ResourceBundleViewResolver 的替代品

问题描述 投票:0回答:1

我有一个带有views.properties 文件的旧版jsp 应用程序,我使用

ResourceBundleViewResolver
加载所有视图。由于
ResourceBundleViewResolver
已从 Spring 5.3 中弃用,我检查了 spring 文档,它说“从 5.3 开始,支持 Spring 的通用视图解析器变体和/或自定义解析器实现”。我想知道使用 BundleViewResolver 的替代方法。提前道歉,我对处理遗留应用程序非常陌生。

编辑

配置.java

public class Configuration_place extends WebMvcConfigurerAdapter {

@Bean
public ResourceBundleViewResolver viewResolver(){
ResourceBundleViewResolver tilesViewResolver = new ResourceBundleViewResolver();
tilesViewResolver.setBasename("views");
return tilesViewResolver;
}

@Bean
    public TilesConfigurer tilesConfigurer() {
        TilesConfigurer tilesConfigurer = new TilesConfigurer();
        tilesConfigurer.setDefinitions("/tiles-def.xml");
        return tilesConfigurer;
    }

视图.属性

index.(class) = org.springframework.web.servlet.view.tiles3.TilesView
index.url = index

login.(class) = org.springframework.web.servlet.view.tiles3.TilesView
login.url = login

tiles-def.xml

<?xml version="1.0" encoding="UTF-8" ?>
<!DOCTYPE tiles-definitions PUBLIC
        "-//Apache Software Foundation//DTD Tiles Configuration 3.0//EN"
"http://tiles.apache.org/dtds/tiles-config_3.0.dtd">

<tiles-definitions>
<definition name = "master.page" template = "/views/templates/layout_main.jsp">
<put-attribute name = "title" value = "JSP Application" />
<put-attribute name = "header" value = "/views/templates/header.jsp" />
<put-attribute name = "menu" value = "/views/templates/menubar.jsp" />
<put-attribute name = "error" value = "/views/templates/errorMessage.jsp" />
<put-attribute name = "content" value = "" />
<put-attribute name = "footer" value = "/views/templates/footer.jsp" />
</definition>

<definition name = "index" extends="master.page">
<put-attribute name="content" value="/views/intro.jsp" />

</definition>

<definition name = "login" extends="master.page">
<put-attribute name="content" value="/views/login.jsp" />
</definition>

</tiles-definitions>
spring spring-mvc jsp
1个回答
0
投票

Ravi,您能解决上述问题吗?我也面临同样的问题。

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