我尝试在表格中输出这两个食谱,但我不能,它只是首先显示我。如何在表格中输出两个配方
这是我的xml:
<region id="southwest">
<name>Southwest</name>
<recipe id="1" author="" kitchen="en" origin="en">
<title>Chicken Fiesta Salad</title>
<type>main</type>
<difficulty>middle</difficulty>
<time>90 min.</time>
<portions>8</portions>
</recipe>
<recipe id="2" author="" kitchen="en" origin="en">
<title>Cupcakes</title>
<type>dessert</type>
<difficulty>hard</difficulty>
<time>60 min.</time>
<portions>8</portions>
</recipe>
</region>
这是我的xsl:
<table border="1">
<tr>
<th>Title</th>
<th>Region</th>
<th>Type</th>
<th>Difficulty</th>
<th>Time</th>
<th>Portions</th>
</tr>
<xsl:for-each select="region">
<tr>
<td><xsl:value-of select="recipe/title"/></td>
<td><xsl:value-of select="name"/></td>
<td><xsl:value-of select="recipe/type"/></td>
<td><xsl:value-of select="recipe/difficulty"/></td>
<td><xsl:value-of select="recipe/time"/></td>
<td><xsl:value-of select="recipe/portions"/></td>
</tr>
</xsl:for-each>
</table>
有人可以帮助我吗?
答案 0 :(得分:0)
您需要for-each
超过配方元素而不是区域元素
<xsl:for-each select="region/recipe">
并适当调整xpath表达式,例如
<td><xsl:value-of select="title"/></td>
<td><xsl:value-of select="../name"/></td>