Question

我将以下输入分配给变量：

echo $var1
abc001: text goes here yyy003: text goes here uuuu004: text goes here

输出应该如下（没有冒号:)基本上我想在没有冒号的新行上打印以主机名开头的每个列表

abc001 text goes here
yyy003 text goes here
uuuu004 text goes here

Answer 1

awk版本：

awk '{for (i=1;i<=NF;i++) printf $i~":"?"\n"$i" ":$i" "}' <<< "$var1"| awk 'NF {sub(/:/,x);print}'
abc001 text goes here
yyy003 text goes here
uuuu004 text goes here

另一个版本，基于Fredriks正则表达式：

awk '{gsub(/[a-zA-Z]+[0-9]+/,"\n&");gsub(/:|^\n/,x)}1' <<< "$var1"
abc001 text goes here
yyy003 text goes here
uuuu004 text goes here

sed版本：

sed -E 's/://;s/(\w+):/\n\1/g' <<< "$var1"
abc001 text goes here
yyy003 text goes here
uuuu004 text goes here

Answer 2

使用GNU awk的一种方式：

$ gawk 'NR>1{print host, $0}{host=RT}' RS='[[:alnum:]]+:' OFS='\b' <<< $var1
abc001 text goes here
yyy003 text goes here 
uuuu004 text goes here

Answer 3

或许这样的事情：

$ echo $var1
host1: text goes here host2: text goes here Host3:text goes here

$ sed -e 's/\([Hh]ost\)/\n\1/g' -e 's/://g' <<< $var1

host1 text goes here 
host2 text goes here 
Host3text goes here

更新

$ echo $var1
aaa1: text goes here bbbb0002: text goes here AAAA0012: text goes here

$ sed -e 's/\([a-zA-Z]\+[0-9]\+\)/\n\1/g' -e 's/://g' <<< $var1

aaa1 text goes here 
bbbb0002 text goes here 
AAAA0012 text goes here

Answer 4

grep -Po '.*?(?=( [^: ]*:|$))' file|sed 's/://'

以你的例子：

kent$ echo "abc001: text goes here yyy003: text goes here uuuu004: text goes here"|grep -Po '.*?(?=( [^: ]*:|$))'|sed 's/://'
abc001 text goes here
yyy003 text goes here
uuuu004 text goes here

Sed / Awk命令用于格式化分配给变量的文本

4 个答案: