Extract xml tag value using awk command

You can use awk as shown below, however, this is NOT a robust solution and will fail if the xml is not formatted correctly e.g. if there are multiple elements on the same line.

$ dt=$(awk -F '[<>]' '/IntrBkSttlmDt/{print $3}' file)$ echo $dt1967-08-13

I suggest you use a proper xml processing tool, like xmllint.

$ dt=$(xmllint --shell file <<< "cat //IntrBkSttlmDt/text()" | grep -v "^/ >")$ echo $dt1967-08-13

The following gawk command uses a record separator regex pattern to match the XML tags. Anything starting with a < followed by at least one non-> and terminated by a > is considered to be a tag. Gawk assigns each RS match into the RT variable. Anything between the tags will be parsed as the record text which gawk assigns to $0.

gawk 'BEGIN { RS="<[^>]+>" } { print RT, $0 }' myfile

below code stores all the tag values in an array!hope this helps.But i still belive this is not an optimal way to do it.

> perl -lne 'if(/>[^<]*</){$_=~m/>([^<]*)</;push(@a,$1)}if(eof){foreach(@a){print $_}}' tempA2001-12-17T09:30:4700.01967-08-13CLRGxxAAAAAAAAAAA