Split xml node formatwise

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XML

<?xml version="1.0"?>
-<DocumentElement>
-<test>
<c0>manas</c0>
<c1>100</c1>
<c2>20</c2>
<c3>33</c3>
<c4>54</c4>
 <c5>65</c5>
<c6>78</c6>
</test>
 -<test>
<c0>manasi</c0>
<c1>101</c1>
<c2>25</c2>
 <c3>76</c3>
<c4>90</c4>
<c5>12</c5>
<c6>74</c6>

</test>
 -<test>
<c0>mani</c0>
<c1>115</c1>
<c2>29</c2>
<c3>56</c3>
<c4>11</c4>
<c5>95</c5>
<c6>78</c6>
</test>
 </DocumentElement>

Console Application
For first Node Test:
manas 100 20 33
manas 54 65 78

For Second Node Test:
manasi 101 25 76
manasi 90 12 74

For Third Node Test:
mani 115 29 56
mani 11 95 78

Sir,
Aim is to split multiple number of columns .

Actually 7 columns values for First node Test like manas 100 20 33 54 65 78
My requirement is to display first column(manas) with 3 other row, again it will create a break and write like first column(manas) with next 3 columns.
Then same type it will create for second Node Test.
Then same type it will create for Third Node Test.

Posted 2-Jul-13 8:22am

connect2manas

Updated 2-Jul-13 8:28am

H.Brydon

v3

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H.Brydon · Answer 1 · 2013-07-02T08:35:00

Solution 1

It sounds like you want to write a C# console app and transform your data using xsl. This is straightforward; just create a console app. Use the System.Xml and System.Xml.Xsl library namespaces.

C#

XslTransform xslt = new XslTransform();
xslt.Load("mytransform.xsl");
XPathDocument mydata = new XPathDocument("inputdata.xml");
XmlWriter writer = new XmlTextWriter(Console.Out);
xslt.Transform(mydata,null,writer, null);

Put the formatting details in the .xsl file.

Posted 2-Jul-13 8:35am

H.Brydon

Comments

connect2manas 2-Jul-13 14:56pm

thanks for your answer.

kindly help me to split the below like:

-<documentelement>
-<test>
<c0>manas
<c1>100
<c2>20
<c3>33
<c4>54
<c5>65
<c6>78

- plz try to load this file.
-print the 2 values in a single row
like
manas 100
20 33
54 65
78

plz help me.

H.Brydon 2-Jul-13 15:05pm

This looks too much like homework, which we don't do here. You need to put in some effort to learn this yourself. "Please do this for me" is not a valid request.

Have a look at xslt techniques in a book or on the web. A good reference is http://www.w3schools.com/xsl/default.asp

Mayank Dubey · Answer 2 · 2013-07-02T08:49:00

Hope it will solve your problem as per my understanding.

DECLARE @XML XML='<documentelement>
<test><c0>manas</c0><c1>100</c1><c2>20</c2><c3>33</c3><c4>54</c4><c5>65</c5><c6>78</c6></test>
<test><c0>manasi</c0><c1>101</c1><c2>25</c2><c3>76</c3><c4>90</c4><c5>12</c5><c6>74</c6></test>
<test><c0>mani</c0><c1>115</c1><c2>29</c2><c3>56</c3><c4>11</c4><c5>95</c5><c6>78</c6></test>
</documentelement>'
 
 Select Column1= x.value('c0[1]','varchar(100)') + ' '+ x.value('c1[1]','varchar(100)')+' ' +x.value('c2[1]','varchar(100)') +' '+x.value('c3[1]','varchar(100)'),
 Column2= x.value('c0[1]','varchar(100)') + ' '+ x.value('c4[1]','varchar(100)')+' ' +x.value('c5[1]','varchar(100)') +' '+x.value('c6[1]','varchar(100)')     
 from @XML.nodes('/DocumentElement/test') e(x)