将与模式匹配的行移至上一行 [英] move line which matches pattern to previous line

问题描述

我有一个结构为

12312
desc: bleh...
9938
desc: blah...
desc: bloh...
desc: blih...
desc: bluh...
9912
desc: blah...

我想将匹配模式"desc:"的行移至上一行，或删除每个模式"desc:"之前的行中的"\ n".

and i want to move line which matches pattern "desc:" to previous line or delete '\n' in line which goes before every pattern "desc:".

所需的输出:

12312 desc: bleh...
9938 desc: blah... desc: bloh... desc: blih... desc: bluh... 
9912 desc: blah...

我尝试过

awk '!/desc:/{
 printf "%s ",$0
 getline
 printf "%s \n",$0
}
/desc/{print}' file

没有结果.

实际上所有数据都是 awk -F \'{print $ 4" \ t"$ 6}'的输出也许我首先可以做点什么?

actually all the data is the output of awk -F\" '{print $4 "\t" $6}' maybe i can do something in the first place?

推荐答案

sed oneliner

sed oneliner

sed ':a $!N;s/\ndesc/ desc/;ta P;D'

将输出

12312 desc: bleh...
9938 desc: blah... desc: bloh... desc: blih... desc: bluh...
9912 desc: blah...

这篇关于将与模式匹配的行移至上一行的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！