unix shell while loop doesn't see the variable

Question

do you know why in this command, $x is visible

ls -d */ $1 | while read x; do echo $x; done

return the sub directories in the current directory.

However, when I put in another command, $x seems to have no value

ls -d */ $1 | while read x; do ls -lat /order | grep $x; done

return nothing

Well, the second loop lists the files in /order directory, which is not self-evidently related to */ or $1. Did you mean to use something else in place of /order? — Jonathan Leffler
– Jonathan Leffler, Commented Feb 24, 2016 at 20:14
My intention is to use the result from the ls -d */, say 123, and ls -lat /order | grep 123 — timpham
– timpham, Commented Feb 24, 2016 at 20:19
Is there a file /order/123? Or a file in /order with 123 in any of the characteristics shown by ls -lat? (length including digits 123, mainly). If not, the grep won't find anything. — Jonathan Leffler
– Jonathan Leffler, Commented Feb 24, 2016 at 20:21
Oh I see it now. the outcome of ls -d */ contains the / character, so it is 123/. Of course when I grep for 123/ it won't find anything — timpham
– timpham, Commented Feb 24, 2016 at 20:23
Each pipe | separates commands that are run in their own separate subshells. You cannot split a loop across two separate processes. — David C. Rankin
– David C. Rankin, Commented Feb 24, 2016 at 20:51

chenyb999 · Accepted Answer · 2016-02-24 23:23:29Z

1

there is a / at the end of output before the pipe, you should filter it.

ls -d */ $1 | while read xx; do ls -l /order |grep `echo ${xx}|sed  's/.$//'`  ; done

answered Feb 24, 2016 at 23:23

chenyb999

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1 Comment

ls -d */ $1 | tr / ' ' | while read x; do ls -l /order | grep $x; done