在 XSLT 中，我有来自 XML 的状态，并且对应于该状态，我有我自己的状态代码，我想将其放入转换后的 XML 中。

源 XML:

<states>
<state>New York</state>
<state>California</state>
</states>

预期结果:

<states>
    <state>NY</state>
    <state>CA</state>
</states>

因为我有状态列表，所以我不能使用 when 或 If 语句，所以他们是否有任何优化的方法来解决 XSLT 1.0 或 2.0 中的这个问题？提前致谢。

最佳答案

这是您可以查看的一种方式:

XSLT 2.0

<xsl:stylesheet version="2.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:my="http://example.com/my">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>

<xsl:key name="state" match="state" use="." />

<!-- identity transform -->
<xsl:template match="@*|node()">
    <xsl:copy>
        <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="state">
    <xsl:copy>
        <xsl:value-of select="key('state', ., document(''))/@code"/>
    </xsl:copy>
</xsl:template>

<my:states>
    <state code="AL">Alabama</state>
    <state code="AK">Alaska</state>
    <state code="AZ">Arizona</state>
    <state code="AR">Arkansas</state>
    <state code="CA">California</state>
    <state code="CO">Colorado</state>
    <state code="CT">Connecticut</state>
    <state code="DE">Delaware</state>
    <state code="DC">District of Columbia</state>
    <state code="FL">Florida</state>
    <state code="GA">Georgia</state>
    <state code="HI">Hawaii</state>
    <state code="ID">Idaho</state>
    <state code="IL">Illinois</state>
    <state code="IN">Indiana</state>
    <state code="IA">Iowa</state>
    <state code="KS">Kansas</state>
    <state code="KY">Kentucky</state>
    <state code="LA">Louisiana</state>
    <state code="ME">Maine</state>
    <state code="MD">Maryland</state>
    <state code="MA">Massachusetts</state>
    <state code="MI">Michigan</state>
    <state code="MN">Minnesota</state>
    <state code="MS">Mississippi</state>
    <state code="MO">Missouri</state>
    <state code="MT">Montana</state>
    <state code="NE">Nebraska</state>
    <state code="NV">Nevada</state>
    <state code="NH">New Hampshire</state>
    <state code="NJ">New Jersey</state>
    <state code="NM">New Mexico</state>
    <state code="NY">New York</state>
    <state code="NC">North Carolina</state>
    <state code="ND">North Dakota</state>
    <state code="OH">Ohio</state>
    <state code="OK">Oklahoma</state>
    <state code="OR">Oregon</state>
    <state code="PA">Pennsylvania</state>
    <state code="RI">Rhode Island</state>
    <state code="SC">South Carolina</state>
    <state code="SD">South Dakota</state>
    <state code="TN">Tennessee</state>
    <state code="TX">Texas</state>
    <state code="UT">Utah</state>
    <state code="VT">Vermont</state>
    <state code="VA">Virginia</state>
    <state code="WA">Washington</state>
    <state code="WV">West Virginia</state>
    <state code="WI">Wisconsin</state>
    <state code="WY">Wyoming</state>
</my:states>

</xsl:stylesheet>

如果您愿意，可以将状态列表及其代码放在外部 XML 文档中并从那里进行查找。

As I have list of states so I can't use when or If statement

实际上，使用 xsl:choose 也同样有效。

关于xml - 如何在 XSLT 中创建键值对并使用键获取值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/36429302/