CODEWITHSUNDEEP
linuxbash
Mike C
Mike C

Reputation: 65

Linux: read a variable as a file

I have a directory called A , and I would like to know the modification date of one file/directory inside. Here is my code

#!/bin/bash

find A | sort -d
var="$(head -2 | tail -1)"
echo "$var"
date -r $var '+%S'

That's basically how I want my code to be , when I do that, it only shows what's inside my directory A ( the find command), the echo doesn't work , date neither. I have a message saying : date +%s no file or directory of this type.

I've seen some questions about using a variable as a file, but in my case I don't see anything that could cause a problem (like a use of slash). So I wanted to know if there is something wrong.

Upvotes: 1

Views: 139

Answers (1)

Jay jargot
Jay jargot

Reputation: 2868

find A | sort -d print sorted list of files to the standard output.

head -2 | tail -1 makes head wait for data from the standard input.

With nothing coming from the standard input, I think the script should hang forever (?)

var is probably an empty string. So echo print an empty string (nothing), and a new line.

Whatever is stored in var, the filename is not found on the disk. This is why date generated this error.

Give a try to this:

#!/bin/bash --

 var="$(find A | sort -d | head -2 | tail -1)"
 printf "%s " "$var"
 date -r "$var" '+%s'

find can print last modification time of the files with -printf and %T.

Give a try to this:

find A -printf "%p %T@\n" | sort -d | awk 'NR==2{print ; quit}'

-printf "%p %T@\n": prints the filename + (space char) + last modification time of the file as seconds since Jan. 1, 1970, 00:00 GMT, with fractional part.

awk 'NR==2{print ; quit} prints only the second line.

Upvotes: 6

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