awk腳本讀取行匹配模式和結果用逗號分隔輸出

Question

幫助shell腳本以逗號分隔的行讀取模式，最終結果輸出應再次以逗號分隔。在下面的例如read行中用逗號分隔，並只輸出用逗號隔開的木偶字符串。

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

輸出要求：

docker-one,docker-three 

Answer 1

awk來救援！

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

puppet-one,puppet-two,puppet-four 

的碼頭工人，並替換最後一個逗號

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，如果你的意思是非傀儡

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

使用本機bash正則表達式運算符~和csv格式使用GNU paste獨自一人，

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 ',' 

產生一個輸出作爲

puppet-one,puppet-two,puppet-four 

和其他琴絃比開始puppet的那些，做了否定的正則表達式匹配，

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

Answer 3

這聽起來像其中之一可能是什麼您正在查找：

$ 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