首页 \ 问答 \ 在Rails 4中格式化我的JSON输出(Pretty format my JSON output in Rails 4)

在Rails 4中格式化我的JSON输出(Pretty format my JSON output in Rails 4)

我在我的控制器中使用pretty_generate,但是我收到以下错误

'只允许生成一代JSON对象或数组'

@celebrities = Celebrity.includes(:category)    
respond_to do |format|
  format.json { render json: JSON.pretty_generate(@celebrities.to_json(:include =>{:category => {:only => [:category]} })) }
end

我不知道为什么我会收到这个错误


I am using pretty_generate in my controller, but I am getting the following error

'only generation of JSON objects or arrays allowed'

@celebrities = Celebrity.includes(:category)    
respond_to do |format|
  format.json { render json: JSON.pretty_generate(@celebrities.to_json(:include =>{:category => {:only => [:category]} })) }
end

I am not sure why I am getting this error


原文:https://stackoverflow.com/questions/32439655
更新时间:2024-03-26 07:03

最满意答案

我遇到了同样的问题,暴露了两个版本的webservice与不同的网址。

old version within http://hostname/ws.wsdl
new version within http://hostname/version/ws.wsdl

我的解决方案不是使用通用的org.springframework.ws.transport.http.MessageDispatcherServletservlet,而是使用默认的org.springframework.web.servlet.DispatcherServlet,并在我的bean配置中配置url映射到不同的wsdl版本。

我更喜欢这个解决方案,因为它可以在不对任何spring类进行子类化的情

web.xml中:

<servlet>
    <servlet-name>webservice</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value></param-value>
    </init-param>
</servlet>

<servlet-mapping>
    <servlet-name>webservice</servlet-name>
    <url-pattern>/ws</url-pattern>
    <url-pattern>/ws.wsdl</url-pattern>
    <url-pattern>/version/ws</url-pattern>
    <url-pattern>/version/ws.wsdl</url-pattern>
</servlet-mapping>

beans.xml中

<bean id="messageFactory" class="org.springframework.ws.soap.saaj.SaajSoapMessageFactory" />

<bean
    class="org.springframework.ws.transport.http.WebServiceMessageReceiverHandlerAdapter">
    <property name="messageFactory" ref="messageFactory" />
</bean>

<bean id="messageDispatcher"
    class="org.springframework.ws.soap.server.SoapMessageDispatcher" />

<bean
    class="org.springframework.ws.transport.http.WsdlDefinitionHandlerAdapter" />

<bean class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
    <property name="mappings">
        <props>
            <prop key="/ws.wsdl">ws</prop>
            <prop key="/version/ws.wsdl">ws-newversion</prop>
        </props>
    </property>
    <property name="defaultHandler" ref="messageDispatcher" />
</bean>

<bean id="ws"
    class="org.springframework.ws.wsdl.wsdl11.SimpleWsdl11Definition">
    <constructor-arg value="classpath:wsdl/oldversion/Service.wsdl" />
</bean>

<bean id="ws-newversion"
    class="org.springframework.ws.wsdl.wsdl11.SimpleWsdl11Definition">
    <constructor-arg value="classpath:wsdl/newversion/CarService.wsdl" />
</bean>

因此,每个wsdl都在SimpleUrlHandlerMapping-Bean中配置的给定路径上公开。


After poking around spring-ws source, I found there is no support for exposing a multi-node path for static-sourced WSDL configuration.

So I subclassed MessageDispatcherServlet and SimpleWsdl11Definition, and in my servlet, provided my own WSDL-request mapper that supports existing WsdlDefinition beans, as well as my "location-specified" WsdlDefinition bean.

Yields ability to configure in this sort of manner:

<!-- exposes URL:  host/context-root/servlet-name/MyService.wsdl -->
<bean id="MyService" class="org.springframework.ws.wsdl.wsdl11.SimpleWsdl11Definition">
    <property name="wsdl" value="/WEB-INF/wsdl/MyService.wsdl" />
</bean>

<!-- exposes URL:  host/context-root/servlet-name/some/multi/node/taxonomy/path/MyService.wsdl -->
<bean id="MyService.otherVersion" class="path.to.my.EnhancedWsdl11Definition">
    <property name="wsdl" value="/WEB-INF/wsdl/otherVersion/MyService.wsdl" />
    <property name="locationUri" value="some/multi/node/taxonomy/path/MyService.wsdl" />
</bean>

All is well.

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