Bash: recursively find php files, preprocess with php, capture output to html files

Question

I'm looking to scan a directory for files named *.php, and then execute each one with the PHP preprocessor, and capture the output to the same location as the file, with the same filename, only .html extension instead of .php.

so far I have:

find . -type f -name "*.php" | xargs -I {} php {} >> '{ dirname {}; basename {} .php; }'

thanks for your help on this bash puzzle !

imho its what bash does best. Then again the beer tastes the same if you open it with a bottle opener or a hammer. — ficuscr
– ficuscr, Commented Sep 20, 2012 at 22:12

ficuscr · Accepted Answer · 2012-09-20 22:24:32Z

1

Think this works...

#!/bin/bash
FILES=/var/www/*.php
for f in $FILES
do
  echo "Running script $f..."
  php $f > $f.out
done

Maybe needing full path to bin where your php lives.

EDIT:

php $f > ${f%.php}.html

edited Sep 20, 2012 at 22:24

answered Sep 20, 2012 at 21:41

ficuscr

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3 Comments

Marc B Over a year ago

it'll be doing php whatever.php > whatever.php.out. need to strip off the extension for the .out part.

ficuscr Over a year ago

True. But doesn't bother me or cat. But now I do see the OP wants name.html

cars Over a year ago

yes, I'm aiming for the extension .html INSTEAD of .php, preserving directory structure to output folder.

Marc B · Accepted Answer · 2012-09-20 21:42:31Z

0

You'd be better off with using -exec directly in find. xargs is for when you need to pass bulk parameters, while you're looking for something that's more per-file. e.g.

find ... -exec "php {} > $(basename {}).html;"

or something similar. Been a while since I've had to use bash, so that almost certainly won't work, but should get you started.

answered Sep 20, 2012 at 21:42

Marc B

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