如何在bash中使用正则表达式在双引号之间选择字符串？

Question

I really searched this kind of question among the other similar questions here but couldn't find any answer.

I have the following text in a file:

TYPE_A = "1" AND TYPE_B = "6" AND TYPE_C = "8755asd-"

I am trying to grep the following (the double quotes can include numbers, characters and special characters):

TYPE_C = "8755asd-"

The grep command that I am using is this: grep -o "TYPE_C = \\"[0-9a-zA-Z-]*\\"" text.txt

However, I think the regex in this command is not generic.

Thank you for your help.

Answer 1

You could use the below regex which matches the corresponding TYPE only if the double quotes contain all the characters you mentioned . That is, atleast a character, atleast a number, atleast a special character.

grep -oP 'TYPE_[A-Z]\s*=\s*"(?=[^"]*[A-Za-z])(?=[^"]*\d)(?=[^"]*[^"\w])[\W\w]*?"' file

DEMO

Example:

$ cat ri
TYPE_A = "1" AND TYPE_B = "6" AND TYPE_C = "8755asd-"
TYPE_A = "1" AND TYPE_B = "6" AND TYPE_C = "875d"
TYPE_A = "1" AND TYPE_B = "6" AND TYPE_C = "asd-"
TYPE_A = "1" AND TYPE_B = "6" AND TYPE_C = "875-"
$ grep -oP 'TYPE_[A-Z]\s*=\s*"(?=[^"]*[A-Za-z])(?=[^"]*\d)(?=[^"]*[^"\w])[\W\w]*?"' ri
TYPE_C = "8755asd-"

Answer 2

I'd do it with sed :

sed -n 's@.* TYPE_C = "\([0-9a-zA-Z-]*\)".*@\1@p' text.txt

That is, -n to not print lines that don't match. Then replace the entire line with the part in double quotes after TYPE_C, and p to print the result:

8755asd-

Answer 3

如果模式保持不变，则可以使用cut获得相同的输出。

echo 'TYPE_A = "1" AND TYPE_B = "6" AND TYPE_C = "8755asd-"' | cut -d"\"" -f6