在嵌套 For Each 中的 XSLT 中排序
[英] Sorting in XSLT within nested For Each
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问题描述
这是我想要做的:我有一个学院列表，每个学院内都有一个部门列表.我想显示整个部门列表，按部门名称排序，但要指明教职员工.
XML 如下所示:
<学院名称="科学"><部门名称="dept2"><head>X先生</head><building>A座</building>等等...</部门><部门名称="dept3"><head>X先生</head><building>B座</building>等等...</部门></教师><学院名称="教育"><部门名称="dept1"><head>Y先生</head><building>C座</building>等等...</部门></教师></院系>
XSLT 看起来像这样:(为了便于解释，我简化了 XSLT.)
<xsl:sort select="部门名称"><xsl:for-each select="Departments"><xsl:element name="div"><xsl:attribute name="id"><xsl:value-of select="facultName"></xsl:attribute><h3><xsl:value-of select="deptName">- <xsl:value-of select="facultName"></h3>//这里有更多的东西</xsl:element></xsl:for-each></xsl:for-each>
我希望输出如下所示:
Dept1(教育)负责人:Y先生建筑:C座Dept2(科学)负责人:X先生建筑:一个街区Dept3(科学)负责人:X先生建筑:B座
按部门名称排序的位置.
我还希望能够使用 Javascript 对特定教员隐藏所有系，即隐藏 ID 中具有特定教员的所有 div.
我什至不确定我的尝试是否可行(或合乎逻辑).我唯一的其他选择似乎是生成一个全新的部门列表，将教职员工作为要素之一.那么我只需要一个 for-each.不幸的是，我无法真正控制 XML 的生成方式，因此我希望能够以这种方式进行.
感谢您的帮助.谢谢！
解决方案
如果要按名称顺序列出所有系，不考虑系，直接遍历系即可
<xsl:sort select="@deptName"/></xsl:for-each>
然后，要获得系的教员名称，您可以很容易地访问父元素
所以，假设您有以下 XML
<Faculty id="1"facultyName="喝啤酒"><部门 id="1" deptName="Real Ale"/><部门 id="2" deptName="Lager"/></教师><Faculty id="2"facultyName="食物"><部门 id="3" deptName="鱼和薯条"/><部门 id="4" deptName="馅饼"/></教师></院系>
当您应用以下 XSLT 时
<xsl:output method="html" indent="yes"/><xsl:template match="/Faculties"><xsl:apply-templates select="教师/部门"><xsl:sort select="@deptName"/></xsl:apply-templates></xsl:模板><xsl:template match="部门"><div id="{../@facultyName}"><h3><xsl:value-of select="concat(@deptName, ' - ', ../@facultyName)"/></h3>
</xsl:模板></xsl:stylesheet>
以下是输出
<h3>鱼和薯条 - 食品</h3>
<div id="喝啤酒"><h3>拉格啤酒——喝啤酒</h3>
<div id="食物"><h3>馅饼 - 食物</h3>
<div id="喝啤酒"><h3>Real Ale - 喝啤酒</h3>
请注意，通常最好使用 xsl:apply-templates 而不是 xsl:for-each，所以这就是我在 XSLT 中使用的.
Here's what I'm trying to do:
I have a list of Faculties, within each of which is a list of departments. I want to display the entire list of departments, sorted by Department name, but indicating the faculty.
The XML looks like this:
<Faculties>
<Faculty Name="Science">
<Department name="dept2">
<head>Mr X</head>
<building>A Block</building>
etc...
</Department>
<Department name="dept3">
<head>Mr X</head>
<building>B Block</building>
etc...
</Department>
</Faculty>
<Faculty Name="Education">
<Department name="dept1">
<head>Mr Y</head>
<building>C Block</building>
etc...
</Department>
</Faculty>
</Faculties>
The XSLT looks something like this: (I've simplified the XSLT for explanation purposes.)
<xsl:for-each select="Faculties">
<xsl:sort select="DepartmentName">
<xsl:for-each select="Departments">
<xsl:element name="div">
<xsl:attribute name="id"><xsl:value-of select="facultName"></xsl:attribute>
<h3><xsl:value-of select="deptName"> - <xsl:value-of select="facultName"></h3>
//More stuff here
</xsl:element>
</xsl:for-each>
</xsl:for-each>
I'd like the output to look like:
Dept1 (Education)
Head: Mr Y
Building: C Block
Dept2 (Science)
Head: Mr X
Building: A Block
Dept3 (Science)
Head: Mr X
Building: B Block
Where it's sorted by Department Name.
I also want to be able to hide all the departments from a particular faculty using Javascript, i.e. hide all divs which have a particular faculty in the id.
I'm not even sure if what I'm attempting is possible (or logical). My only other option seems to be generating a completely new list of departments, with faculty as one of the elements. Then I'll only need one for-each. Unfortunately, I can't really control how the XML is generated, so I'm hoping to be able to do it this way.
I appreciate any help. Thanks!
解决方案
If you want to list all departments in name order, regardless of the faculty, you can simply iterate over the departments directly
<xsl:for-each select="Faculty/Department">
<xsl:sort select="@deptName" />
</xsl:for-each>
Then, to get the faculty name for the department, you can access the parent element quite easily
<xsl:value-of select="../@facultyName" />
So, assuming you have the following XML
<Faculties>
<Faculty id="1" facultyName="Beer Drinking">
<Department id="1" deptName="Real Ale" />
<Department id="2" deptName="Lager" />
</Faculty>
<Faculty id="2" facultyName="Food">
<Department id="3" deptName="Fish and Chips" />
<Department id="4" deptName="Pies" />
</Faculty>
</Faculties>
When you apply the following XSLT
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="html" indent="yes"/>
<xsl:template match="/Faculties">
<xsl:apply-templates select="Faculty/Department">
<xsl:sort select="@deptName" />
</xsl:apply-templates>
</xsl:template>
<xsl:template match="Department">
<div id="{../@facultyName}">
<h3><xsl:value-of select="concat(@deptName, ' - ', ../@facultyName)" /></h3>
</div>
</xsl:template>
</xsl:stylesheet>
The following is output
<div id="Food">
<h3>Fish and Chips - Food</h3>
</div>
<div id="Beer Drinking">
<h3>Lager - Beer Drinking</h3>
</div>
<div id="Food">
<h3>Pies - Food</h3>
</div>
<div id="Beer Drinking">
<h3>Real Ale - Beer Drinking</h3>
</div>
Do note, it is usually preferably to use xsl:apply-templates over xsl:for-each, and so this is what I have used in the XSLT.
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