bash - How to handle parenthesis in grep? -

bash - How to handle parenthesis in grep? -

str of next pattern:

1 abc (1 <something>)

for example:

1 abc (1 hello) 1 abc (1 shalom) 1 abc (1 hola)

how extract <something> str using egrep?

if want extract <something> i'd suggest grep -p (perl regex):

grep -p -o '(?<=\(1 ).*?(?=\))' inputfile

the -o returns matched portion beingness <something>. regex looks text preceded (1 , followed ).

you wouldn't able egrep doesn't back upwards lookarounds. best you'd extract (1 <something>) with:

egrep -o '\(1 (.*)\)' inputfile [foo@bar ~]$ grep -p -o '(?<=\(1 ).*?(?=\))' inputfile hello shalom hola [foo@bar ~]$ egrep -o '\(1 (.*)\)' inputfile (1 hello) (1 shalom) (1 hola)

bash grep