In my last blog I wrote about Spring transactional services written in Jython, and now I want to write about my way to use Spring MVC with Jython.
I will describe the following steps:
- preparing the web.xml
- Spring servlet configuration
- Jython controller
In the web.xml, beside the ContextLoaderServlet we need a Spring Front Controller that is the entry point to all requests. This Front Controller then dispatches to the specific Task or Action or how ever you want to call it. Here is the servlet declaration and it's mapping for all URLs ending with .py
=== web.xml ===
<servlet>
<servlet-name>example</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>2</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>example</servlet-name>
<url-pattern>*.py</url-pattern>
</servlet-mapping>
|
For the servlet name "example" we need a configuration file in the WEB-INF folder that starts with "example" and the servlet name that always ends with "-servlet.xml". This file is finally called "example-servlet.xml" and holds the bean configurations for the Spring MVC part.
=== example-servlet.xml ===
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE beans PUBLIC "-//SPRING//DTD BEAN//EN" "http://www.springframework.org/dtd/spring-beans.dtd">
<beans>
<bean id="jythonController" class="com.my.view.controller.JythonController">
<property name="mappings">
<props>
<prop key="/index.*">com.my.view.controller.ForwardController(view="home", activeMenu="home")</prop>
<prop key="/login.*">com.my.view.controller.LoginController(view="login", activeMenu="login")</prop>
</props>
</property>
</bean>
<bean id="urlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">
<props>
<prop key="/*.py">jythonController</prop>
</props>
</property>
</bean>
<bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="viewClass"><value>org.springframework.web.servlet.view.JstlView</value></property>
<property name="prefix"><value>/WEB-INF/jsp/</value></property>
<property name="suffix"><value>.jsp</value></property>
</bean>
</beans>
|
This MVC configuration needs some detailed explanation because it wires the Spring and Jython world. The urlMapping part catches all requests that end with *.py and delegates them to the jythonController that is a native Java class. It also could be a java-compatible Jython class, but I simply chose native Java. The jythonController then delegates the specific request with its specific url to the mapped Jython controller class. This mappings are declared as a property element in the jythonController bean where the key is the url pattern like ".../index.py" or ".../login.py" and the value is a Jython expression, that will create a new instance of the Jython controller class. So while Spring initializes the Jython Controller bean, the Jython statements are executed and the resulting Jython controller instances are cached as singletons.
So the trick is simply that a native Java controller instance stores a number of Jython controller classes which will then execute the concrete controller functionality. You can see it as a babushka doll with 'controllers-within-controllers'.
The next advantage is that the Jython controller singletons can be configured declaratively. In this example I say that the LoginController should froward to the "login" view that will result in the "WEB-INF/jsp/login.jsp" page (see viewResolver bean), and that the highlighted menu (activeMenu argument) should be "login" in the shown page's navigation.
Because IMO it is so intuitive, here it is in other words.
The jython Controller executes the Jython statement
com.my.view.controller.LoginController(view="login", activeMenu="login")
|
The corresponding Jython code of the controller would look like the following listing.
=== LoginController.py ===
from org.springframework.web.servlet.mvc import Controller
from org.springframework.web.servlet import ModelAndView
from org.springframework.web.context.support import WebApplicationContextUtils
from java.util import HashMap
from java.lang import Boolean, System, Integer
class LoginController(Controller):
def __init__(self, view=None, activeMenu=None):
self.view = view
self.activeMainMenu = None
self.activeSubMenu = None
if activeMenu != None:
activeMenuTokens = activeMenu.split(".")
self.activeMainMenu = activeMenuTokens[0]
if len(activeMenuTokens) == 2:
self.activeSubMenu = activeMenuTokens[1]
def getApplContext(self, request):
servletContext = request.getAttribute("servletContext")
return WebApplicationContextUtils.getWebApplicationContext(servletContext)
def getLoginService(self, request):
return self.getApplContext(request).getBean("loginService");
def handleRequest(self, req, res):
"""Ensures an existing user session, sets the user's view settings
and calls completeModel().
This method should not be overridden."""
session = req.getSession(Boolean.TRUE)
model = HashMap()
model.put("activeMainMenu", self.activeMainMenu)
model.put("activeSubMenu", self.activeSubMenu)
# do the real processing
self.completeModel(req, session, model)
return ModelAndView(self.view, model)
def completeModel(self, req, session, model):
"""Handles the complete controller logic and fills the model.
The model is a java HashMap.
This method can be overridden."""
self.getTestService(req).checkLogin(req)
# ...
# ...
|
In the previous code the controller accesses the loginService bean, that is also a Spring bean, to get the service instance from the Spring application context. To accomplish this, the request has to hold the servletContetxt instance. This is not a very elegant solution, but better ones are wellcome.
The model is represented by a simple java.util.HashMap which is later on transformed to the request scope by Spring MVC (BTW this is the same way I handle the model in my open source project Flow4J).
You can also simply convert the LoginController in a generic BaseController of which you can subclass and override only the completeModel() method.
For the curious ones, I also support my implementation of the native Java JythonController class
=== JythonController.java ===
package de.my.view.controller;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Properties;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.python.core.PyInstance;
import org.python.core.PyJavaInstance;
import org.python.util.PythonInterpreter;
import org.springframework.util.PathMatcher;
import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.AbstractController;
import org.springframework.web.servlet.mvc.Controller;
import org.springframework.web.util.UrlPathHelper;
public class JythonController extends AbstractController {
private UrlPathHelper urlPathHelper = new UrlPathHelper();
private Map urlMappings;
private Map controllerMappings = new HashMap();
/**
* Spring callback that creates the Jython controller instances and caches
* the singletons.
*
* @param mappings the contrller mappings
*/
public void setMappings(Properties mappings) {
this.urlMappings = mappings;
for (Iterator iter = urlMappings.entrySet().iterator(); iter.hasNext();) {
Map.Entry entry = (Map.Entry) iter.next();
String urlPath = (String) entry.getKey();
String expr = (String) entry.getValue();
System.out.println("register controller [urlPath: " + urlPath
+ ", expr: " + expr + "]");
Controller controller = (Controller) getPyJavaObject(expr,
Controller.class);
controllerMappings.put(urlPath, controller);
}
}
/**
* Evaluates the jython statement and returns the result casted to the given
* type. Uses the JythonInitServlet that stores an sntance of the Jython interpreter.
*
* @param stmnt jython statement
* @param clazz type to which the result should be casted
* @return the statements result casted to the given type
* @see JythonInitServlet
*/
static public Object getPyJavaObject(String stmnt, Class clazz) {
PythonInterpreter interp = JythonInitServlet.getInterpreter();
Object result = interp.eval(stmnt);
if (result instanceof PyJavaInstance) {
PyJavaInstance inst = (PyJavaInstance) result;
return inst.__tojava__(clazz);
} else if (result instanceof PyInstance) {
PyInstance inst = (PyInstance) result;
return inst.__tojava__(clazz);
}
throw new RuntimeException("Cannot evaluate statement '" + stmnt
+ "' and force result to '" + clazz.getName() + "'");
}
/**
* Returns the Jython controller singleton that is mapped to the given url.
*
* @param urlPath the url path for which the controller is looked up
* @return the Jython controller
*/
private Controller getController(String urlPath) {
for (Iterator iter = controllerMappings.entrySet().iterator(); iter
.hasNext();) {
Map.Entry entry = (Map.Entry) iter.next();
String registeredPath = (String) entry.getKey();
if (PathMatcher.match(registeredPath, urlPath)) {
return (Controller) entry.getValue();
}
}
return null;
}
/**
* The Spring MVS callback for executing the controllers functionality.
* Looks up the Jython controller singleton and delegates the request to it.
*
* @return the model and view
* @see org.springframework.web.servlet.mvc.AbstractController#handleRequestInternal(javax.servlet.http.HttpServletRequest,
* javax.servlet.http.HttpServletResponse)
*/
public ModelAndView handleRequestInternal(HttpServletRequest request,
HttpServletResponse response) throws Exception {
request.setAttribute("servletContext", getServletContext());
String lookupPath = urlPathHelper.getLookupPathForRequest(request);
Controller controller = getController(lookupPath);
ModelAndView modelAndView = controller.handleRequest(request, response);
return modelAndView;
}
}
|
That is i for now, and comments on this topic are appreciated.
分享到:
相关推荐
Jython是一种基于Java平台的Python实现,它允许Python代码与Java平台无缝集成。Jython的出现使得Python开发者能够利用Java的强大功能,同时保持Python的简洁性和易读性。它不仅实现了Python标准库,还支持大部分Java...
一个具有Jython集成的简单Java Spring Boot应用程序。 安装 git clone git@github.com:amemifra/Spring-Jython.git 您的系统必须已安装 Java 玛文 VS代码 Java扩展包-VS代码扩展 其余客户端扩展-VS代码扩展 例子 ...
该资源包含jython2.5.4.jar和jython2.7.0.jar,jython的版本要与jdk的版本要对应,这样可以解决“Unsupported major.minor version 51.0”问题。其中,对于jdk1.6.x的版本,请使用jython2.5.4.jar jdk1.7.x的版本,...
spring-batch-jython 能够在 python 中实现一个 spring 批处理作业这是一项正在进行的工作 使用: make deps 创建 deps。 make test 测试python接口
8. **Jython for Web开发**:Jython可以与Java Web框架如Spring、Struts或Play Framework集成,用于构建高效、可扩展的Web应用。 9. **Jython与Java的互操作性**:Jython提供了`__javabridge__`模块,使Python代码...
1. **Python与Jython的区别** Python是一种跨平台的高级编程语言,以其易读性强和代码简洁著称。Jython则是Python的一个变种,它与Java紧密集成,允许开发者使用Python语法编写Java应用程序。 2. **Jython的优势**...
jython官方正版资源 为了省去大家找Jython安装包的时间,附上此Jython版本。 Jython是一种完整的语言,而不是一个Java翻译器或仅仅是一个Python编译器,它是一个Python语言在Java中的完全实现。
1. Web开发:Jython可以结合Java的Web框架,如Spring或Struts,构建Web应用。 2. 数据分析:借助Java的数据处理库,如Apache Commons Math,Jython可进行复杂的数据分析。 3. 自动化测试:利用Jython编写测试脚本,...
标签 "burp" 和 "jython" 确定了这个压缩包的主要用途,即与Burp Suite的Python插件开发有关。Burp Suite的用户通常会用Jython来创建自定义扫描规则、代理拦截逻辑或自动化测试脚本,提高他们的渗透测试效率。 在...
### Jython 各个版本下载地址解析 Jython 是一种Python的实现,它允许Python代码在Java平台上运行。本文将详细介绍Jython不同版本的下载地址及相关文件信息。 #### Jython 2.2 版本 - **发布日期**:2007年8月24...
总的来说,Jython是Java与Python融合的一种有效手段,尤其适用于那些需要利用Python生态但又不希望完全脱离Java环境的项目。无论是进行快速原型开发,还是在现有Java系统中引入Python功能,Jython都是一个值得考虑的...
Jython 与传统的 CPython(Python 的标准实现)有所不同。它不是简单的 Java 翻译器或编译器,而是一个完整的 Python 解释器,实现了 Python 的语法和标准库,并且能够调用 Java 类库。这意味着 Jython 程序员不仅...
Jython 简单配置和使用 Jython 是一种完整的语言,而不是一个 Java 翻译器或仅仅是一个 Python 编译器,它是一个 Python 语言在 Java 中的完全实现。Jython 也有很多从 CPython 中继承的模块库。最有趣的事情是 ...
1. **与Python兼容**:Jython遵循Python的语法和语义,使得熟悉Python的开发者能够无缝地过渡到Jython环境。它支持大部分标准库,同时也提供了对Java类库的访问。 2. **运行在Java平台上**:Jython程序可以直接在...
Jython 安装和使用方法 Jython 是一个基于 Java 语言的 Python 实现,它可以让 Python 语言运行在 Java 平台上。下面将详细介绍 Jython 的安装和使用方法。 安装 Jython 首先,需要从 Jython 官方网站下载 Jython...
这份文档主要介绍了两个核心概念:一个是可以与Jython对象映射的Java接口类,以及一个用于从Jython文件中获取Java对象、Jython对象和Jython函数对象的工厂方法类。 ### Jython与Java接口的映射 首先,我们关注到的...
标题 "Jython" 提到的是一个与Python编程语言相关的开源项目——Jython。Jython是Python的一个实现,它允许Python代码在Java平台上运行。Jython将Python语言的灵活性和简洁性与Java平台的强大功能相结合,使得开发者...
5. **使用Jython库**:虽然Jython与标准Python 2.7兼容,但需要注意的是,并非所有Python库都支持Jython。因此,在编写脚本时,应确保使用的库是Jython兼容的,或者寻找相应的Java替代品。 **标签"python"关联知识...