blocks|key|3053192|text|您应该为您可以返回的两个对象返回一个超类或接口。如果你还没有为它们定义一个超类，你仍然可以返回Object，并进行不太好的instanceof验证，以知道返回了哪个类。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3053193|代码示例：|3053194|class+SuperClass+{
}

class+MyClass1+extends+SuperClass+{
}

class+MyClass2+extends+SuperClass+{
}

private+SuperClass+myMethod(String+parameter)+{
++++SuperClass+sc+=+null;
++++//just+trying+some+code,+this+could+be+replaced+for+your+logic
++++if+(parameter.equals("1"))+{
++++++++sc+=+new+MyClass1();
++++}+else+{
++++++++sc+=+new+MyClass2();
++++}
++++return+sc;
}

private+void+myOtherMethod(String+parameter)+{
++++//some+code+here...
++++SuperClass+sc+=+myMethod(parameter);
++++if+(sc+!=+null)+{
++++++++//securing+the+upcast...
++++++++if+(sc+instanceof+MyClass1)+{
++++++++++++//logic+of+MyClass1+instance...
++++++++}
++++++++if+(sc+instanceof+MyClass2)+{
++++++++++++//logic+of+MyClass2+instance...
++++++++}
++++}
++++//more+logic+over+here...
}|code-block|syntax|javascript|3053195|entityMap^0|1B|6|1P|A|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

You should return a super class or interface for both objects you could return. If you haven´t defined a super class for them, you can still return <code>Object</code> and do the not so good <code>instanceof</code> validation to know which class has been returned.

Code Example:

<pre><code>class SuperClass {
}

class MyClass1 extends SuperClass {
}

class MyClass2 extends SuperClass {
}

private SuperClass myMethod(String parameter) {
 SuperClass sc = null;
 //just trying some code, this could be replaced for your logic
 if (parameter.equals("1")) {
 sc = new MyClass1();
 } else {
 sc = new MyClass2();
 }
 return sc;
}

private void myOtherMethod(String parameter) {
 //some code here...
 SuperClass sc = myMethod(parameter);
 if (sc != null) {
 //securing the upcast...
 if (sc instanceof MyClass1) {
 //logic of MyClass1 instance...
 }
 if (sc instanceof MyClass2) {
 //logic of MyClass2 instance...
 }
 }
 //more logic over here...
}
</code></pre>

blocks|key|3053213|text|在这种情况下，两个类的超类或接口必须相同，然后从函数返回超类引用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3053214|例如：|3053215|class+SuperClass+{
//+Variables

//+Functions
}

class+example1+extends+SuperClass{

//+Variables

//+Functions

}

class+example2+extends+SuperClass{

//+Variables

//+Functions

}


private+SuperClass+d(String+example)+{

++++++++if(example.startsWith("example1"))
++++++++++++return+new+example1();
++++++++else+if((example.startsWith("example2"))
++++++++++++return+new+example2();

}|code-block|syntax|javascript|3053216|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

In this case supper class or interface of both the class must be same and then return that super class reference from the function.

For example - 

<pre><code>class SuperClass {
// Variables

// Functions
}

class example1 extends SuperClass{

// Variables

// Functions

}

class example2 extends SuperClass{

// Variables

// Functions

}


private SuperClass d(String example) {

 if(example.startsWith("example1"))
 return new example1();
 else if((example.startsWith("example2"))
 return new example2();

}
</code></pre>

blocks|key|3053240|text|private+Object+getObject(String+example)+throws+ClassNotFoundException,+IllegalAccessException,+InstantiationException+{
++++if+(example.startsWith("example1"))
++++++++return+Class.forName("example1ClassName").newInstance();
++++if+(example.startsWith("example2"))+{
++++++++return+Class.forName("example2ClassName").newInstance();
++++}
++++return+null;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3053241|但是正如您注意到的，这个方法抛出了很多异常，所以要小心。您的类可能没有公共默认构造函数或一些类路径问题。它很敏感，需要强制转换操作才能与返回值一起操作。这样做的好处是您可以避免任何继承/超类问题，并且可以使用任何要实例化的类。|unstyled|3053242|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>private Object getObject(String example) throws ClassNotFoundException, IllegalAccessException, InstantiationException {
 if (example.startsWith("example1"))
 return Class.forName("example1ClassName").newInstance();
 if (example.startsWith("example2")) {
 return Class.forName("example2ClassName").newInstance();
 }
 return null;
}
</code></pre>

But as you noticed this method throws a lot of exceptions so be careful. You classes might have no public default constructors or some classpath issues. It's sensitive and requires cast operation to operate with returned value. The benefit is that you are free of any inheritance/superclass issues and can use any class to be instantiated.

I need to check for an condition and if it is satisfied i need to return a class if it is not satisfied i need to return another class 

<pre><code>private &lt;T&gt; d(String example) {

 if(example.startsWith("example1"))
 return new example1();
 else if((example.startsWith("example2"))
 return new example2();

 }
</code></pre>

I am not sure how to return it can u please explain.

Detecting which class to be returned in java

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我需要检查一个条件，如果它满足，我需要返回一个类，如果它不满足，我需要返回另一个类private <T> d(String example) { if(example.startsWith("example1")) return new example1(); else...

问检测java中要返回的类
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