XSLT:如何找到节点的唯一子节点的数量? [英] XSLT: How to find the count of unique children of a node?

问题描述

我的 XML 如下所示:

My XML looks like this:

<foo>
    <bar name="a">
        <baz name="xyz">
            <time>2</time>
            <date>3</date>
        </baz>
    </bar>
    <bar name="b">
        <baz name="xyz">
            <time>2</time>
            <date>3</date>
        </baz>
    </bar>
    <bar name="c">
        <baz name="xyz">
            <time>2</time>
            <date>3</date>
        </baz>
    </bar>
</foo>

我正在编写一个 XSL，它需要像这样运行:如果所有 baz 子项都相同，则 doSomething else doSomethingElse.我当前的节点是 foo.

I am writing an XSL that needs to function like this: If all the baz children are same then doSomething else doSomethingElse. My current node is foo.

我是 XSLT 的新手，我知道 XSL 中的条件.现在看起来像这样:

I am new to XSLT and I am aware of the conditionals in XSL. It looks something like this as of now:

<xsl:template match="foo">   
<xsl:choose>
    <xsl:when test="[My condition]"> 
        doSomething()
    </xsl:when>
    <xsl:otherwise>
        doSomethingElse()
    </xsl:otherwise>
</xsl:choose>
</xsl:template>

在当前示例中，它应该 doSomething() 因为所有 baz 元素都相同.

In the current example, it should doSomething() as all the baz elements are the same.

如果我找出唯一的 baz 元素的数量，我可以测试它是否等于 1.如果是，那么我会doSomething() else doSomethingElse()

If I find out the number of unique baz elements, I can test whether it is equal to one. If it is, then I will doSomething() else doSomethingElse()

我应该如何实现这一点?MyCondition 应该是什么?

How should I implement this? What should MyCondition be?

PS:我的 XSL 版本是 1.0

PS: My XSL version is 1.0

推荐答案

如果所有 baz 子节点都相同，则 doSomething elsedoSomethingElse.我当前的节点是 foo.

If all the baz children are same then doSomething else doSomethingElse. My current node is foo.

这令人困惑，因为:

• baz 不是 foo 的孩子；
• 您的标题说找到唯一孩子的数量" - 但事实并非如此必须找到它才能知道它们是否相同.

• baz are not children of foo;
• your title says "find the count of unique children" - but it is not necessary to find it in order to know if they are same.

尝试类似:

<xsl:template match="foo">
    <xsl:variable name="first-baz" select="(bar/baz)[1]" />
    <xsl:choose>
        <xsl:when test="bar/baz[date!=$first-baz/date or time!=$first-baz/time]">
            <!-- THEY ARE NOT ALL SAME -->
        </xsl:when>
        <xsl:otherwise>
            <!-- THEY ARE ALL SAME -->
        </xsl:otherwise>
    </xsl:choose>
</xsl:template>

请注意，这里假设每个 baz 都有一个 date 和一个 time.否则你需要测试 not(date=$first-baz/date) 等

Note that this assumes that every baz has a date and a time. Otherwise you need to test for not(date=$first-baz/date) etc.

另见:http://www.jenitennison.com/xslt/grouping/muenchian.html

现在，假设所有 bar/baz 元素都有相同的标签(不是必须 date 和 time 但说 a, b 和 c)，什么会是那个案例的测试属性?

Now, assuming all the bar/baz elements have the same tags (not necessarily date and time but say a, b and c), what would be the test attribute for that case?

这使得它变得更加复杂.你仍然可以构造一个密钥:

This makes it significantly more complex. Still you could construct a key:

<xsl:key name="first-baz" match="foo/bar[1]/baz[1]/*" use="name()" />

然后进行测试:

<xsl:when test="bar/baz/*[. != key('first-baz', name())]">

如果存在 baz 的任何子节点，其字符串值与作为第一个 baz 子节点的同名节点不同，则返回 true.

This returns true if any child of baz exists whose string-value is different from an equally named node that is child of the first baz.

请注意，此处定义的键是文档范围的.如果要将测试限制为当前 foo 祖先，则必须在键中包含其 id.

Note that the key, as defined here, is document-wide. If you want to restrict the test to the current foo ancestor, then you must include its id in the key.

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