Question

起初我想知道使用xslt是否可以实现这一点，如果是这样的话怎么做：

我有两个xmls。第一个是关于发票的信息

$(document).ready(function(){
    $('#selector').datetimepicker({
        viewMode: 'days',
        format: 'DD/MM/YYYY'
    });     
});

，第二个是关于发票项目的具体信息

<INVOICES>
    <INVOICE>
        <INVOICE_NUMBER>FV2014000572</INVOICE_NUMBER>
        <SUPPLIER_NAME>Supplier 1</SUPPLIER_NAME>
    </INVOICE>
    <INVOICE>
        <INVOICE_NUMBER>FV2014000573</INVOICE_NUMBER>
        <SUPPLIER_NAME>Supplier 1</SUPPLIER_NAME>
    </INVOICE>
</INVOICES>

我想以下列格式将这两个xmls转换为一个：

<INVOICE_ITEMS>
    <INVOICE_ITEM>
        <ITEM_INVOICE_NUMBER>FV2014000572</ITEM_INVOICE_NUMBER>
        <ITEM_NAME>ITEM 1</ITEM_NAME>
    </INVOICE_ITEM>
    <INVOICE_ITEM>
        <ITEM_INVOICE_NUMBER>FV2014000573</ITEM_INVOICE_NUMBER>
        <ITEM_NAME>ITEM 1</ITEM_NAME>
    </INVOICE_ITEM>
    <INVOICE_ITEM>
        <ITEM_INVOICE_NUMBER>FV2014000573</ITEM_INVOICE_NUMBER>
        <ITEM_NAME>ITEM 2</ITEM_NAME>
    </INVOICE_ITEM>
<INVOICE_ITEMS>

如果可能的话，从最终的xml中删除ITEM_INVOICE_NUMBER个元素。

由于

Answer 1

这个非常基本的样式表应该为您提供如何继续的线索。它将应用于您在输入中提供的第一个XML，并假定存储其他信息的第二个XML名为 data.xml 。

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="1.0">

    <xsl:variable name="invoiceitems" select="document('data.xml')" />

    <!-- just copy the element, and process contents. Warning: attributes (if any) won't be processed. -->
    <xsl:template match="*">
        <xsl:copy>
            <xsl:apply-templates />
        </xsl:copy>
    </xsl:template>

    <xsl:template match="INVOICE">
        <xsl:variable name="current.invoice.number" select="INVOICE_NUMBER" />
        <xsl:copy>
            <xsl:apply-templates />
            <!-- create an INVOICE_ITEMS element, and start to retrieve information from the other XML -->
            <INVOICE_ITEMS>
                <xsl:apply-templates select="$invoiceitems//INVOICE_ITEM[ITEM_INVOICE_NUMBER/text() = $current.invoice.number]"></xsl:apply-templates>
            </INVOICE_ITEMS>
        </xsl:copy>
    </xsl:template>

</xsl:stylesheet>

带条件的XSLT复制元素

1 个答案: