使用ognl in表達式可能會遇到的問題

package cn;

public class A {
	
	private Integer id;

	public Integer getId() {
		return id;
	}

	public void setId(Integer id) {
		this.id = id;
	}

	@Override
	public int hashCode() {
		final int prime = 31;
		int result = 1;
		result = prime * result + ((id == null) ? 0 : id.hashCode());
		return result;
	}

	@Override
	public boolean equals(Object obj) {
		if (this == obj)
			return true;
		if (obj == null)
			return false;
		if (getClass() != obj.getClass())
			return false;
		A other = (A) obj;
		if (id == null) {
			if (other.id != null)
				return false;
		} else if (!id.equals(other.id))
			return false;
		return true;
	}
	
	

}

 

<%

	A a1 = new A();
	a1.setId(1);
	A a2 = new A();
	a2.setId(2);
	
	List<A> list = new ArrayList();
	list.add(a1);
	list.add(a2);
	
	request.setAttribute("list", list);
%>

<s:iterator value="#request.list" var="a">
	<s:property value="#a in #request.list"/><br/>
</s:iterator>

</body>
</html>

 

期望輸出：
true
true

但是實際是：
true

如果配置struts.el.throwExceptionOnFailure=true，那麼會得到如下異常: 

invalid comparison: cn.A and cn.A - Class: ognl.OgnlOps
File: OgnlOps.java
Method: compareWithConversion

 

原因在其ognl LanguageGuide 中也說了：

寫道

The ordering operators compare with compareTo() if their arguments are non-numeric and implement Comparable; otherwise, the arguments are interpreted as numbers and compared numerically. The in operator is not from Java; it tests for inclusion of e1 in e2, where e2 is interpreted as a collection. This test is not efficient: it iterates the collection. However, it uses the standard OGNL equality test.

它通過對象的compareTo去比較，且此處的in不是來自Java語言的。

 

所以如果你的對象沒有實現Comparable，那麼比較就會出問題，解決方案是：

1、實現Comparable來比較

2、使用集合本身自帶的contains

 

其實ognl完全可以規避這個問題，通過改造ognl.OgnlOps.compareWithConversion方法：

原來是：

case NONNUMERIC:
                    if ( ( t1 == NONNUMERIC ) && ( t2 == NONNUMERIC ) )
                    {
                    	
                        if ( ( v1 instanceof Comparable ) && v1.getClass().isAssignableFrom( v2.getClass() ) )
                        {
                            result = ( (Comparable) v1 ).compareTo( v2 );
                            break;
                        }
                        throw new IllegalArgumentException( "invalid comparison: " + v1.getClass().getName()
                            + " and " + v2.getClass().getName() );
                    }
                    // else fall through

 

可以改造爲如下來解決：

case NONNUMERIC:
                    if ( ( t1 == NONNUMERIC ) && ( t2 == NONNUMERIC ) )
                    {
                        if ( ( v1 instanceof Comparable ) && v1.getClass().isAssignableFrom( v2.getClass() ) )
                        {
                            result = ( (Comparable) v1 ).compareTo( v2 );
                            break;
                        }
                        result = -1;
                    }

 即最後不拋出異常，而是爲-1。

 

提交問題後（https://issues.apache.org/jira/browse/WW-4230），人家覺得沒改的必要，那就無所謂了，知道問題即可。