java JSF：有条件地呈现列表项 (<li>)
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JSF: conditionally render a list item (<li>)
提问by George Armhold
I just inherited a project implemented in JSF. I have the following code which looks fine in Chrome, but Firefox renders the borders on the "empty" list items:
我刚刚继承了一个在 JSF 中实现的项目。我有以下代码在 Chrome 中看起来不错，但 Firefox 在“空”列表项上呈现边框：
<ul>
<li><a href="/home">Home</li>
<li>
<s:link view="/signup.xhtml" rendered="#{someCondition}">Sign Up</s:link>
</li>
<!-- etc... -->
</ul>
Which ends up looking like:
最终看起来像：
Is there a JSF tag to conditionally render the <li>?
是否有 JSF 标记来有条件地呈现<li>?
回答by mrembisz
If you do it as in @CoolBeans example, you will get a <span> around your <li>. In some cases it might disrupt your layout, besides you don't really want an extra tag under <ul>. To get rid of it, use <ui:fragment rendered="#{condition}" />around your item instead of <h:panelGroup>.
如果您按照@CoolBeans 示例进行操作，您将在 <li> 周围得到一个 <span>。在某些情况下，它可能会破坏您的布局，此外您真的不希望 <ul> 下有一个额外的标签。要摆脱它，请<ui:fragment rendered="#{condition}" />在您的项目周围使用而不是 <h:panelGroup>。
Also you can use style attribute to hide an item:
您也可以使用 style 属性来隐藏项目：
<li style="display: #{condition ? 'list-item' : 'none'};" />
回答by CoolBeans
No liis vanilla html, not a jsf component.
不，li是 vanilla html，而不是 jsf 组件。
You can hackit by putting a <h:panelGroup />around the lielement and setting the renderedproperty on it.
您可以通过在元素周围放置 a并在其上设置属性来破解它。<h:panelGroup />lirendered
ie.
IE。
<h:panelGroup rendered="#{someCondition}">
<li>
<s:link view="/signup.xhtml">Sign Up</s:link>
</li>
</h:panelGroup>
Another option is to use <f:verbatim rendered="#{someCondition}" >, but keep in mind that it has been deprecated in JSF 2.0.
另一种选择是使用<f:verbatim rendered="#{someCondition}" >，但请记住，它在 JSF 2.0 中已被弃用。
回答by Antoine Fontaine
You could also use the core JSTL library:
您还可以使用核心 JSTL 库：
<c:if test="#{someCondition}">
<li>
<s:link view="/signup.xhtml">Sign Up</s:link>
</li>
</c:if>
I think using this core tag makes the code easier to understand than using the panelGroup tag.
我认为使用这个核心标签比使用 panelGroup 标签更容易理解代码。
