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Hi,
and
Regards,
Geurt Lagemaat
[xsl] Copy a binary file using Java extension
Subject: [xsl] Copy a binary file using Java extension From: Geurt Lagemaat <lagemaat@xxxxxxxxxxxxxxxxxxxxxxxxx> Date: Sat, 06 Nov 2010 17:12:26 +0100 |
Hi,
I'm trying to copy a binary file using XSLT (2) (Saxon8) with java extensions.
I know I can use xmlns:fos="java.io.FileOutputStream", xmlns:file="java.io.File" to write functions like
<xsl:variable name="testfile" select="file:new(resolve-uri($filename))"/> <xsl:choose> <xsl:when test="not(file:exists($testfile))"> <xsl:value-of select="false()"/> </xsl:when> <xsl:otherwise> <xsl:value-of select="true()"/> </xsl:otherwise> </xsl:choose>
and
<xsl:variable name="fos" select="fos:new($ImgFilePath)"/> <!-- de base64 content as string --> <xsl:variable name="b64" select="b64:new(string(image/binair))"/> <!-- write to filestream while converting base64 to binary --> <xsl:value-of select="fos:write($fos, b64:getBinaryValue($b64))"/> <xsl:value-of select="fos:close($fos)"/>
But I cant rework these fragments to a functional copyFile. Problem is there is no simple binary fileCopy available without opening streams etc. So the most obvious way is to work with the fos, but how to read a binary file in a way that fos can write it?
Another idea was to use the java.io.File renameTo , but I cant translate the Java code is a XSLT way to do it.
Anyone has a idea? I don't want to use a Java class because the target environment is a sort of restricted area with no access to classpaths etc (I'm suspecting trouble with the classpath if I want to use a Java class to get this functionality).
Regards,
Geurt Lagemaat
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