Question

因此,我有Xml。

<depot id="D7">
    <address> 
        <number>6000</number> 
        <street>Yonge</street> 
        <city>Toronto</city> 
        <province>ON</province> 
        <postal_code>M2M 2E4</postal_code> 
    </address> 
</depot>

我拥有数百座锡尔的仓库。

now in my xsl, I have defined a variable called locale that stores a postal code like "M1C". After this I want to select only those depot where the postal_code is like locale . In other words, If I specify locale to be "M1C", then I should get all the depot whose postal_code contains "M1C", so depot with "M1C A18", "M1C B2C", etc all should be in the result.

目前,我行文如下:

< xsl:for-each select="depot[address[postal_code=$locale]]">

这只给我贴上准确的邮政编码,而不是“M1C A18”、“M1C B2C”等。我想用的是这样的东西。

<xsl:for-each select="depot[address[postal_code=*$locale*]]">

与野心相联,但并不奏效。建议

Answer 1

<>Use:

depot[starts-with(address/postal_code, $locale)]

这里我们假设,任何<代码><>depot 都有一个单一的<代码>address/postal_code descendent,而$ locale的任何可能价值都不是$ locale的其他任何可能价值的先决条件。

If, for example, the second assumption isn t true, then use:

depot[starts-with(address/postal_code, concat($locale,    ))]

在XPath 2.0(例如matches(<>/code>)功能)中,可提供真正的定期表达能力,但对于一个简单的问题而言,这种能力是必要的。

Answer 2


使用:

<xsl:for-each select="depot[contains(address/postal_code,$locale)]" />

仅与<代码>depot中包含下定义的碎块的内容相匹配。

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