I have a spring application and deploying in external tomcat(version-8.0.0).Using STS ide for development.
applicationContext.xml
<bean id="Manager" class="com.data.managers.Manager" p:hostName="${hostName}"
p:userName="${username}" p:pWord="${password}">
</bean>
Manager Class
public class Manager {
private String pWord;
public void setpWord(String pWord) {
this.pWord = pWord;
}
}
When i build the application,applicationContext file showing error
No setter found for property 'pWord' in class com.data.managers.Manager
what was wrong here.
Help me to solve this issue
thanks in advance
Getter and Setter for pWord should be getPWord() and setPWord(). The first letter after set and get should be uppercase.
Better change the property to password and getter setter to getPassword and setPassword
Related
i have a component that reads a configuration value from application.properties and accepts a string parameter in its constructor as such..
#Component
public class Person
{
#Value("${greeting}")
String greeting;
String name;
public Person(String name)
{
this.name = name;
onGreet( greeting + ", " + name );
}
public void onGreet(String message)
{
}
}
I need to instantiate this component as follows and override its "onGreet" event in the calling code as follows:
Person jack = new Person("jack")
{
public void onGreet(String message)
{
System.out.println( message );
}
};
However I end up getting this..
Parameter 0 of constructor in demo11.Person required a bean of type 'java.lang.String' that could not be found.
My application.properties is as follows:
greeting=hello
What am I missing here? Thank you.
It is literally telling you that the only constructor that you have requires a parameter that Spring knows nothing about.
Add a #Value to that String name in the constructor (right before the parameter) like so public Person(#Value("${name}") String name) if you want Spring to initalize it or remove that constructor
EDIT: some more explanation:
Spring is a dependency injection container. Meaning you define beans and let Spring create and inject them for you. Defining beans can be done in several ways (Java configuration, annotations or xml) here you are using annotation way via #Component.
Now that you have defined your bean (aka component) for Spring it will create it. For it to create it it needs to call a constructor. For that you need to provide it with all information necessary for constructor call - meaning all parameters. If parameters are other classes they need to be defined as beans as well (For example via #Component) if they are simple types like String you need to provide #Value for them.
Lastly if you ever use new ... to define Spring managed beans then the whole Spring magic disappears since Spring doesnt know about this bean instantiation anymore and will not autowire anything into it. For all intenses and purposes Spring is not aware of any objects you create with new.
I've defined an application name using the bootstrap.yml file in my spring boot application.
spring:
application:
name: abc
How can i get this application name during runtime/programmatically ?
You should be able to use the #Value annotation to access any property you set in a properties/YAML file:
#Value("${spring.application.name}")
private String appName;
#Autowired
private ApplicationContext applicationContext;
...
this.applicationContext.getId();
Please, find this:
# IDENTITY (ContextIdApplicationContextInitializer)
spring.application.name=
spring.application.index=
In Spring Boot Reference Manual.
And follow with source code for that ContextIdApplicationContextInitializer class:
#Override
public void initialize(ConfigurableApplicationContext applicationContext) {
applicationContext.setId(getApplicationId(applicationContext.getEnvironment()));
}
Where the default behavior is with this:
/**
* Placeholder pattern to resolve for application name
*/
private static final String NAME_PATTERN = "${vcap.application.name:${spring.application.name:${spring.config.name:application}}}";
Since the #Value annotation is discouraged in Spring Boot when referencing configuration properties, and because applicationContext.getId(); doesn't always return the value of spring.application.name another way is to get the value from the Environment directly
private final Environment environment;
...
public MyBean(final Environment environment) {
this.environment = environment;
}
...
private getApplicationName() {
return this.environment.get("spring.application.name");
}
Another possible way would be to create your own ConfigurationProperties class to get access to the value.
I'm not saying these are the best ways, and I hope/wish that there is a better way, but it is a way.
Note! If your using a SpringBootTest, you need to suplly the properties/yml.
Otherwise, the environment/appcontext does not load the config files.
The, your app name is not set.
Like so:
#PropertySource("classpath:application.properties")
#RunWith(SpringRunner.class)
#SpringBootTest
....
This post is aged but I hate unanswered questions.
So use the following snippet:
#Value("${spring.application.name [: defaultValue]}")
private String appName;
What is between [] is optional.
So I found a really ugly way to do this, but it works so I'm not searching further. Maybe this will help someone.
The basic premise is that spring Environment stores the value inside a propertySource.. It appears that bootstrap config is stored in the ResourcePropertySource and so you can get it from that. For me it is currently throwing an exception, but then I can get the value out of the exception, so I haven't looked any further:
try {
this.environment.getProperty("name", ResourcePropertySource.class);
} catch (ConversionFailedException e) {
String res = (String)e.getValue();
}
And then you can just do this for every property you are interested in.
Like I said ugly, but it works.
I am new for spring security. I've seen many posts on how to inject values via annotation from external property file. I've tried many ways, but I always end up with java.lang.IllegalArgumentException: Could not resolve placeholder 'val.id' exception.
Can you provide me some tips how to handle this exception please?
My java class is the following one:
#Controller
public class Employee {
#Value("${val.id}")
public String valId;
public String getValId() {
return valId;
}
public void setValId(String valId) {
this.valId = valId;
}
My property file is called val.properties which is located under WEB-INF, and its content is
val.id=xyz
I put the following in my main context bean.
<context:property-placeholder location="/WEB-INF/*.properties" />
<bean id="valProp" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer"
p:location="/WEB-INF/val.properties"/>
A continuous question:
The injecting values from properties file to annotated beans works fine as I accepted the answer above. However, I cannot able to inject it to #PreAuthorize(...) annotation by following the same procedure.
Assume I want to secure a method called 'update'. This method is allowed if and only if valId is equal to empId. values of valId and empId are initialized in the val.properties file.
my java bean is:
public class Employee {
public String valId;
public String empId;
public String getValId() {
return valId;
}
public void setValId(String valId) {
this.valId = valId;
}
public String getEmpId() {
return empId;
}
public void setEmpId(String empId) {
this.empId = empId;
}
}
my property file contains:
val.id=nn
emp.id=nn
I have the place holder configuration in my main context file:
<context:property-placeholder location="/WEB-INF/*.properties" />
<bean id="valProp" class="org.springframework.beans.factory.config.PropertiesFactoryBean"
p:location="/WEB-INF/val.properties"/>
My PreAuthorize annotation (method security) is:
#PreAuthorize("(#{valProp['val.id']} == #{valProp['emp.id']})")
public boolean update(){
//if accessable
return true;
}
But the expression #{valProp['val.id']} == #{valProp['emp.id']} is not evaluated.
Did I do any mistake to inject values? It was worked when I annotate member variables, but it doesn't work here. Any idea please? Thanks in advance.
try to consider the following
1). change your annotation to:
#Value("#{valProp['val.id']}")
2). Replace PropertyPlaceholderConfigurer by PropertiesFactoryBean.
Hope this will resolve the exception.
The reason why the exception is thrown is, because the property placeholder by default throws an exception when a values cannot be resolved.
Furthermore you have two property placeholders, via which probably not all values can be resolved.
You can change this behaviour via setting the ignore-unresolvable property:
<context:property-placeholder location="/WEB-INF/*.properties" ignore-unresolvable="true" />
<bean id="valProp" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer"
p:location="/WEB-INF/val.properties" p:ignoreUnresolvablePlaceholders="true" />
Note however that b< turning off this feature typos in a property file will not be detected.
I've never worked with Spring before, and I've run into a configuration object that looks somewhat like this
public class Config {
#Value("${app.module.config1}")
private String config1;
#Value("${app.module.config2}")
private String config2
...
public String getConfig1() {
return config1;
}
...
Can anyone explain what is happening here? I'm assuming this is some type of code injection, but I can't find where these values are coming from!
They allow you to direct inject a Value from a properties file (system or declared property) in the variable. Using the util:properties tag you can add something like this in your applicationContext.xml
<util:properties id="message" location="classpath:com/your/program/resources/message.properties" />
Pointing for a properties file named "message.properties" with some content:
application.hello.message = Hello World!
And then, in your java source file, inject a direct value from this properties file using the #Value annotation:
#Value("#{message['application.hello.message']}")
private String helloWorldMessage;
#Value("${app.module.config1}")
This is part of the spring expression language where the spring framework would look for app.module.config1 JVM property from System.getProperties() and injects the value of that property into config1 attribute in that class. Please see this reference for more details in Spring 3.0.x and this reference for the current docs.
I am developing a web application using spring MVC. I just want a simple example of how to do session management in this. I have seen lot of forums but I am not able to get a clear picture of this
My requirement is
I have an object, which I would like to be accessible in all controllers and JSP's I
would like to set that in the controller and get that in JSP
I am looking for something like
Session.setAtribute();
Could you please let me know a very simple instance . Thank you
There are different ways of accessing servlet session in Spring MVC. But I think this one is the one that best suits your problem. You can create a session scoped bean, which holds your desired info:
#Component("myObjectHolder")
#Scope(WebApplicationContext.SCOPE_SESSION)
public class MyObjectHolderImpl implements MyObjectHolder {
private long userId;
private String username;
private Theme theme;
// Getters & Setter
}
Then, you can access to it from other beans:
#Controller
public class MyController {
#Autowired private MyObjectHolder myObjectHolder;
#RequestMapping
public ModelAndView switchTheme(String themeId) {
...
Theme newTheme = themeService.get(themeId);
myObjectHolder.setTheme(newTheme);
...
}
}
You can access directly from your view too, but you must configure it:
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
...
<property name="exposedContextBeanNames" value="myObjectHolder" />
</bean>
And in your JSP:
Hi ${myObjectHolder.username}, you switched
application theme to ${myObjectHolder.theme.name}
The simplest approach is to access HttpSession directly by injecting it into your handler method:
#RequestMapping("/page")
public ModelAndView page(HttpSession session) {
session.getAttribute("foo");
}