I have list of employees in following format
Name:
Phone No:
Email:
Name:
Address:
Phone No:
Email:
Name:
Country:
Address:
Phone No:
Email:
Name:
Address:
Name :
Email:
Address:
All I want is that to get only Name and Email address.
I tried something like cat filename | grep -e ^Name: -e ^Email: but was not 100% successfull.
Use | which is "or" operator for regex in the grep command. Also, no need for cat. grep can read from file directly.
grep '^Name:\|^Email:' filename
grep -E '^(Name|Email):' filename. ^ and : are common to both sub-expressions, leaving only Name and Email as the non-fixed portion of the extended regular expression.
– cas
Mar 22 '16 at 22:21
If by "not 100% successful" you mean that you want only the values of the fields, without the field-names, then you need to use awk rather than grep.
e.g.
awk -F: '$1 ~ /^(Name|Email)$/ {print $2}' filename
Or, if you still want each record separated by a blank line:
awk -F: '$1 ~ /^(Name|Email)$/ {print $2}; /^[[:space:]]*$/ {print "\n"}' filename
cat | grep) – Archemar Mar 22 '16 at 15:12