用ajax渲染一个div [英] Rendering a div with ajax

问题描述

我有这个div

<div id="name">
    ....
</div>

我还有一个按钮可以做到这一点:

<h:commandLink id="bttn" action="#">
    <f:ajax render="name"/>
</h:commandLink>

问题是执行时会说没有所谓的名称"组件.有没有一种方法可以使ajax 呈现div而不会将其封装在jsf组件中?

解决方案

不行.在JSF的幕后，Ajax渲染的工作原理大致如下:

for (String clientIdToRender : clientIdsToRender) {
    UIComponent componentToRender = viewRoot.findComponent(clientIdToRender);
    // ...
}

但是，由于纯HTML <div>不是 作为

<h:panelGroup layout="block" id="name">
    ...
</h:panelGroup>

另请参见:

• 是否可以使用以下命令更新非JSF组件(纯HTML) JSF ajax?

I have this div

<div id="name">
    ....
</div>

I also have a button that does this:

<h:commandLink id="bttn" action="#">
    <f:ajax render="name"/>
</h:commandLink>

The problem is that when executing, it says that there is no component called "name". Is there a way I can render a div with an ajax without enclosing it within a jsf component?

解决方案

No, there's no way. Ajax rendering works roughly as follows under JSF's hoods:

for (String clientIdToRender : clientIdsToRender) {
    UIComponent componentToRender = viewRoot.findComponent(clientIdToRender);
    // ...
}

However, as the plain HTML <div> is not as a fullworthy JSF component available by UIViewRoot#findComponent(), JSF can't find anything to generate the updated HTML output for.

You do not necessarily need to wrap it in another JSF component, you can also just use a JSF component which generates already a HTML <div> element. That is the <h:panelGroup> component whose layout attribute is set to block (as in, "render a block level element").

<h:panelGroup layout="block" id="name">
    ...
</h:panelGroup>

See also:

• Is it possible to update non-JSF components (plain HTML) with JSF ajax?

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