BASH Command save to variable with a variable in the command

Question

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

Did you try path_pci_white=$(ll $(path_white) | xargs | cut -d / -f 8 | cut -b 6-13) path_withe between parenthesis ? — Luc M
– Luc M, Commented Feb 9, 2012 at 23:20
when i do that i get a different error such that: path_white: command not found, and ll: command not found — EST
– EST, Commented Feb 9, 2012 at 23:22
That's because Luc wrote $(path_white) when he meant ${path_white} (at least I hope he meant that). $(path_white) tries to run a program called path_white, as you have seen, which does not work. — John Zwinck
– John Zwinck, Commented Feb 9, 2012 at 23:27
@bing281: have you noticed how you have a trailing slash on the path in one case and not the other? Try changing that around and see if it matters. — John Zwinck
– John Zwinck, Commented Feb 9, 2012 at 23:28

John Kugelman · Accepted Answer · 2012-02-09 23:27:06Z

2

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

answered Feb 9, 2012 at 23:27

John Kugelman

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2 Comments

John Zwinck Over a year ago

The OP didn't even state that he was running these things in a script--good psychic debugging!

Barry Jones Over a year ago

Or source your rc file in your script, though I don't think this would be considered best practice.

John Zwinck · Accepted Answer · 2012-02-09 23:19:53Z

1

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

answered Feb 9, 2012 at 23:19

John Zwinck

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2 Comments

EST Over a year ago

I don't actually follow you, however the last command above works perfectly in the terminal. ll /sys/block/sdb | xargs | cut -d / -f 8 | cut -b 6-13

John Zwinck Over a year ago

Oh, then the xargs part is useless and should be removed, as all it is doing is echoing on stdout what it gets on stdin. Can you show us the output of that ll command on your system, and the desired output of the pipeline?

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