我有一个这样的名单赶上的Joomla版本打印线+接下来的两行使用awk如果下一行匹配
./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';
我要的是,只有行+如果public
是两个下一行显示在下一行。否则线必须被忽略
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 $DEV_LEVEL = '9';
我曾尝试使用awk和这个awk脚本
BEGIN{ RS=""; FS="\n" }
/public/ {
for (i=1; i<=NF; i++) {
if (! (($i ~ /./) && ($(i+1) !~ /public/) && ($(i+2) !~ /public/))) {
print $i
}
}
print ""
}
但是这会导致:
./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
此。两次。 (三次由于你知道什么) –
@JamesBrown对不起,我不明白...过了漫长的一天... –
三个字符填充。 –