How to print the content inside the tag using grep command? [closed]

Question

In linux, how can we use grep command to print the contents that comes inside this tag?

<errorPayload>XXXXXXXX</errorPayload>

I tried grep -Po '<errorPayload>' abc.log, but it only prints <errorPayload>

Answer 1

Don't use grep to parse XML or HTML.

Instead, use a proper parser:

xidel -s -e '//errorPayload/text()' file
XXXXXXXX

Can work with xmllint and xmlstarlet as well too, in some less cases (weak HTML support):

xmlstarlet sel -t -v '//errorPayload/text()' file
xmllint --xpath '//errorPayload/text()' file

Answer 2

It only prints <errorPayload> because that's what you told it to do by using the -o (--only-matching) option. From the man page, that means "Print only the matched (non-empty) parts of a matching line..."

If you want to see just the content of the tag, you need to create a regular expression that matches only the content, but not the start/end tag.

This should do it:

grep -Po '(?<=<errorPayload>).*(?=</errorPayload>)' abc.log

Given you sample input in abc.log, this produces:

XXXXXXXX

The expression (?<=<errorPayload>) is a "positive look-behind assertion": it means that the given pattern needs to match before our target expression, but is not considered part of the "matched content". The expression (?=</errorPayload>) is a "positive look-ahead assertion", which does the same thing but for a following pattern.

See e.g. this article for more details about look-ahead and look-behind assertions.

---

Caveat: grep is a rotten tool for parsing XML. The above will work as long as the XML formatting in your log files is consistent.