在单词字符和逗号的模式之后grep for contents

Question

echo  "this is a test:foo,bar,baz']" | grep -o -E "test:.*" | awk -F: '{ print $2 }'
foo,bar,baz']

我最后打印']，如何只打印字符和普通字，没有别的，在这种情况下我只需要提取foo,bar,baz

Answer 1

您可以使用单个awk：

echo "this is a test:foo,bar,baz']" | awk -F 'test:' '{sub(/[^,[:alnum:]].*/, "", $2); print $2}'

foo,bar,baz

或者，您可以使用单个sed：

echo  "this is a test:foo,bar,baz']" | sed 's/.*test://; s/[^,[:alnum:]].*//'

foo,bar,baz

Answer 2

echo "this is a test:foo,bar,baz']"| awk -F: '{sub(/baz../,"baz"); print $2}'

输出

  

FOO，酒吧，巴兹

Answer 3

使用gnu grep 珍珠正则表达式

$ echo  "this is a test:foo,bar,baz']" | grep -oP "(?<=test:)(\w,*)+"
foo,bar,baz