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Old March 1st, 2013, 08:08 PM
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Default How to convert unfixed row and column from xml to xml

Hello,

I have the below xml file that I would like to convert. I need to form a complete record by extracting the column data from each row. For example: the below will found a complete record.

first row.col_2, second row.col_0, second row.col_1, second row.col_2, third row.col_0, etc.

I am trying to accomplish this in xslt 1.0. I know it might be easier in 2.0 but this is what we currently have in house.

Please help if you know how to do this...

<root>
<row>
<col_0>address</col_0>
<col_1/>
<col_2>a</col_2>
<col_3>b</col_3>
<col_4>c</col_4>
<col_5>d</col_5>
<col_6>e</col_6>
<col_7>f</col_7>
<col_n>this could be the last data</col_n>
</row>
<row>
<col_0>03/21/2012 0:00</col_0>
<col_1>1</col_1>
<col_2>0.369</col_2>
<col_3>0</col_3>
<col_4>0.026</col_4>
<col_5>1.125</col_5>
<col_6>0</col_6>
<col_7>0.015</col_7>
</row>
<row>
<col_0>03/21/2012 0:00</col_0>
<col_1>48</col_1>
<col_2>k 00 00</col_2>
<col_3>k 00 00</col_3>
<col_4>k 00 00</col_4>
<col_5>k 00 00</col_5>
<col_6>k 00 00</col_6>
<col_7>k 00 20</col_7>
</row>
</
root>
 
Old March 2nd, 2013, 06:48 AM
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Default

You can use simple xpath to pick any particular item you want from a row or column.

e.g. For the first row, and third column you would use "/root/row[1]/*[3]" (this would select col_2 in your example above, as your first column is col_0).

Without seeing how you want your output formatted etc it hard to give more advice than that.
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Old March 2nd, 2013, 03:43 PM
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Default How to convert unfixed row and column from xml to xml

Samjudson,

Thank you so much for responding. Below is my desire output. I would like to loop through the rows to extract the information then for a complete record in the xml.

<ROOT>
<RECORDS>
<RECORD ID="first row.col_2 DATA" CN="second row.col_1 data">
<VALUES>
<VAL V="second row.col_2 data" BDATE="second row.col_0 data"/>
<VAL V="third row.col_2" BDATE="third row.col_0 data"/>
</VALUES>
</RECORD>
</RECORDS>
</
ROOT>





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