Match a string using grep

Question

I want to match the below string using a regular expression in grep command.

File name is test.txt ,

Unknown Unknown

Jessica Patiño

Althea Dubravsky 45622

Monique Outlaw 49473

April Zwearcan 45758

Tania Horne 45467

I want to match the lines containing special characters alone from the above list of lines; the line which I exactly need is 'Jessica Patiño', which contains a non-ASCII character.

I used,

grep '[^0-9a-zA-Z]' test.txt

But it returns all lines.

Answer 1

The following command should return the lines you want:

grep -v '^[0-9a-zA-Z ]*$' test.txt

Explanation

• [0-9a-zA-Z ] matches a space or any alphanumeric character.
• Adding the asterisk matches any string containing only these characters.
• Prepending the pattern with ^ and appending it with $ anchors the string to the beginning and end of line so that the pattern matches only the lines which contain only the desired characters.
• Finally, the -v or --invert-match option to grep inverts the sense of matching, ie, select non-matching lines.

Answer 2

The provided answers should work for the example text given. However, you're likely to come across people with hyphens or apostrophes in their names, etc. To search for all non-ASCII characters, this should do the trick:

grep -P "[\x00-\x1F\x7F-\xFF]" test.txt

-P enables "Perl" mode and allows use of character code searches. \\x00-\\x1F are control characters, and \\x7F-\\xFF is everything above 126.

Answer 3

I would use:

grep [^0-9a-zA-Z\s]+ test.txt

live example

Or, even better:

 grep -i "[^\da-z\s]" test.txt