I want to replace the version in my code using git rev-parse HEAD with template string %VERSION% in a source file.
For simplicity I will use date as version command in this question.
Given test.txt
$ echo "This is test-%VERSION%." > test.txt
$ cat test.txt
This is test-%VERSION%.
Expect
This is test-Sat Dec 2 16:48:59 +07 2017.
These are failed try
$ echo "This is test-%VERSION%." > test.txt
$ sed -i 's/%VERSION/`date`/' test.txt && cat test.txt
This is test-`date`%.
$ echo "This is test-%VERSION%." > test.txt
$ DD=`date` sed -i 's/%VERSION/$DD/' test.txt && cat test.txt
This is test-$DD%.
$ echo "This is test-%VERSION%." > test.txt
$ DD=`date` sed -i "s/%VERSION/$DD/" test.txt && cat test.txt
This is test-%.
Do I really need to use xargs ?
You can embed $(...) within double-quotes, but not in single-quotes:
sed -i "s/%VERSION%/$(date)/" test.txt && cat test.txt
(Same as `...` but you shouldn't use that obsolete syntax, $(...) is better.)
Btw, for testing purposes, it's better to use sed without -i,
so the original file is not modified:
sed "s/%VERSION%/$(date)/" test.txt
As a side note, and this is a completely different discussion, but worth mentioning here. This may look like it should work but it doesn't, and you may wonder why:
DD=$(date) sed -i "s/%VERSION%/$DD/" test.txt && cat test.txt
Why it doesn't work?
Because the $DD embedded in the "..." is evaluated at the time the command is executed.
At that time the value of DD is not set to the output of $(date).
In the "..." it will have whatever value it had before executing the command.
For the sed process, the value DD with the output of $(date) is visible,
but sed doesn't use that, because why would it.
The "..." passed to sed is evaluated by the shell, not by sed.