Subject: Re: [xsl] replacing of characters based on position From: "Imsieke, Gerrit, le-tex" <gerrit.imsieke@xxxxxxxxx> Date: Tue, 16 Nov 2010 07:25:38 +0100 |
well, i am using xslt 2.0. i need the position based replacement rather than regexp. The reason is i want reverse back '0' with - later at some point , so thaat i can guarente that same uuid is going to be generated after reversal.
step1:550e8400-e29b-41d4-a716-446655440000 to 550e84000e29b041d40a7160446655440000
step2:550e84000e29b041d40a7160446655440000 to 550e8400-e29b-41d4-a716-446655440000
--- On Tue, 16/11/10, David Carlisle<davidc@xxxxxxxxx> wrote:
From: David Carlisle<davidc@xxxxxxxxx> Subject: Re: [xsl] replacing of characters based on position To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Cc: "ram"<ram_kurra@xxxxxxxxxxx> Date: Tuesday, 16 November, 2010, 4:40 AM On 15/11/2010 22:44, ram wrote:Hi, I have xslvariable which holds the following uuid value (550e8400-e29b-41d4-a716-446655440000) which has '-' at 9-14-19-24 positions.I want to go these positons9-14-19-24 and replace the characters '-' with zero's.Is there any functionsuch as replace(position values,character,replace character) available in xslt.
you should always say whether you are using xslt 1 or 2, especially if asking about string handling,
do you actually need position based replace, (that would be easy using xslt 2 regexp, or only slightly more verbose using a mixture of substring and concat) but your original problem statement could be solved by translate(.,'-','0')
David
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