Get Node value from a XML using xmllint

Question

I have a xml called Det.xml like this :

<?xml version="1.0" encoding="UTF-8"?>
    <S:Envelope xmlns:S="http://schemas.xmlsoap.org/soap/envelope/">
        <S:Body>
            <ns4:grtHgetRed xmlns:ns2="http://object" xmlns:ns3="http://object" xmlns:ns4="http://object">
                <RequestId>lol</RequestId>
                <MessageDateTime>54.009</MessageDateTime>
                <SenderId>UH</SenderId>
                <ReceiverId>GER</ReceiverId>
                <TrackingNumber>45</TrackingNumber>
                <ServerName>trewds</ServerName>
                <ResponseType>success</ResponseType>
                <StatusInfo>
                <Status>success</Status>
                <SystemMessage>Hagert</SystemMessage>
                <UserMessage>Hgert</UserMessage>
                <Origination>htref</Origination>
                </StatusInfo>
            </ns4:grtHgetRed>
        </S:Body>
    </S:Envelope>

I am trying to get the ResponseType node value success from it using xmllint in Unix shell script and so i tried the following :

echo "cat //*[local-name()='S:Envelope'/*[local-name()='S:Body']/*[local-name()='ns4:grtHgetRed']/*[local-name()='ResponseType']" | xmllint --shell Det
.xml | sed '/^\/ >/d' | sed 's/<[^>]*.//g'

But it's not working . Also i don't have xpath in my unix environment . Can any one tell me what am i doing wrong here ?

Answer 1

The local-name() is just the bit after the colon, so instead of e.g. local-name()='S:Envelope' try just local-name()='Envelope'.

/*[local-name()='Envelope']/*[local-name()='Body']/*[local-name()='grtHgetRed']/*[local-name()='ResponseType']

Or you may want to consider an alternative tool such as xmlstarlet which has better support for this kind of thing.