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XSL for converting all rows and columns of XSL for Export to Excel. dynamic looping

Answer 1

A:

What is the code you currently have (tried) ? What does the input code like? What should the ouput code look like?

It's going to be rather difficult to answer if you don't supply some details. ;)

2010-06-30 22:12:03

Hi, I added the input XML data. The output code works the xsl:for-each ./*[count(child::*) = 0] and displayes proper column name in the result, but I don't know how to get subsequent columns in the result. Please provide me a solution if you have. Thanks in advance to all

Enggr 2010-07-01 17:51:27

Answer 2

+1 A:

Preserving your for-each driven process:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
    <xsl:template match="/">
        <HTML>
            <HEAD>
                <STYLE type="text/css"> TABLE{table-layout: automatic; width:100%} .tblHeader{background-color:RGB(192,192,192);font-weight:bold} .row1{background-color:RGB(204,204,255)} .row2{background-color:RGB(153,204,255)} </STYLE>
            </HEAD>
            <BODY>
                <TABLE border="1">
                    <THEAD>
                        <TR class="tblHeader">
                            <xsl:for-each select="*/*[1]/*">
                                <TH>
                                    <xsl:value-of select="name()"/>
                                </TH>
                                <!-- {Getting the xml column header here} -->
                            </xsl:for-each>
                        </TR>
                    </THEAD>
                    <TBODY>
                        <xsl:for-each select="ClientArray/Client">
                            <TR>
                                <xsl:attribute name="class">
                                    <xsl:choose>
                                        <xsl:when test="position() mod 2 = 1">row1</xsl:when>
                                        <xsl:otherwise>row2</xsl:otherwise>
                                    </xsl:choose>
                                </xsl:attribute>
                                <xsl:for-each select="*">
                                    <TD>
                                        <xsl:value-of select="."/>
                                    </TD>
                                </xsl:for-each>
                            </TR>
                        </xsl:for-each>
                    </TBODY>
                </TABLE>
            </BODY>
        </HTML>
    </xsl:template>
</xsl:stylesheet>

Note: I'll come back later with an "XSLT style" stylesheet.

Edit: I was in a rush. The key was for my answer to come. Sorry.

Edit 2: Added stylesheet in "XSLT style".

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
    <xsl:template match="/ClientArray">
        <HTML>
            <HEAD>
                <STYLE type="text/css"> TABLE{table-layout: automatic; width:100%} .tblHeader{background-color:RGB(192,192,192);font-weight:bold} .row1{background-color:RGB(204,204,255)} .row2{background-color:RGB(153,204,255)} </STYLE>
            </HEAD>
            <BODY>
                <TABLE border="1">
                    <THEAD>
                        <TR class="tblHeader">
                            <xsl:apply-templates select="Client[1]/*" mode="headers"/>
                        </TR>
                    </THEAD>
                    <TBODY>
                        <xsl:apply-templates/>
                    </TBODY>
                </TABLE>
            </BODY>
        </HTML>
    </xsl:template>
    <xsl:template match="Client">
    <TR class="row{2 - position() mod 2}">
            <xsl:apply-templates/>
        </TR>
    </xsl:template>
    <xsl:template match="Client/*" mode="headers">
        <TH>
            <xsl:value-of select="name()"/>
        </TH>
    </xsl:template>
    <xsl:template match="Client/*">
        <TD>
            <xsl:value-of select="."/>
        </TD>
    </xsl:template>
</xsl:stylesheet>

Note: These stylesheets assume that all the children of Client elements are the same and are in the same order. I'd come back with a more general solution if you want to.

Alejandro 2010-07-01 20:17:30

Hi Alejandro, The solution you gave me was excellent/great. It worked 100% for me. Thanks a lot. You saved me a lifetime. I will let you know if I need more refinement. I think I need. Thanks again

Enggr 2010-07-06 16:05:49

@Enggr: I'm glad this helps you. Remember to provide input and output in your questions. Also, if this is the correct solution for you, you should mark as answer in order to help other people.

Alejandro 2010-07-06 16:17:12

Ok, I will provide input/output from now. I will mark this as answer to help other people. (Also people will know you are very kind to put the effort and provide the answer :) )

Enggr 2010-07-07 14:37:30

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XSL for converting all rows and columns of XSL for Export to Excel. dynamic looping

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