Perl 仅打印从起点到终点的内容

要只打印从开始到结束的内容吗

例如：

Started at Mon Nov 25 14:30:02 ALMT 2013
Load average for db1 is 10 which is less than 100
Load average for db2 is 3 which is less than 100
Load average for db3 is 6 which is less than 100
Load average for db4 is 5 which is less than 100
Load average for db5 is 5 which is less than 100
Mon Nov 25                                                                                 page
some other data
some other data
some other data

预期：

Started at Mon Nov 25 14:30:02 ALMT 2013
Load average for db1 is 10 which is less than 100
Load average for db2 is 3 which is less than 100
Load average for db3 is 6 which is less than 100
Load average for db4 is 5 which is less than 100
Load average for db5 is 5 which is less than 100
Mon Nov 25                                                                                 page

我试着这样做：

perl -0777 -lpe 's|Started .*? page||xsg' 1

但这会删除我要打印的整个区域。

请尝试以下操作：

$ perl -0777 -lpe 's|.*(Started .*? page).*|$1|xsg' 1

---

像这样的事情可能会发生：

$ awk '/^Started/ {p=1} p; /page$/ {p=0}' file
Started at Mon Nov 25 14:30:02 ALMT 2013
Load average for db1 is 10 which is less than 100
Load average for db2 is 3 which is less than 100
Load average for db3 is 6 which is less than 100
Load average for db4 is 5 which is less than 100
Load average for db5 is 5 which is less than 100
Mon Nov 25                                                                                 page

• /^Started/{p=1}

  当它找到以“Started”开头的字符串时，则标志

  p

  设置为1

• p

  在标志

  p

  为真时打印该行

• /page$/{p=0}

  当它找到以“page”结尾的字符串时，则标志

  p

  设置为0

---

还有，（）：

或与。但请注意，如果遇到“开始”标记而不是“结束”，它将打印从开始标记到文件结尾的所有内容

sed -n '/^Started/, /page/p' file

---

呃，我看不到任何结束点。只需向右滚动，有“开始于…”，结束是“25页”，所以记录的“结束”是“开始”行中的日期，后面是字符串“页”？“页面”前有大量空白？你必须明确，因为我们看不懂你的心思：你必须告诉我们你需要什么。Strange在我的控制台上不工作，只需打印所有输出：/n检查你的示例输入是否匹配以下内容：一行以“Started”开头，一行以“page”结尾。哇，这太短了，@1_CR！非常感谢，我添加它作为另一个选项，并+1您的答案，因为它也使用这种逻辑。

sed -n '/^Started/, /page/p' file

perl -0777 -ne 'print /(Started .*? page)/xsg' 1