Question

我有以下xml

## 13 Aug 2013 14:53:44, 390 [INFO] OrderId 100 otherInfo
<someXML>details
<info>details<info>
<info1>details<info1>
</someXML>
## 13 Aug 2013 14:53:44, 390 [INFO] OrderId 105 otherInfo
<someXML>details
<info>details<info>
<info1>details<info1>
</someXML>
## 13 Aug 2013 14:55:45, 490 [INFO] OrderId 100 otherInfo  
<someXML>details
<info>details<info>
<info1>details<info1>
</someXML>
## 13 Aug 2013 14:53:44, 390 [INFO] OrderId 105 otherInfo
<someXML>details
<info>details<info>
<info1>details<info1>
</someXML>

我想用orderid“示例OrderId 100”搜索特定行并打印该行及其下面的所有内容，直到下一个以doulbe hash（##）开头的订单行所以，如果我用orderid 100搜索，我应该得到以下

## 13 Aug 2013 14:53:44, 390 [INFO] OrderId 100 otherInfo
<someXML>details
<info>details<info>
<info1>details<info1>
</someXML>
## 13 Aug 2013 14:55:45, 490 [INFO] OrderId 100 otherInfo 
<someXML>details
<info>details<info>
<info1>details<info1>
</someXML>

Answer 1

您的记录很好地分开，因此请将记录分隔符设置为##并搜索您想要的OrderId：

$ awk '/OrderId 100/{print RS $0}' RS='##' ORS=''  file
## 13 Aug 2013 14:53:44, 390 [INFO] OrderId 100 otherInfo
<someXML>details
<info>details<info>
<info1>details<info1>
</someXML>
## 13 Aug 2013 14:55:45, 490 [INFO] OrderId 100 otherInfo
<someXML>details
<info>details<info>
<info1>details<info1>
</someXML>

Answer 2

这可能适合你（GNU sed）：

sed '/OrderId 100/!d;:a;$!{n;/^##/{s/^/\n/;D};ba}' file

使用sed，awk在模式之间提取数据行，直到具有特殊字符的行的开头

2 个答案: