ars*_*nal 2 java jsp spring-mvc restful-url
我最近开始使用Spring MVC框架.我在网上阅读大量教程时取得了很大的进步.
关于我申请的背景 -
我必须使用表单中提供的详细信息对其他服务(已在tomcat上部署)进行REST URL调用.所以我已经使用JSP创建了一个表单,其内容如下图所示 - 我不知道如何通过从表单条目中创建url然后显示该URL的响应来进行REST URL调用下一个屏幕.
所以在上面的表格中,如果我已经写了User Id as 1000012848,并且checkbox is selected (means true) for Debug Flag在Attribute Name I have selected first row(通常我们也可以选择所有三个)Machine Name is localhost,Port Number is 8080然后url应该看起来像这样 -
http://localhost:8080/service/newservice/v1/get/PP.USERID=1000012848,debugflag=true/host.profile.ACCOUNT
因此,在我们将从表单条目中创建的所有URL中,下面的行总是在同一个位置 - 然后在每个表单条目开始附加后
service/newservice/v1/get/
现在在创建了上面的url之后,只要我点击提交,就会调用上面的url,无论从URL获得什么响应,它都会显示在下一个屏幕(result.jsp文件)中我不知道怎么做？以下是我创建的文件.任何人都可以帮助我解决我的问题吗？我需要做什么代码才能解决这个问题？
student.jsp文件(制作表格)
<%@taglib uri="http://www.springframework.org/tags/form" prefix="form"%>
<html>
<res:head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<meta name="layout" content="main" />
<title>First Tutorial</title>
</res:head>
<res:body>
<form:form method="POST" action="/_hostnewapp/addStudent">
<table>
<tr>
<td><form:label path="userId">User Id</form:label></td>
<td><form:input path="userId" /></td>
</tr>
<tr>
<td>Debug Flag :</td>
<td><form:checkbox path="debugFlag" /></td>
</tr>
<tr>
<td>Attribute Name</td>
<td><form:select path="attributeNames" items="${attributeNamesList}"
multiple="true" /></td>
</tr>
<!-- <tr>
<td>Environment</td>
<td><form:checkboxes items="${environmentList}"
path="environments" /></td>
</tr>
-->
<tr>
<td><form:label path="machineName">Machine Name</form:label></td>
<td><form:input path="machineName" /></td>
</tr>
<tr>
<td><form:label path="portNumber">Port Number</form:label></td>
<td><form:input path="portNumber" /></td>
</tr>
<tr>
<td colspan="2"><input type="submit" value="Submit" /></td>
</tr>
</table>
</form:form>
</res:body>
</html>
result.jsp文件(我打算在点击该url后显示结果)
<%@taglib uri="http://www.springframework.org/tags/form" prefix="form"%>
<html>
<res:head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<meta name="layout" content="main" />
<title>HostDomain</title>
</res:head>
<res:body>
<h2>Response after submitting the result</h2>
// Not sure what I need to add here to show the result after hitting the url
</res:body>
</html>
控制器类 -
@Controller
public class SampleRaptorController {
@RequestMapping(value = "/student", method = RequestMethod.GET)
public ModelAndView student() {
return new ModelAndView("student", "command", new Student());
}
@RequestMapping(value = "/addStudent", method = RequestMethod.POST)
public String addStudent(@ModelAttribute("SpringWeb") Student student,
ModelMap model) {
model.addAttribute("userId", student.getUserId());
return "result";
}
@ModelAttribute("attributeNamesList")
public Map<String,String> populateSkillList() {
//Data referencing for java skills list box
Map<String,String> attributeNamesList = new LinkedHashMap<String,String>();
attributeNamesList.put("ACCOUNT","host.profile.ACC");
attributeNamesList.put("ADVERTISING","host.profile.ADV");
attributeNamesList.put("SEGMENTATION","host.profile.SEG");
return attributeNamesList;
}
}
您可以使用RestTemplate从Spring组件调用RESTful URL
所以,你的控制器方法可以如下
@Controller
public class SampleRaptorController {
@Autowired
RestTemplate restTemplate;
@RequestMapping(value = "/addStudent", method = RequestMethod.POST)
public String addStudent( @ModelAttribute("SpringWeb") Student student,
Model model){
// Build URL
StringBuilder url = new StringBuilder().
append("http://localhost:8080/service/newservice/v1/get").
append("?PP.USERID=" + student.getUserId).
append("&debugflag=" + student.isDebugFlag);// so on
// Call service
String result = restTemplate.getForObject(url.toString(), String.class);
model.addAttribute("result", result);
return "result";
}
}
您的spring配置应该注册restTemplate,如下所示:
<bean id="restTemplate" class="org.springframework.web.client.RestTemplate"/>
有关更多详细信息,请参阅RestTemplate文档.
以上应该做.
一个建议..您的RESTful URL(http://localhost:8080/service/newservice/v1/get/PP.USERID=1000012848, debugflag=true/host.profile.ACCOUNT)非常糟糕.一旦您解决了问题,我建议您谷歌搜索一个好的RESTful URL的样子.
干杯,Vinay