Question

帮助shell脚本读取逗号分隔行中的模式，最终结果输出应再次以逗号分隔。在下面，例如，读取行以逗号分隔，并仅输出以逗号分隔的木偶字符串。

echo "docker-one,puppet-one,puppet-two,docker-three,puppet-four" | script

需要输出：

docker-one,docker-three

Answer 1

awk救援！

echo "docker-one,puppet-one,puppet-two,docker-three,puppet-four" | 
awk 'BEGIN{RS=ORS=","} /puppet/'

puppet-one,puppet-two,puppet-four

表示docker，并替换最后一个逗号

echo "docker-one,puppet-one,puppet-two,docker-three,puppet-four" | 
awk 'BEGIN{RS=ORS=","} /docker/' | 
sed 's/,$/\n/'

docker-one,docker-three

或者，如果你的意思是 non puppet

echo "docker-one,puppet-one,puppet-two,docker-three,puppet-four" | 
awk 'BEGIN{RS=ORS=","} !/puppet/' | 
sed 's/,$/\n/'

docker-one,docker-three

Answer 2

使用原生bash regEx运算符~并仅使用GNU paste进行csv格式设置，

IFS="," read -ra myArray <<<"docker-one,puppet-one,puppet-two,docker-three,puppet-four"
for i in "${myArray[@]}";do [[ $i =~ ^puppet ]] && echo "$i" ; done | paste -sd ','

生成输出

puppet-one,puppet-two,puppet-four

对于除puppet之外的其他字符串，执行负正则表达式匹配，

for i in "${myArray[@]}";do [[ ! $i =~ ^puppet ]] && echo "$i" ; done | paste -sd ','
docker-one,docker-three

Answer 3

听起来其中一个可能是你正在寻找的东西：

$ echo "docker-one,puppet-one,puppet-two,docker-three,puppet-four" |
awk -F, '{for (i=1;i<=NF;i++) if ($i ~ /puppet/) printf "%s%s", (c++?FS:""), $i; print ""}'
puppet-one,puppet-two,puppet-four

$ echo "docker-one,puppet-one,puppet-two,docker-three,puppet-four" |
awk -F, '{for (i=1;i<=NF;i++) if ($i !~ /puppet/) printf "%s%s", (c++?FS:""), $i; print ""}'
docker-one,docker-three

$ echo "docker-one,puppet-one,puppet-two,docker-three,puppet-four" |
awk -F, '{for (i=1;i<=NF;i++) if ($i ~ /docker/) printf "%s%s", (c++?FS:""), $i; print ""}'
docker-one,docker-three

Answer 4

使用tr，grep和paste：

$ echo "docker-one,puppet-one,puppet-two,docker-three,puppet-four" \
       | tr , '\n' | grep -v puppet | paste -s -d , -
docker-one,docker-three

awk脚本读取行匹配模式和结果输出与逗号分隔

4 个答案: