i want to save current day file daily
for this is am using below command.
in order to script this part, i am saving date output in a file using below command
thru below command i want to list the file of specific date but below command is listing all the content.
o/p
What you are doing is cumbersome - so, consider using the find command.
To directly answer what I can:
This gives the most recent file name only , assuming you have already executed a cd command to be in the right directory --
I do not understand your code very well - so this is the only contribution I can make at this point.
Trying to understand and optimize your approach - do consider jim mcnamaras comments, though - I think this might do what you want:
Please note that
- no grep is needed when you're using awk anyhow
- no awk for date is needed, use its format option
- ls does not need the -rt options i your case
The error in post#1 is, you're writing the results to a file but try to expand a variable of the same name which is empty.
first of all i am very sorry if i was not able to explain my question
here i am again explaining it.
My motive is to save all the file which are generating daily.
as i do not want to do this manually i am trying to write a script for this.
Now the command which i am using here, to list current day file is mentioned as below. asuming current date is 10th May 2015
using above command i can list all the files of 10th May.
now to script this here i am saving date in file using below command
now here i am checking if it is working or not for this here i am using below command
but what i see here is that it is listing all the files.
what i expect from the command to do is to list only 10th may file.
here what i am expecting from command No.3 is, to work similar to the below command
ls -lrt | grep "May 10"
I hope this time i am able to explain my concern.
regards
scriptor
---------- Post updated at 08:50 PM ---------- Previous update was at 08:36 PM ----------
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