仅保留与模式匹配的每个连续行序列中的第一行

Question

如果 2 个或更多连续行包含特定模式，则删除所有匹配的行并仅保留第一行。

在下面的示例中，当 2 个或多个连续行包含“逻辑 IO”时，我们需要删除所有匹配的行但保留第一行。

输入文件：

select * from test1 where 1=1
testing logical IO 24
select * from test2 where condition=4
parsing logical IO 45
testing logical IO 500
handling logical IO 49
select * from test5 where 1=1
testing logical IO 24
select * from test5 where condition=78
parsing logical IO 346
testing logical IO 12

输出文件：

select * from test1 where 1=1
testing logical IO 24
select * from test2 where condition=4
parsing logical IO 45
select * from test5 where 1=1
testing logical IO 24
select * from test5 where condition=78
parsing logical IO 346

Answer 1

使用awk：

awk '/logical IO/ {if (!seen) {print; seen=1}; next}; {print; seen=0}' file.txt 

• /logical IO/ {if (!seen) {print; seen=1}; next}检查该行是否包含logical IO，如果找到并且变量seen为假，即前一行不包含logical IO，然后打印该行，设置seen=1并转到下一行，否则转到下一行，就像上一行一样logical IO

• 对于任何其他行，{print; seen=0}, 打印该行和集合seen=0

例子：

$ cat file.txt 
select * from test1 where 1=1
testing logical IO 24
select * from test2 where condition=4
parsing logical IO 45
testing logical IO 500
select * from test5 where 1=1
testing logical IO 24
select * from test5 where condition=78
parsing logical IO 346
parsing logical IO 346
testing logical IO 12

$ awk '/logical IO/ {if (!seen) {print; seen=1}; next}; {print; seen=0}' file.txt 
select * from test1 where 1=1
testing logical IO 24
select * from test2 where condition=4
parsing logical IO 45
select * from test5 where 1=1
testing logical IO 24
select * from test5 where condition=78
parsing logical IO 346