在bash中包含和排除globbing模式

Question

I am trying to use the PHPCompatibility standard for PHP CodeSniffer to test a specific set of files.

I'm currently using find to do this.

find ./path1 ./path2 ./path3 ./path4                                                \
    -type f                                                                         \
    -name '*.php'                                                                   \
    -not -path './path4/dontwantthis/*'                                             \
    -exec ./vendor/bin/phpcs                                                        \
    --standard=PHPCompatibility --runtime-set testVersion 5.6 {}                    \;

However, this is really slow and inefficient because the startup for the script runs for every single file.

The phpcs script takes in a path like ./vendor/bin/phpcs --standard=PHPCompatibility --runtime-set testVersion 5.6 <path-of-your-php-files> and I'd like to find a way to replicate the find stuff with a globbing pattern in place of the <path-of-your-php-files>

One of the major problems I was having was including ./path4/*.php while also excluding ./path4/dontwantthis/*

Answer 1

End your -exec option with + instead of \;. That tells find to run the command once with all the filenames, rather than separately for each filename.

find ./path1 ./path2 ./path3 ./path4                                                \
    -type f                                                                         \
    -name '*.php'                                                                   \
    -not -path './path4/dontwantthis/*'                                             \
    -exec ./vendor/bin/phpcs                                                        \
    --standard=PHPCompatibility --runtime-set testVersion 5.6 {} +