How to use arguments with 'find -exec'?

Question

I'm trying to generate static html files for all php scripts inside dir:

find . -name "*.php" -exec php {} > {}.html \;

But it just dumps all results to {}.html. How to do it properly? And how to remove '.php' from filenames?

The first part is answered here: [*Redirection or pipe inside `find -exec`*](https://superuser.com/q/1411712/432690). — Kamil Maciorowski, Jan 22 '21 at 17:50
I’m voting to close this question because [it was cross-posted to U&L SE](https://unix.stackexchange.com/q/78316/108618) and answered there. Cross-posting is [discouraged](https://meta.stackexchange.com/q/64068/355310). — Kamil Maciorowski, Jan 22 '21 at 18:23

score 0 · Answer 1 · answered Jun 05 '13 at 12:04

0

Got it

find . -name "*.php" | sed "s/\.php//" | xargs -i sh -c 'php {}.php > {}.html'

answered Jun 05 '13 at 12:04

Poma

1 Answers1