How do I use grep to extract a specific field value from lines

I have lines in a file which look like the following

....... DisplayName="john" ..........

where .... represents variable number of other fields.

Using the following grep command, I am able to extract all the lines which have a valid 'DisplayName' field:

grep DisplayName="[0-9A-Za-z[:space:]]*" e:\test

However, I wish to extract just the name (ie "john") f开发者_开发技巧rom each line instead of the whole line returned by grep. I tried piping the output into the cut command but it does not accept string delimiters.

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This works for me:

awk -F "=" '/DisplayName/ {print $2}'

which returns "john". To remove the quotes for john use:

awk -F "=" '/DisplayName/ {gsub("\"","");print $2}'

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Specifically:

sed 's/.*DisplayName="\(.*\)".*/\1/' 

Should do, sed semantics is s/subsitutethis/forthis/ where "/" is delimiter. The escaped parentheses in combination with escaped 1 are used to keep the part of the pattern designated by parentheses. This expression keeps everything inside the parentheses after displayname and throws away the rest.

This can also work without first using grep, if you use:

sed -n 's/.*DisplayName="\(.*\)".*/\1/p'

The -n option and p flag tells sed to print just the changed lines.

More in: http://www.grymoire.com/Unix/Sed.html