CODEWITHSUNDEEP
bashvariablescommand
EST
EST

Reputation: 517

BASH Command save to variable with a variable in the command

I have a command I want to run first I ran another command to get a directory which is saved in a variable: 

path_white="/sys/block/sdb"

Then I want to run another command using this variable and store the output in a variable. I get errors and don't know what I am doing wrong. Any help will be appreciated. 

path_pci_white=$(ll $path_white | xargs | cut -d / -f 8 | cut -b 6-13)

it seems that it is not running the entire command below is the error

/sys/block/sdb : is a directory

when i run 

ll /sys/block/sdb | xargs | cut -d / -f 8 | cut -b 6-13

in the terminal i get what i want output I just want to use a variable and put the output into a variable

Thanks

Upvotes: 0

Views: 418

Answers (2)

John Kugelman
John Kugelman

Reputation: 361615

ll is an alias for ls -l, and aliases aren't defined in shell scripts. Use an explicit ls -l instead.

Upvotes: 2

John Zwinck
John Zwinck

Reputation: 249153

There should not be a pipe after xargs. xargs takes as arguments the command it will run. Otherwise there is no point to it.

Upvotes: 1

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