嗨伙计们,我试图禁止输入密钥在textarea中提交表单
使用以下功能:
function noenter() {
return !(window.event && window.event.keyCode == 13);
}
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HTML代码:
<form:form modelAttribute="myObject" method="post" action="${myUrl}">
<form:textarea path="name" class="new_obj submits_on_return" cols="40" rows="3" onkeypress="return noenter()"></form:textarea>
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但它根本不起作用,总是提交,任何想法为什么?
我正在使用Spring的webapp工作,我总是使用JSP和Jquery.在最近的一次谈话中,有人评论道:你应该使用Flex代替.Flex会成为Web应用程序的不错选择吗?使用它与仅使用JSP和Jquery有什么好处?
我正在从控制器中获取外部网页(作为字符串)。如何将包含HTML的字符串作为返回ModelAndView?
这是我的代码。我得到了一些错误。谁能帮我解决问题?我收到了一些转换错误,比如ClassPathResource转换到资源Resource和资源对象r到资源。
package firstspring;
import javax.annotation.Resource;
import org.springframework.beans.factory.BeanFactory;
import org.springframework.beans.factory.xml.XmlBeanFactory;
import org.springframework.core.io.ClassPathResource;
import org.springframework.core.io.Resource;
public class test {
public static void main(String[] args){
Resource r =new ClassPathResource("applicationContext.xml");
BeanFactory bf=new XmlBeanFactory(r);
Student stud=(Student)bf.getBean("first");
stud.display();
}
}
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这是我的 XML 文件;
<?xml version="1.0" encoding="UTF-8"?>
<beans
xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:p="http://www.springframework.org/schema/p"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd">
<bean id="first" class="Student">
<property name="name" value="sneha"></property>
</bean>
</beans>
Run Code Online (Sandbox Code Playgroud) 我们有一个Web应用程序中实现Spring MVC 3.2使用JPA作为ORM框架。现在的问题是EntityManager正在创建与数据库的大量开放连接。我们希望以这样一种方式处理它:对于每个查询,都应该在完成后建立和关闭连接。
根据 spring 实现 EntityManager 创建一次。但这里的问题是我们希望以某种方式处理 EntityManager 为查询数据库而创建的客户端连接。
每当在数据库中完成查询时,该连接就会进入睡眠状态,相反,我们希望在查询返回结果后关闭它。
数据库类型: MySQL
我的 JPA 配置是:
<tx:annotation-driven transaction-manager="transactionManager" />
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
<bean class="org.springframework.orm.jpa.support.PersistenceAnnotationBeanPostProcessor" />
<bean id="entityManagerFactory"
class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="packagesToScan" value="com.reppify" />
<property name="jpaPropertyMap" ref="jpaPropertyMap" />
<property name="dataSource" ref="dataSourceLocal" />
<property name="persistenceUnitName" value="cron-jpa" />
<property name="jpaVendorAdapter">
<bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter" />
</property>
</bean>
<bean id="dataSourceLocal"
class="org.springframework.jdbc.datasource.DriverManagerDataSource">
<property name="driverClassName" value="${database.driver}" />
<property name="url" value="${database.url}" />
<property name="username" value="${database.user}" />
<property name="password" value="${database.password}" …Run Code Online (Sandbox Code Playgroud) 我正在编写一个 Spring RESTful 服务,它使用 JSON 并执行一些操作。由于请求包含大量参数,我想到了使用 Spring 的 Jackson 映射将我的请求参数映射到 Java 对象。
我的POJO
public class RequestInput {
private int id;
private String name;
// parameters follow
// getter and setter
}
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我的控制器
@Controller
public class RequestController{
@RequestMapping(method=RequestMethod.POST, value="/rest/postRequest")
public void handleRequest(@RequestBody RequestInput input){
// code follows
}
}
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数据发布时一切正常
{“id”:1,“名称”:“ABCD”}
但是当数据发布为
{“id”:1,“first_name”:“ABCD”}
对象中名称的值作为 返回NULL。
你能帮我理解如何将请求中的 first_name 映射到 Java POJO 中的 name param
线路1:
public ModelAndView viewCustomerDetails(@RequestParam("custId") Integer customerId, @RequestParam("categoryName") String categoryName, HttpServletRequest request) throws BusinessException{
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线路2:
public ModelAndView viewCustomerDetails(@RequestMapping("custId") Integer customerId, @RequestMapping("categoryName") String categoryName, HttpServletRequest request) throws BusinessException{
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我要去深入我的项目代码,并具有一定的混乱@RequestParam和@RequestMapping一些时间,我发现@RequestParam有的时候@RequestMapping.在我的理解中,两者都将分配custId给customerId数据成员的值.
我的jsp文件的一部分:
<form:input mandatory="true" id="CustNameTxt" path="custName" cssClass="txtfield controlWidth" readonly="false" />
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为了更好地理解我的问题,我已经在Line2中进行了编辑
我正在尝试实现Spring Boot并设置我的用户服务.我无法找到无法找到UserService bean的错误.任何建议或方向将不胜感激.
UserController的
@RestController
public class UserController {
private final IUserService userService;
@Inject
public UserController(final IUserService userService) {
this.userService = userService;
}
@RequestMapping(value = "/user", method = RequestMethod.POST)
public FBUser createUser(@RequestBody @Valid final FBUser fBUser) {
return userService.save(fBUser);
}
}
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IUserService
public interface IUserService {
FBUser save(FBUser fBUser);
}
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UserService
public class UserService implements IUserService {
private final IUserRepository repository;
@Inject
public UserService(final IUserRepository repository) {
this.repository = repository;
}
@Override
@Transactional
public FBUser save(final FBUser fBUser) {
return …Run Code Online (Sandbox Code Playgroud) 我无法展示可能的餐厅.我得到了我的控制器类:
@Controller
public class RestaurantController extends MultiActionController{
private RestaurantDAO restaurantDAO;
public void setRestaurantDAO(RestaurantDAO restaurantDAO) {
this.restaurantDAO = restaurantDAO;
}
@RequestMapping("/restaurant/{restaurantId}")
public ModelAndView restaurantid(@PathVariable("contactId") int id,
HttpServletRequest request,
HttpServletResponse response) throws Exception {
Restaurant restaurant = restaurantDAO.findRestaurantById(id);
ModelMap modelMap = new ModelMap();
modelMap.addAttribute("restaurant", restaurant);
return new ModelAndView("restaurant", modelMap);
}
}
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我的jsp只是:
<c:out value="${restaurant.name }"
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在我的spring-servlet.xml中:
<bean name="/restaurant/**" class="web.RestaurantController" >
<property name="restaurantDAO" ref="myRestaurantDAO"/>
</bean>
Run Code Online (Sandbox Code Playgroud) 我使用带有spring和hibernate的tapestry 5,我尝试构建我的应用程序抛出我的这个异常??? !!! ???什么是问题,任何建议都会是这样的relif ????
[WARNING] Some problems were encountered while building the effective model for com.fit:rent-a-car:war:1.0.0-SNAPSHOT
[WARNING] 'build.plugins.plugin.version' for org.apache.maven.plugins:maven-compiler-plugin is missing. @ line 194, column 18
[WARNING] 'reporting.plugins.plugin.version' for org.apache.maven.plugins:maven-surefire-plugin is missing. @ line 235, column 17
[WARNING]
[WARNING] It is highly recommended to fix these problems because they threaten the stability of your build.
[WARNING]
[WARNING] For this reason, future Maven versions might no longer support building such malformed projects.
.....
.....
[ERROR] ioc.Registry No service implements …Run Code Online (Sandbox Code Playgroud) spring-mvc ×10
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java-ee ×1
javascript ×1
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json ×1
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