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  • Spring 3 JSON with MVC

    - by stevedbrown
    Is there a way to build Spring Web calls that consume and produce application/json formatted requests and responses respectively? Maybe this isn't Spring MVC, I'm not sure. I'm looking for Spring libraries that behave in a similar fashion to Jersey/JSON. The best case would be if there was an annotation that I could add to the Controller classes that would turn them into JSON service calls. A tutorial showing how to build Spring Web Services with JSON would be great. EDIT: I'm looking for an annotation based approach (similar to Jersey). EDIT2: Like Jersey, I am looking for REST support (POST,GET,DELETE,PUT). EDIT3: Most preferably, this will be the pom.xml entries and some information on using the spring-js with jackson Spring native version of things.

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  • Using spring:nestedPath tag

    - by Ravi
    Hello All, I have this code, I know I'm missing something but don't know what. It would be great if you help me out. I'm new to Spring MVC. I'm trying out a simple Login application in Spring MVC. This is my web.xml <?xml version="1.0" encoding="UTF-8"?> <web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"> <listener> <listener-class> org.springframework.web.context.ContextLoaderListener </listener-class> </listener> <servlet> <servlet-name>springapp</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> </servlet> <servlet-mapping> <servlet-name>springapp</servlet-name> <url-pattern>/app/*</url-pattern> </servlet-mapping> <session-config> <session-timeout> 30 </session-timeout> </session-config> <welcome-file-list> <welcome-file>index.jsp</welcome-file> </welcome-file-list> here is my springapp-servlet.xml file <?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd"> <bean name="/login" class="springapp.web.LoginController"/> <bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver"> <property name="prefix" value="/WEB-INF/jsp/"></property> <property name="suffix" value=".jsp"></property> </bean> This is my applicationContext.xml file <?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="employee" class="springapp.domain.Employee" /> </beans> Here is my LoginController.java file package springapp.web; import org.springframework.web.servlet.ModelAndView; import org.springframework.web.servlet.mvc.SimpleFormController; import springapp.domain.Employee; public class LoginController extends SimpleFormController{ public LoginController(){ setCommandName("loginEmployee"); setCommandClass(Employee.class); setFormView("login"); setSuccessView("welcome"); } @Override protected ModelAndView onSubmit(Object command) throws Exception { return super.onSubmit(command); } } And finally my login.jsp file <%@ taglib uri="http://www.springframework.org/tags" prefix="spring" %> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Timesheet Login</title> </head> <body> <spring:nestedPath path="employee"> <form action="" method="POST"> <table width="200" border="1" align="center" cellpadding="10" cellspacing="0"> <tr> <td>Username:</td> <td> <spring:bind path="userName"> <input type="text" name="${status.expression}" id="${status.value}" /> </spring:bind> </td> </tr> <tr> <td>Password</td> <td> <spring:bind path="password"> <input type="password" name="${status.expression}" id="${status.value}" /> </spring:bind> </td> </tr> <tr> <td colspan="2"><input type="submit" value="Login" /></td> </tr> </table> </form> </spring:nestedPath> </body> </html> But when I try to run the code I get this error javax.servlet.ServletException: javax.servlet.jsp.JspTagException: Neither BindingResult nor plain target object for bean name 'employee' available as request attribute

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  • How to propagate spring security login to EJBs?

    - by tangens
    Context I have a J2EE application running on a JBoss 4.2.3 application server. The application is reachabe through a web interface. The authentication is done with basic authentication. Inside of the EJBs I ask the security context of the bean for the principal (the name of the logged in user) and do some authorization checks if this user is allowed to access this method of the EJB. The EJBs life inside a different ear than the servlets handling the web frontend, so I can't access the spring application context directly. Required change I want to switch to Spring Security for handling the user login. Question How can I propagate the spring login information to the JBoss security context so I can still use my EJBs without having to rewrite them? Ideas and links I already found a page talking about "Propagating Identity from Spring Security to the EJB Layer", but unfortunatelly it refers to an older version of Spring Security (Acegi) and I'm not familiar enough with Spring Security to make this work with the actual version (3.0.2).

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  • deep injection - spring

    - by Bob
    What is the best way (or options) for accessing spring components at layers deep within the application that aren't managed by spring? For example, I have a low level utility POJO class into which I need to autowire/inject a spring component. I'll call it LowLevelHelper. There are multiple classes that use LowLevelHelper - most are layers away from anything that is hooked up with spring. One option would be to make all the layers in to spring components, but that seems like I'm hacking my design to force spring to help me. I have some complex things going on that won't be nearly as clean if I have to @Autowire all the pieces and don't new anything. Another option might be to manually inject the component in the low level class, but I'm not really sure if this is possible or the right solution.

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  • Is it possible to transfer data between html pages driven by spring web flow?

    - by Easwaramoorthy Kanagaraj
    I am aware of passing data between jsp in spring web flow. Is it possible to transfer data between html pages driven by spring web flow. I don't want to use the HTML5 local storage capabilities. Example: Page 1: Search box for an employee id. Page 2: Search result for the employee details. Two ways that I could think: Get the employee details in page 1 by ajax and pass the result to the page two. Pass the employee id to page 2 and get the result by ajax in onload. In both case I need to pass any variable/data. I am confused in doing this. Is there anything in the Spring webflow using which I could do this? Thanks in advance, Easwar

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  • SpringFramework3.0: How to create interceptors that only apply to requests that map to certain contr

    - by Fusion2004
    In it's simplest form, I want an interceptor that checks session data to see if a user is logged in, and if not redirects them to the login page. Obviously, I wouldn't want this interceptor to be used on say the welcome page or the login page itself. I've seen a design that uses a listing of every url to one of two interceptors, one doing nothing and the other being the actual interceptor you want implemented, but this design seems very clunky and limits the ease of extensibility of the application. It makes sense to me that there should be an annotation-based way of using interceptors, but this doesn't seem to exist. My friend has the idea of actually modifying the handler class so that during each request it checks the Controller it is mapping the request to for a new annotation we would create (ex @Interceptor("loginInterceptor") ). A major point of my thinking is the extensibility, because I'd like to later implement similar interceptors for role-based authentication and/or administration authentication. Does it sound like my friend's approach would work for this? Or what is a proper way of going about doing this?

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  • Use spring tag in XSLT

    - by X-Pippes
    I have a XSL/XML parser to produce jsp/html code. Using MVC model I need to accees spring library in order to perform i18n translation. Thus, given the xml <a> ... <country>EN</country> ... </a> and using <spring:message code="table_country_code.EN"/> tag, choose based on the browser language, the transalation into England, Inglaterra, etc... However, the XSL do not support <spring:message> tag. The idea is to have a XSLT with something like this <spring:message code="table_country_code.><xsl:value-of select="country"/>"/>` I also tried to create the spring tag in Java when I make a parse to create the XML but I sill have the same error. ERROR [STDERR] (http-0.0.0.0-8080-1) file:///C:/Software/Jboss/jboss-soa-p-5/jboss-as/bin/jstl:; Line #5; Column #58; The prefix "spring" for element "spring:message" is not bound. How can I resolve?

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  • How to create TestContext for Spring Test?

    - by HDave
    Newcomer to Spring here, so pardon me if this is a stupid question. I have a relatively small Java library that implements a few dozen beans (no database or GUI). I have created a Spring Bean configuration file that other Java projects use to inject my beans into their stuff. I am now for the first time trying to use Spring Test to inject some of these beans into my junit test classes (rather than simply instantiating them). I am doing this partly to learn Spring Test and partly to force the tests to use the same bean configuration file I provide for others. In the Spring documentation is says I need to create an application context using the "TestContext" class that comes with Spring. I believe this should be done in a spring XML file that I reference via the @ContextConfiguration annotation on my test class. @ContextConfiguration({"/test-applicationContext.xml"}) However, there is no hint as to what to put in the file! When I go to run my tests from within Eclipse it errors out saying "failed to load Application Context"....of course.

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  • Spring validation has error in XML document from ServletContext resource

    - by user1441404
    I applied spring validation in my registration page .but the follwing error are shown in my server log of my app engine server. javax.servlet.UnavailableException: org.springframework.beans.factory.xml.XmlBeanDefinitionStoreException: Line 22 in XML document from ServletContext resource [/WEB-INF/spring/appServlet/servlet-context.xml] is invalid; nested exception is org.xml.sax.SAXParseException; lineNumber: 22; columnNumber: 30; cvc-complex-type.2.4.c: The matching wildcard is strict, but no declaration can be found for element 'property'. My code is given below : <?xml version="1.0" encoding="UTF-8"?> <beans:beans xmlns="http://www.springframework.org/schema/mvc" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:beans="http://www.springframework.org/schema/beans" xmlns:context="http://www.springframework.org/schema/context" xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.0.xsd http://www.springframework.org/schema/jee http://www.springframework.org/schema/jee/spring-jee-3.0.xsd > <beans:bean name="/register" class="com.my.registration.NewUserRegistration"> <property name="validator"> <bean class="com.my.validation.UserValidator" /> </property> <beans:property name="formView" value="newuser"></beans:property> <beans:property name="successView" value="home"></beans:property> </beans:bean> <beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver"> <beans:property name="prefix" value="/WEB-INF/views/" /> <beans:property name="suffix" value=".jsp" /> </beans:bean> </beans:beans>

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  • How to use SQLErrorCodeSQLExceptionTranslator and DAO class with @Repository in Spring?

    - by GuidoMB
    I'm using Spring 3.0.2 and I have a class called MovieDAO that uses JDBC to handle the db. I have set the @Repository annotations and I want to convert the SQLException to the Spring's DataAccessException I have the following example: @Repository public class JDBCCommentDAO implements CommentDAO { static JDBCCommentDAO instance; ConnectionManager connectionManager; private JDBCCommentDAO() { connectionManager = new ConnectionManager("org.postgresql.Driver", "postgres", "postgres"); } static public synchronized JDBCCommentDAO getInstance() { if (instance == null) instance = new JDBCCommentDAO(); return instance; } @Override public Collection<Comment> getComments(User user) throws DAOException { Collection<Comment> comments = new ArrayList<Comment>(); try { String query = "SELECT * FROM Comments WHERE Comments.userId = ?"; Connection conn = connectionManager.getConnection(); PreparedStatement stmt = conn.prepareStatement(query); stmt = conn.prepareStatement(query); stmt.setInt(1, user.getId()); ResultSet result = stmt.executeQuery(); while (result.next()) { Movie movie = JDBCMovieDAO.getInstance().getLightMovie(result.getInt("movie")); comments.add(new Comment(result.getString("text"), result.getInt("score"), user, result.getDate("date"), movie)); } connectionManager.closeConnection(conn); } catch (SQLException e) { e.printStackTrace(); //CONVERT TO DATAACCESSEXCEPTION } return comments; } } I Don't know how to get the Translator and I don't want to extends any Spring class, because that is why I'm using the @Repository annotation

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  • How to display downloaded youtube annotations ( not captions) on offline video player?

    - by kowalsk
    i need to display the annotations for some of my downloaded youtube videos. It's also important to make them closeable. Making them clickable would also be nice. What i found out so far: From what i understand there there is a vlc plugin/extension that could also render the annotations but i'm having a hard time finding it. Mplayer might also be an option but i'd have to convert the xml files to .bmp and then use a bmov filter to play them. Any suggestions welcome. Edit: to further clarify i would like to display/overlay annotations from a youtube xml file (i'm willing to go through a conversion step if i have to) pretty much the same way i can display subtitles from a srt or sub file. Edit2: I'm open to using other players too ( if vlc is not a feasible option for what i want)

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  • spring classloader isolation problem on jboss

    - by mkosm
    Hi, i have two ears deployed on jboss with proper classloaders isolation settings. When seam bean call spring beans which make some queries on database everything works fine, but when spring quartz job bean is invoked and execute tries to execute database queries then there is a problem because spring tries too use hibernate jar from other ear and exception is thrown! It is clearily spring classloader isolation problem. Did anyone meet such a problem? How to ensure isolation?

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  • Where to put default-servlet-handler in Spring MVC configuration

    - by gigadot
    In my web.xml, the default servlet mapping, i.e. /, is mapped to Spring dispatcher. In my Spring dispatcher configuration, I have DefaultAnnotationHandlerMapping, ControllerClassNameHandlerMapping and AnnotationMethodHandlerAdapter which allows me to map url to controllers either by its class name or its @Requestmapping annotation. However, there are some static resources under the web root which I also want spring dispatcher to serve using default servlet. According to Spring documentation, this can be done using <mvc:default-servlet-handler/> tag. In the configuration below, there are 4 candidate locations that I marked which are possible to insert this tag. Inserting the tag in different location causes the dispatcher to behave differently as following : Case 1 : If I insert it at location 1, the dispatcher will no longer be able to handle mapping by the @RequestMapping and controller class name but it will be serving the static content normally. Cas 2, 3 : It will be able to handle mapping by the @RequestMapping and controller class name as well as serving the static content if other mapping cannot be done successfully. Case 4 : It will not be able to serve the static contents. Therefore, Case 2 and 3 are desirable .According to Spring documentation, this tag configures a handler which precedence order is given to lowest so why the position matters? and Which is the best position to put this tag? <?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:mvc="http://www.springframework.org/schema/mvc" xmlns:context="http://www.springframework.org/schema/context" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd"> <context:annotation-config/> <context:component-scan base-package="webapp.controller"/> <!-- Location 1 --> <!-- Enable annotation-based controllers using @Controller annotations --> <bean id="annotationUrlMapping" class="org.springframework.web.servlet.mvc.annotation.DefaultAnnotationHandlerMapping"/> <!-- Location 2 --> <bean id="controllerClassNameHandlerMapping" class="org.springframework.web.servlet.mvc.support.ControllerClassNameHandlerMapping"/> <!-- Location 3 --> <bean id="annotationMethodHandlerAdapter" class="org.springframework.web.servlet.mvc.annotation.AnnotationMethodHandlerAdapter"/> <!-- Location 4 --> <mvc:default-servlet-handler/> <!-- All views are JSPs loaded from /WEB-INF/jsp --> <bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver"> <property name="viewClass" value="org.springframework.web.servlet.view.JstlView"/> <property name="prefix" value="/WEB-INF/jsp/"/> <property name="suffix" value=".jsp"/> </bean> </beans>

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  • How can I get access to spring container?

    - by Antony
    I have a spring container running, and I have class with which I want to have access to the bean created inside spring container. The class I have is not registered with the spring container. One thing I can do is that I can use MethodInvoker to call a static method, so I will have access to static field (this would be a bean from spring container) in my class always. Do we have class like WebapplicationContextUtils for a application that is not web?

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  • How to setup an hibernate project using annotations compliant with JPA2.0

    - by phmr
    I would like to setup an hibernate project, for this I use the lastest hibernate 3.5-Final. BTW my IDE is Netbeans. The problem is that each time a run the application, it seems to start from a fresh database whatever db backend I use (I tried hsqldb & sqlite): here is my persistence.xml <?xml version="1.0" encoding="UTF-8"?> <persistence version="1.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd"> <persistence-unit name="PMMPU" transaction-type="RESOURCE_LOCAL"> <provider>org.hibernate.ejb.HibernatePersistence</provider> <class>model.Extrait</class> <class>model.Mot</class> <class>model.Prefixe</class> <class>model.Suffixe</class> <class>model.Texte</class> <properties> <property name="hibernate.dialect" value="dialect.SQLiteDialect"/> <property name="hibernate.connection.username" value=""/> <property name="hibernate.connection.driver_class" value="org.sqlite.JDBC"/> <property name="hibernate.connection.password" value=""/> <property name="hibernate.connection.url" value="jdbc:sqlite:test.db"/> <property name="hibernate.cache.provider_class" value="org.hibernate.cache.NoCacheProvider"/> <property name="hibernate.hbm2ddl.auto" value="update"/> </properties> </persistence-unit> </persistence> I tried to change hibernate.hbm2ddl.auto value. I got a HibernateUtil class which takes care of creating the emf & em: public class HibernateUtil { private static EntityManagerFactory emf = null; private static EntityManager em = null; public static EntityManagerFactory getEmf() { if(emf == null) emf = Persistence.createEntityManagerFactory("PMMPU"); return emf; } public static EntityManager getEm() { if(em == null) em = getEmf().createEntityManager(); return em; } What did a I do wrong ? edit1: further research with mysql lead me to think that the problem is related to sqlite & hsqldb interaction with hibernate 3.5

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  • Interfaces with hibernate annotations

    - by molleman
    Hello i am wondering how i would be able to annotate an interface @Entity @Table(name = "FOLDER_TABLE") public class Folder implements Serializable, Hierarchy { @Id @GeneratedValue(strategy = GenerationType.AUTO) @Column(name = "folder_id", updatable = false, nullable = false) private int fId; @Column(name = "folder_name") private String folderName; @OneToMany(cascade = CascadeType.ALL) @JoinTable(name = "FOLDER_JOIN_FILE_INFORMATION_TABLE", joinColumns = { @JoinColumn(name = "folder_id") }, inverseJoinColumns = { @JoinColumn(name = "file_information_id") }) private List< Hierarchy > fileInformation = new ArrayList< Hierarchy >(); above and below are 2 classes that implement an interface called Hierarchy, the folder class has a list of Hierarchyies being a folder or a fileinformation class @Entity @Table(name = "FILE_INFORMATION_TABLE") public class FileInformation implements Serializable, Hierarchy { @Id @GeneratedValue(strategy = GenerationType.AUTO) @Column(name = "file_information_id", updatable = false, nullable = false) private int ieId; @Column (name = "location") private String location; i have serached the web for someway to annotate or a workaround but i cannot map the interface which is simply this public interface Hierarchy { } i get a mapping exeception on the List of hierarchyies with a folder but i dont know how to map the class correctly

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  • Setting column length of a Long value with JPA annotations

    - by Gearóid
    Hi, I'm performing a little database optimisation at the moment and would like to set the column lengths in my table through JPA. So far I have no problem setting the String (varchar) lengths using JPA as follows: @Column(unique=true, nullable=false, length=99) public String getEmail() { return email; } However, when I want to do the same for a column which is of type Long (bigint), it doesn't work. For example, if I write: @Id @Column(length=7) @GeneratedValue(strategy = GenerationType.AUTO) public Long getId() { return id; } The column size is still set as the default of 20. Are we able to set these lengths in JPA or am I barking up the wrong tree? Thanks, Gearoid.

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  • using the ASP.NET Caching API via method annotations in C#

    - by craigmoliver
    In C#, is it possible to decorate a method with an annotation to populate the cache object with the return value of the method? Currently I'm using the following class to cache data objects: public class SiteCache { // 7 days + 6 hours (offset to avoid repeats peak time) private const int KeepForHours = 174; public static void Set(string cacheKey, Object o) { if (o != null) HttpContext.Current.Cache.Insert(cacheKey, o, null, DateTime.Now.AddHours(KeepForHours), TimeSpan.Zero); } public static object Get(string cacheKey) { return HttpContext.Current.Cache[cacheKey]; } public static void Clear(string sKey) { HttpContext.Current.Cache.Remove(sKey); } public static void Clear() { foreach (DictionaryEntry item in HttpContext.Current.Cache) { Clear(item.Key.ToString()); } } } In methods I want to cache I do this: [DataObjectMethod(DataObjectMethodType.Select)] public static SiteSettingsInfo SiteSettings_SelectOne_Name(string Name) { var ck = string.Format("SiteSettings_SelectOne_Name-Name_{0}-", Name.ToLower()); var dt = (DataTable)SiteCache.Get(ck); if (dt == null) { dt = new DataTable(); dt.Load(ModelProvider.SiteSettings_SelectOne_Name(Name)); SiteCache.Set(ck, dt); } var info = new SiteSettingsInfo(); foreach (DataRowView dr in dt.DefaultView) info = SiteSettingsInfo_Load(dr); return info; } Is it possible to separate those concerns like so: (notice the new annotation) [CacheReturnValue] [DataObjectMethod(DataObjectMethodType.Select)] public static SiteSettingsInfo SiteSettings_SelectOne_Name(string Name) { var dt = new DataTable(); dt.Load(ModelProvider.SiteSettings_SelectOne_Name(Name)); var info = new SiteSettingsInfo(); foreach (DataRowView dr in dt.DefaultView) info = SiteSettingsInfo_Load(dr); return info; }

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  • question on database query using hibernate in java with annotations

    - by molleman
    Hello, simple question regarding HQL(Hibernate query language) so i have user class , that can hold a list of Projects, how do i take this out of the database depending on a username, this is how i take out my user String username = "stephen"; YFUser user = (YFUser) session.createQuery("select u FROM YFUser u where u.username = :username").setParameter("username", name).uniqueResult(); but i want to take out the list of projects here is the projects list within the class YFUser(my user class); how would i query the database to get this list of projects @Entity @Table(name = "yf_user_table") public class YFUser implements Serializable,ILightEntity { ......... @OneToMany(cascade = CascadeType.ALL,fetch = FetchType.LAZY) @JoinTable(name = "YFUSER_JOIN_PROJECT", joinColumns = { @JoinColumn(name = "user_id") }, inverseJoinColumns = { @JoinColumn(name = "project_id") }) private List<Project> projects = new ArrayList<Project>();

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  • Enums and Annotations

    - by PeterMmm
    I want to use an Annotation in compile-safe form. To pass the value() to the Annotation i want to use the String representation of an enum. Is there a way to use @A with a value from enum E ? public class T { public enum E { a,b; } // C1: i want this, but it won't compile @A(E.a) void bar() { // C2: no chance, it won't compile @A(E.a.toString()) void bar2() { } // C3: this is ok @A("a"+"b") void bar3() { } // C4: is constant like C3, is'nt it ? @A(""+E.a) void bar4() { } } @interface A { String value(); }

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  • What does the Spring framework do? Should I use it? Why or why not?

    - by sangfroid
    So, I'm starting a brand-new project in Java, and am considering using Spring. Why am I considering Spring? Because lots of people tell me I should use Spring! Seriously, any time I've tried to get people to explain what exactly Spring is or what it does, they can never give me a straight answer. I've checked the intros on the SpringSource site, and they're either really complicated or really tutorial-focused, and none of them give me a good idea of why I should be using it, or how it will make my life easier. Sometimes people throw around the term "dependency injection", which just confuses me even more, because I think I have a different understanding of what that term means. Anyway, here's a little about my background and my app : Been developing in Java for a while, doing back-end web development. Yes, I do a ton of unit testing. To facilitate this, I typically make (at least) two versions of a method : one that uses instance variables, and one that only uses variables that are passed in to the method. The one that uses instance variables calls the other one, supplying the instance variables. When it comes time to unit test, I use Mockito to mock up the objects and then make calls to the method that doesn't use instance variables. This is what I've always understood "dependency injection" to be. My app is pretty simple, from a CS perspective. Small project, 1-2 developers to start with. Mostly CRUD-type operations with a a bunch of search thrown in. Basically a bunch of RESTful web services, plus a web front-end and then eventually some mobile clients. I'm thinking of doing the front-end in straight HTML/CSS/JS/JQuery, so no real plans to use JSP. Using Hibernate as an ORM, and Jersey to implement the webservices. I've already started coding, and am really eager to get a demo out there that I can shop around and see if anyone wants to invest. So obviously time is of the essence. I understand Spring has quite the learning curve, plus it looks like it necessitates a whole bunch of XML configuration, which I typically try to avoid like the plague. But if it can make my life easier and (especially) if make it can make development and testing faster, I'm willing to bite the bullet and learn Spring. So please. Educate me. Should I use Spring? Why or why not?

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  • View Body not showing in Spring 3.2

    - by Dennis Röttger
    I'm currently trying to get into Java Web Development in general in Spring more specifically. I've set up my project as follows - hello.jsp: <html> <head> <title>Spring 3.0 MVC Series: Hello World - ViralPatel.net</title> </head> <body> <p>ABC ${message}</p> </body> </html> HelloWorldController.java: package controllers; import org.springframework.stereotype.Controller; import org.springframework.web.bind.annotation.RequestMapping; import org.springframework.web.servlet.ModelAndView; @Controller public class HelloWorldController { @RequestMapping("/hello") public ModelAndView helloWorld() { String message = "Hello World, Spring 3.0!"; System.out.println(message); return new ModelAndView("hello", "message", message); } } web.xml: <?xml version="1.0" encoding="UTF-8"?> <web-app xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" version="2.5"> <display-name>Spring3MVC</display-name> <welcome-file-list> <welcome-file>index.jsp</welcome-file> </welcome-file-list> <servlet> <servlet-name>spring</servlet-name> <servlet-class> org.springframework.web.servlet.DispatcherServlet </servlet-class> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>spring</servlet-name> <url-pattern>*.html</url-pattern> </servlet-mapping> </web-app> spring-servlet.xml: <?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" xmlns:context="http://www.springframework.org/schema/context" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd"> <context:component-scan base-package="controllers" /> <bean id="viewResolver" class="org.springframework.web.servlet.view.UrlBasedViewResolver"> <property name="viewClass" value="org.springframework.web.servlet.view.JstlView" /> <property name="prefix" value="/WEB-INF/jsp/" /> <property name="suffix" value=".jsp" /> </bean> </beans> I can start up the Server just fine and navigate to hello.html, which is resolved by the servlet to give me hello.jsp, the title of the .jsp shows (Spring 3.0 MVC Series: etc. etc.), alas, the body does not. Not the JSTL-Variable and not the "ABC" either. I've implemented jstl-1.2 in my lib-folder.

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