Hi Phil,

I'm not aware of any XQuery standard functionality to solve your problem.

However, a simple work-around could be to create an XML file (say "docs.xml") that lists for each doument you need to load its location (URL / absolute filename) and desired name in the collection, e.g., as follows

<docs> <doc loc="...URL-1..." name="...name-1..."/> ... <doc loc="...URL-n..." name="...name-n..."/> </docs>

an then query this XML file using XQuery to load all document in a single collection:

for $doc in fn:doc("[abolute_path_to]docs.xml")/docs/doc return pf:add-doc($doc/@loc, $doc/@name, "MyCollection")

That should be it.

On you favorite unix-like system, you could easily create the docs.xml, e.g., by executing the following shell (bash) commands in the top-level directory of your document collection, asuming all files are called *.xml and all filenames are unique

echo "<docs>" > /tmp/docs.xml for i in `find $PWD -name *.xml` ; do echo "<doc loc="$i" name="${i##*/}"/>" done >> /tmp/docs.xml echo "</docs>" >> /tmp/docs.xml

I have no idea how to do somehting similar on Windows.

Hope this helps you further.

Stefan

On Mon, Sep 10, 2007 at 06:34:59AM -0700, Philip Webster wrote:

Hi, I have a directory containing XML files, all sharing a common structure. I want to load them all into a collection so I can run queries against them using the XQuery interface in MonetDB4.

The tutorials give useful examples that show how to load a single file into a collection, but I need an automated process to load all of the XML files I have. There are over 60,000 files, currently spread over 16 directories to make them more managable.

I've been looking around on the Web, but I haven't found any XQuery examples that do what I want. Is it possible? If so, can anyone give an example to show me how to load my data?

Thanks,

Phil

-- View this message in context: http://www.nabble.com/-newbie--Loading-a-directory-of-XML-files-into-a-colle... Sent from the monetdb-users mailing list archive at Nabble.com.