如何在Spring XML中将通用参数传递给构造函数？

Question

I have a class

package com.foo;
public class Foo<T extends Number>{

    private T value;

    public Foo(T value){
        this.value=value;
    }

}

Which I'm trying to instantiate in Spring XML.

<bean id="myFoo" class="com.foo.Foo">
    <constructor-arg type="java.lang.Number" value="1" />
</bean>

But when I run my App I get this error...

How can I solve this?

Answer 1

There is no constructor accepting an argument for type `java.lang.Number'. Check the JavaDoc: java.lang.Number

You can set a literal, but in your case, it appears you should choose Integer, which extends from Number. In order to set the literal value, you would use Spring's expression language. You can either do: "#{new Integer(10)}" or the shorthand: "#{10}"

<bean id="myFoo" class="com.foo.Foo">
    <constructor-arg type="java.lang.Number" value="#{new Integer(10)}" />
</bean>