Subject: Urgent..please advise From: Charles Fong <cfong@xxxxxxxxxxxxx> Date: Mon, 26 Jul 1999 01:00:08 -0700 |
This is a part of the source xml file.
<CNTROLAREA>
<SENDER>
<LOGICALID>XX141HG09</LOGICALID>
<COMPONENT>PUR</COMPONENT>
<TASK>MAINT</TASK>
<REFERENCEID>95129945823449</REFERENCEID>
<CONFIRMATION>1</CONFIRMATION>
<LANGUAGE>EN</LANGUAGE>
<CODEPAGE>test</CODEPAGE>
<AUTHID>CMKURT</AUTHID>
</SENDER>
<DATETIME qualifier="CREATION">
<YEAR>1998</YEAR>
<MONTH>11</MONTH>
<DAY>21</DAY>
<HOUR>17</HOUR>
<MINUTE>11</MINUTE>
<SECOND>45</SECOND>
<SUBSECOND>0000</SUBSECOND>
<TIMEZONE>-0600</TIMEZONE>
</DATETIME>
</CNTROLAREA>
I am still trying to figure out a xsl file which can produce
the following
resulting xml:
<CNTROLAREA>
<SENDER>
<LOGICALID>XX141HG09</LOGICALID>
<COMPONENT>PUR</COMPONENT>
<TASK>MAINT</TASK>
<REFERENCEID>95129945823449</REFERENCEID>
<CONFIRMATION>1</CONFIRMATION>
<LANGUAGE>EN</LANGUAGE>
<CODEPAGE>test</CODEPAGE>
<AUTHID>CMKURT</AUTHID>
</SENDER>
<CREATION_DATE>1998/11/21 17:11:45</CREATION_DATE>
</CNTROLAREA>
Basically, I want to try to identify the DATETIME structure in the source
xml file and collapse the structure to a simple data structure depending
on
the "qualifier" value of the DATETIME as shown above. All the other
xml
is untouched and need to appear in the resulting xml. Also this DATETIME
structure
can appear at any level of the tree in the source xml (above only a
simple example).
So my questions is : How can I achieve this in XSLT? i.e. extracting
the
DATETIME structure at whatever level from the source xml and collapsing
it
as shown above with all other xml untouched in the source. If
this can be done,
could you kindly provide me the xsl command to do it? Thanks a lot
in advance.
charles
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: re-import with match="*", Jon Smirl | Thread | RE: Urgent..please advise, Maxime Levesque |
result-encoding in XT 19990725, Vidar B. Gundersen | Date | Re: result-encoding in XT 19990725, Miloslav Nic |
Month |