Question

我是xslt的新手。下面是rdlc报告中的xml片段。我想写一个xsl转换，它告诉我“Field”属性“Name”不等于“DataField”子元素的情况。在下面的示例中，这将是最后一个字段（Name =“Url”）。

<?xml version="1.0" encoding="utf-8"?>
<Report xmlns="http://schemas.microsoft.com/sqlserver/reporting/2008/01/reportdefinition" xmlns:rd="http://schemas.microsoft.com/SQLServer/reporting/reportdesigner">
    <DataSets>
        <DataSet Name="TestReportModel">              
            <Fields>
                <Field Name="Name">
                    <DataField>Name</DataField>
                    <rd:TypeName>System.String</rd:TypeName>
                </Field>
                <Field Name="Description">
                    <DataField>Description</DataField>
                    <rd:TypeName>System.String</rd:TypeName>
                </Field>
                <Field Name="Url">
                    <DataField>DataFieldDoesNotMatchAttribute</DataField>
                    <rd:TypeName>System.String</rd:TypeName>
                </Field>
            </Fields>
        </DataSet>
    </DataSets>
</Report>

Answer 1

请尝试下面的样式表：

<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
    xmlns:def="http://schemas.microsoft.com/sqlserver/reporting/2008/01/reportdefinition"
    xmlns:rd="http://schemas.microsoft.com/SQLServer/reporting/reportdesigner"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    exclude-result-prefixes="def rd">

    <xsl:strip-space elements="*"/>

    <xsl:output method="xml" indent="yes" omit-xml-declaration="yes" />

    <xsl:template match="/">
        <xsl:apply-templates select="def:Report/def:DataSets/def:DataSet/def:Fields/def:Field[@Name != def:DataField]"/>
    </xsl:template>

    <xsl:template match="def:Field">
        <Field>
            <xsl:copy-of select="@*"/>
            <DataField><xsl:value-of select="def:DataField"/></DataField>
            <xsl:copy-of select="rd:TypeName"/>
        </Field>
    </xsl:template>

</xsl:stylesheet>

如何使用xslt在rdlc xml中查找错误

1 个答案: