Spring MVC - inject context-param from web.xml

Question

I have defined the context params in web.xml

<context-param>
        <param-name>apikey</param-name>
        <param-value>45370652</param-value>
    </context-param>
    <context-param>
        <param-name>secretkey</param-name>
        <param-value>3eada72ef0ae12e15b138ae098c268c087f08ca8</param-value>
    </context-param>
</web-app>

I have enabled @Component and @Value annotations in Class and field levels in the bean class. But, it doesn't seem to read them. It is always null

@Component public class TokBoxSettings {

@Value("${apikey}")
private  String apikey; 

@Value("${secretkey}")
private  String secretkey;

I have also added the below bean mapping to spring-servlet.xml for configuring PropertyPlaceholder

<bean class="org.springframework.web.context.support.ServletContextPropertyPlaceholderConfigurer">
</bean>

Please let me know what i have missed

Generally bean definition and settings are contained in @Configuration annotated classes. Did you try changing @Component to @Configuration for TokBoxSettings? Those two stereotypes may differ in ordering. — px5x2
– px5x2, Commented Nov 5, 2015 at 12:46

sanjay · Accepted Answer · 2015-11-06 05:55:49Z

1

I am junior spring + hibernate developer,I have one alternative solution for this question,may be you know

I used bean name and its property to set direct context-param value

<bean ...>
   <property name="apikey" value="${apikey}" />
</bean>

answered Nov 6, 2015 at 5:55

sanjay

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Varun · Accepted Answer · 2015-12-22 10:13:15Z

1

Set property ignoreUnresolvablePlaceholders to true inside bean declaration. i.e. add below tags in spring-servlet.xml

<bean class="org.springframework.web.context.support.ServletContextPropertyPlaceholderConfigurer">
    <property name="ignoreUnresolvablePlaceholders" value="true"/>
</bean>

Its worked for me.

answered Dec 22, 2015 at 10:13

Varun

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