如果下一行匹配，则使用awk打印行+下两行

Question

我有一个这样的列表来捕捉joomla版本

./somedir/bla/old/libraries/cms/version/version.php
        public $RELEASE = '2.5';
        public $DEV_LEVEL = '24';
./somedir/bla3/www/libraries/cms/version/version.php
        public $RELEASE = '2.5';
        public $DEV_LEVEL = '9';
./somedir/bla4/www/libraries/cms/version/version.php
./somedir/bla5/www/w/scripts/version.php
./somedir/bla6/www/libraries/cms/version/version.php
./somedir/bla7/www/libraries/cms/version/version.php
        public $RELEASE = '2.5';
        public $DEV_LEVEL = '9';

我想要的是，如果public在下一行，那么只显示这一行+下两行。必须忽略Else行

所以结果应该是：

./somedir/bla/old/libraries/cms/version/version.php
        public $RELEASE = '2.5';
        public $DEV_LEVEL = '24';
./somedir/bla3/www/libraries/cms/version/version.php
        public $RELEASE = '2.5';
        public $DEV_LEVEL = '9';
./somedir/bla7/www/libraries/cms/version/version.php
        public $RELEASE = '2.5';
        public $DEV_LEVEL = '9';

我已经尝试使用Awk和这个awk脚本

BEGIN{ RS=""; FS="\n" }
 /public/ {
    for (i=1; i<=NF; i++) {
        if ( ! (($i ~ /./) && ($(i+1) !~ /public/) && ($(i+2) !~ /public/) ) ) {
            print $i
        }
    }
    print ""
}

但这会导致：

./somedir/bla/old/libraries/cms/version/version.php
        public $RELEASE = '2.5';
./somedir/bla3/www/libraries/cms/version/version.php
        public $RELEASE = '2.5';
./somedir/bla7/www/libraries/cms/version/version.php
        public $RELEASE = '2.5';

我错过了dev_level的第二条公共线路