all the words are separated. As an example, echo ${names[1]} returns

Stack 

and not

Stack Overflow 

Answer 1

Stack Overflow has a space so it's being treated as two different variables when assigned to the array. In BASH and even other shells like KSH, etc, variable names cannot have spaces in them so a variable like Stack Overflow wouldn't work even if it were surrounded with single or double quotes.

You can see this with the following commands:

export Stack Overflow=name echo $Stack Overflow 

The output will be Overflow as the argument for echo is, in effect, $Stack, which hasn't been declared or assigned a value, and the word Overflow.

echo $Overflow 

That will output name as the export command assigned that value to the variable Stack with Overflow=name.

If you try either of these:

export "Stack Overflow'=name export 'Stack Overflow'=name 

You'll get the error -bash: export: Stack Overflow=name': not a valid identifier or -bash: export: Stack Overflow=name': not a valid identifier.

One thing that you can do instead is to place underscores between the two strings:

export Stack_Overlow=name 

That way, echo $Stack_Overflow will output name. That's going to require you to do more work because what you have in the CSV file contains spaces.