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Xslt 如何使用XSL 1.0在重复上下文中显示当前日期?_Xslt_Xslt 1.0_Repeat - Fatal编程技术网

Xslt 如何使用XSL 1.0在重复上下文中显示当前日期?

Xslt 如何使用XSL 1.0在重复上下文中显示当前日期?,xslt,xslt-1.0,repeat,Xslt,Xslt 1.0,Repeat,我试图在重复上下文中显示当前日期,其中我的元素位于重复上下文之外 示例XML: <?xml version="1.0" encoding="UTF-8"?> <root> <AllLines> <Versions> <field1>V1.0</field1> <field2>A1.0</field

我试图在重复上下文中显示当前日期,其中我的元素位于重复上下文之外

示例XML:

<?xml version="1.0" encoding="UTF-8"?>
<root>
    <AllLines>
        <Versions>
            <field1>V1.0</field1>
            <field2>A1.0</field2>
        </Versions>
        <Versions>
            <field1>V2.0</field1>
            <field2>A2.0</field2>
        </Versions>
        <Versions>
            <field1>V3.0</field1>
            <field2>A3.0</field2>
        </Versions>
        <Versions>
            <field1>V4.0</field1>
            <field2>A4.0</field2>
        </Versions>
        <Versions>
            <field1>V5.0</field1>
            <field2>A5.0</field2>
        </Versions>
    </AllLines>
    <CurrentLines>
        <CurrentLine date="21 Aug">
            <field1>X1.0</field1>
            <field2>Y1.0</field2>
            <field3>Y1.0</field3>
        </CurrentLine>
        <CurrentLine date="30 Jan">
            <field1>X2.0</field1>
            <field2>Y2.0</field2>
            <field3>Y2.0</field3>
        </CurrentLine>
        <CurrentLine date="02 Feb">
            <field1>X3.0</field1>
            <field2>Y3.0</field2>
            <field3>Y3.0</field3>
        </CurrentLine>
        <CurrentLine date="21 Aug">
            <field1>X4.0</field1>
            <field2>Y4.0</field2>
            <field3>Y4.0</field3>
        </CurrentLine>
        <CurrentLine date="03 Jan">
            <field1>X5.0</field1>
            <field2>Y5.0</field2>
            <field3>Y5.0</field3>
        </CurrentLine>
    </CurrentLines>
</root>
每个重复组中的第三个元素总是显示第一个日期属性,来自
/root/CurrentLines/CurrentLine

预期结果将是:

V1.0
A1.0
21 Aug
--------
V2.0
A2.0
30 Jan
-------
V3.0
A3.0
02 Feb
------
V4.0
A4.0
21 Aug
------
V5.0
A5.0
03 Jan
有没有办法在
版本之后继续重复并显示当前字段,但根据相应的当前
CurrentLine
元素显示当前
@date
属性

谢谢大家!

试试看:

<xsl:for-each select="/root/AllLines/Versions">
    <xsl:variable name="i" select="position()" />
    <fo:block>
        <xsl:value-of select="field1" />
    </fo:block>
    <fo:block>
        <xsl:value-of select="field2" />
    </fo:block>
    <fo:block border-bottom="1pt solid black" margin-bottom="4pt">
        <xsl:value-of select="/root/CurrentLines/CurrentLine[$i]/@date" />
    </fo:block>
</xsl:for-each>

谢谢!这解决了我的问题
V1.0
A1.0
21 Aug
--------
V2.0
A2.0
30 Jan
-------
V3.0
A3.0
02 Feb
------
V4.0
A4.0
21 Aug
------
V5.0
A5.0
03 Jan
<xsl:for-each select="/root/AllLines/Versions">
    <xsl:variable name="i" select="position()" />
    <fo:block>
        <xsl:value-of select="field1" />
    </fo:block>
    <fo:block>
        <xsl:value-of select="field2" />
    </fo:block>
    <fo:block border-bottom="1pt solid black" margin-bottom="4pt">
        <xsl:value-of select="/root/CurrentLines/CurrentLine[$i]/@date" />
    </fo:block>
</xsl:for-each>