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
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