如何从 JSF 生成 JSON 响应？
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How to generate JSON response from JSF?
提问by Jitendra
I have created a page where I want to get the JSON response from a JSF page, but when i try to get page it shows me whole html page.
我创建了一个页面，我想在其中从 JSF 页面获取 JSON 响应，但是当我尝试获取页面时，它显示了整个 html 页面。
<html xmlns="http://www.w3.org/1999/xhtml"><head>
<title>Facelet Title</title></head><body>
[{"value": "21", "name": "Mick Jagger"},{"value": "43", "name": "Johnny Storm"},{"value": "46", "name": "Richard Hatch"},{"value": "54", "name": "Kelly Slater"},{"value": "55", "name": "Rudy Hamilton"},{"value": "79", "name": "Michael Jordan"}]
</body></html>
回答by BalusC
JSF is a MVC framework generating HTML, not some kind of a REST web service framework. You're essentially abusing JSF as a web service. Your concrete problem is simply caused by placing <html>tags and so on in the view file yourself.
JSF 是一个生成 HTML 的 MVC 框架，而不是某种 REST Web 服务框架。您实际上是在滥用 JSF 作为 Web 服务。您的具体问题只是由您自己<html>在视图文件中放置标签等引起的。
If you really insist, then you can always achieve this by using <ui:composition>instead of <html>. You also need to make sure that the right content type of application/jsonis been used, this defaults in JSF namely to text/html.
如果您真的坚持，那么您始终可以通过使用<ui:composition>代替<html>. 您还需要确保使用了正确的内容类型application/json，这在 JSF 中默认为text/html。
<ui:composition
xmlns:f="http://java.sun.com/jsf/core"
xmlns:ui="http://java.sun.com/jsf/facelets">
<f:event type="preRenderView" listener="#{bean.renderJson}" />
</ui:composition>
with
和
public void renderJson() throws IOException {
FacesContext facesContext = FacesContext.getCurrentInstance();
ExternalContext externalContext = facesContext.getExternalContext();
externalContext.setResponseContentType("application/json");
externalContext.setResponseCharacterEncoding("UTF-8");
externalContext.getResponseOutputWriter().write(someJsonString);
facesContext.responseComplete();
}
But I strongly recommend to look at JAX-RS or JAX-WS instead of abusing JSF as a JSON web service. Use the right tool for the job.
但我强烈建议查看 JAX-RS 或 JAX-WS，而不是将 JSF 滥用为 JSON Web 服务。为工作使用正确的工具。
See also:
也可以看看：
回答by Oleg
I'm even using contentType="text/xhtml" with JSF 2.2 and it works great. No needs in renderJson() from the BalusC's answer above
我什至在 JSF 2.2 中使用 contentType="text/xhtml" 并且效果很好。上面 BalusC 的回答中不需要 renderJson()
<f:view encoding="UTF-8" contentType="text/html"
xmlns="http://www.w3.org/1999/xhtml"
xmlns:h="http://xmlns.jcp.org/jsf/html"
xmlns:f="http://xmlns.jcp.org/jsf/core">
<h:outputText value="#{stationView.getClosestStationen(param.longitude, param.latitude)}" escape="false"/>
</f:view>
Ajax call:
阿贾克斯调用：
$.ajax({
url: requestContextPath + '/rest/stationen.xhtml',
type: "GET",
data: {
"longitude": x,
"latitude": y
},
dataType: "json",
success: function (data) {
$.each(data, function (i, station) {
...
});
},
error: function () {
...
}
})
回答by webstrap
I used just a facelet for the output (but JSF 1.2)
我只使用了一个 facelet 作为输出（但是 JSF 1.2）
<f:view contentType="application/json" xmlns:f="http://java.sun.com/jsf/core" xmlns:h="http://java.sun.com/jsf/html">
<h:outputText value="#{someBean.getJson()}" escape="false"/>
</f:view>
回答by Joeri Hendrickx
Why are you using jsf for that? Use a servlet for serving your JSON. My suggestion would be to use a jaxrs implementation, like cxf.
你为什么要使用 jsf 呢？使用 servlet 为您的 JSON 提供服务。我的建议是使用 jaxrs 实现，比如 cxf。
