Replace file if and only if stdin differs from file contents

Question

I am looking for a program that takes a file name (e.g. as an argument), reads from stdin into a buffer, and compares this with the contents of the named file.

If the contents differ, the program (ideally atomically) overwrites the file with the contents of the input buffer.

Otherwise, it leaves the file unchanged - no write occurs, no change of last-modified timestamp, and without generating a temporary file.

Example Usage: echo "test" | myprogram myfile.txt

Does this program exist as a common Unix command line tool? Is there some nice way to achieve this tersely in the shell without a temporary file? (rsync do otherwise)

Hauke Laging · Accepted Answer · 2017-08-07 07:06:46Z

1

#! /bin/bash

TMP_FILE=/tmp/myprogram.$$.tmp
COMP_FILE="$1"

test -f "$COMP_FILE" && test -r "$COMP_FILE" && test -w "$COMP_FILE" || exit 2

cat >"$TMP_FILE"
cmp "$COMP_FILE" "$TMP_FILE" || cp "$TMP_FILE" "$COMP_FILE"
rm "$TMP_FILE"

answered Aug 7, 2017 at 7:06

Hauke Laging

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Gohu · Accepted Answer · 2017-08-07 09:54:27Z

0

As far as I know there is not an existing tool that does what you want. The following script myprogram should work:

#!/bin/bash

var="$(cat)"
md5_stdin=$(echo "$var" | md5sum | cut -d" " -f 1)
md5_file=$(md5sum myfile.txt | cut -d" " -f1)
[[ "$md5_stdin" != "$md5_file" ]] && echo "$var" > myfile.txt

answered Aug 7, 2017 at 9:54

Gohu

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