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如何在迭代xsl:for-each 10行时选择第一次出现的数据

我想知道如何在xml文档中迭代xsl:for-each时选择第一次出现的行数据.

如果假设xml文档是这样的:

<catalog>
    <cd>
        <title>Empire Burlesque</title>
        <artist>Bob Dylan</artist>
        <country>USA</country>
        <company>Columbia</company>
        <price>10.90</price>
        <year>1985</year>
    </cd>
    <cd>
        <title>Hide your heart</title>
        <artist>Bonnie Tyler</artist>
        <country>UK</country>
        <company>CBS Records</company>
        <price>9.90</price>
        <year>1988</year>
    </cd>
    <cd>
        <title>Greatest Hits</title>
        <artist>Dolly Parton</artist>
        <country>USA</country>
        <company>RCA</company>
        <price>9.90</price>
        <year>1982</year>
    </cd>
    <cd>
        <title>Still got the blues</title>
        <artist>Gary Moore</artist>
        <country>UK</country>
        <company>Virgin records</company>
        <price>10.20</price>
        <year>1990</year>
    </cd>
    <cd>
        <title>Eros</title>
        <artist>Eros Ramazzotti</artist>
        <country>EU</country>
        <company>BMG</company>
        <price>9.90</price>
        <year>1997</year>
    </cd>
    <cd>
        <title>One night only</title>
        <artist>Bee Gees</artist>
        <country>UK</country>
        <company>Polydor</company>
        <price>10.90</price>
        <year>1998</year>
    </cd>
    <cd>
        <title>Sylvias Mother</title>
        <artist>Dr.Hook</artist>
        <country>UK</country>
        <company>CBS</company>
        <price>8.10</price>
        <year>1973</year>
    </cd>
    <cd>
        <title>Maggie May</title> …
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xml xslt

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