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Spring properties (property-placeholder) autowiring

开发者 https://www.devze.com 2023-01-01 02:37 出处:网络
I have in my applicationContext.xml <context:property-placeholder location=\"classpath*:*.properties\" />

I have in my applicationContext.xml

<context:property-placeholder location="classpath*:*.properties" />


<bean id="clientPreferencesManager" class="pl.bildpresse.bildchat2.business.ClientPreferencesManager" >
    <property name="clientApiUrl" val开发者_运维问答ue="${clientapi.url}" />     
</bean>

Is it possible to do the same by autowire ? Something like :

@Autowired
@Qualifier("${clientapi.url}")
public void setClientApiUrl(String clientApiUrl) {
    this.clientApiUrl = clientApiUrl;
}


You can use @Value:

@Value("${clientapi.url}") 
public void setClientApiUrl(String clientApiUrl) { 
    this.clientApiUrl = clientApiUrl; 
}


It took me some time to understand why it didn't work. I always used a # instead of a $. I always got the message:

EL1008E:(pos 0): Field or property 'secretkey' cannot be found on object of type 'org.springframework.beans.factory.config.BeanExpressionContext'

Just had to change it from:

@Value("#{secretkey}')

to

@Value('${secretkey}')

I hope this saves somebody's time.


Ok. Just got it. You need to add @Autowired Something like:

@Autowired
@Value("${clientapi.url}") 
private StringValueResolver resolver;

I'm using spring 3.0.0.RELEASE

Cheers


For spring 3.0, the correct way is the one shown - using @Value("${expression}")

For spring pre-3.0, you can try:

@Autowired
private StringValueResolver resolver;

There were no context initialization problems here, but I'm not sure it will work. Using the resolver you can resolve properties.


My solution is to use

<context:property-override location="classpath:clientapi.properties" />

and then in clientapi.properties file

clientPreferencesManager.clientApiUrl=http://localhost:8084/ClientAPI/resources/

This one is good too

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