Posted by Peymirat Damien on
URL: https://discuss.orbeon.com/Uploading-files-tp30091p30104.html

Hi Erik,

   All the files (main.xsd and base.xsd) are on a client's computer. So I upload the main.xsd like this :

View.xhtml:

<xforms:upload ref="file" xhtml:size="30">
        <xforms:label></xforms:label>
        <xforms:filename ref="@filename"/>
        <xforms:mediatype ref="@mediatype"/>
        <xxforms:size ref="@size"/>
        <xforms:hint>Choisissez le Schema XML à importer</xforms:hint>
</xforms:upload>

Model.xpl
...
<p:processor name="oxf:xslt">
        <p:input name="data" href="#instance"/>
        <p:input name="config">
                <config xsl:version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
                        <url>
                                <xsl:value-of select="/*/file"/>
                        </url>
                        <content-type>application/xml</content-type>
                        <force-content-type>true</force-content-type>
                        <cache-control>
                                <use-local-cache>false</use-local-cache>
                        </cache-control>
                </config>
        </p:input>
        <p:output name="data" id="url-config"/>
</p:processor>
<p:processor name="oxf:url-generator">
        <p:input name="config" href="#url-config"/>
        <p:output name="data" ref="data" id="output-file" debug="true"/>
</p:processor>
...


My problem is that I can't see how to upload the import element of main.xsd without asking the user. When I have uploaded the main.xsd, the only informations I have is <xs:import namespace="http://www.example.com/SYS/base/2.0/  schemaLocation="SYS/base/2.0/base.xsd"/> . And I don't know how to upload the file base.xsd (which is on client's computer) to the server ???


Thanks


Damien.




-----Message d'origine-----
De : Erik Bruchez [mailto:[hidden email]] De la part de Erik Bruchez
Envoyé : samedi 19 août 2006 03:24
À : [hidden email]
Objet : Re: [ops-users] Uploading files


Damien,

> How can I get the full url of my uploading file ?? Because to upload
> the file include in the schemaLocation, I need to have the path before
> SYS/base/2.0/base.xsd .

I don't quite understand the question. I assume the file is initially on
a client's computer, where its "URL" doesn't matter. Then you will
upload it into a database, and it is up to you to figure out through
which URL that schema will be accessible, for example through the eXist
REST API, or through your own code that serves the schema to whoever
uses it.

-Erik

--
Orbeon - XForms Everywhere:
http://www.orbeon.com/blog/


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