Struts 2中的Json插件问题

我有以下代码，我将实现/ getJson将返回用户对象的function，因为json和/ getJson2将返回user2作为Json对象。

@ParentPackage("json-default") public class JsonAction extends ActionSupport{ private User user = new User("John","Smith"); private User user2 = new User("Smith","John"); public String populate(){ return "populate"; } @Action(value="/getJson", results = { @Result(name="success", type="json")}) public String test(){ return "success"; } @Action(value="/getJson2", results = { @Result(name="success", type="json")}) public String test2(){ return "success"; } @JSON(name="user") public User getUser() { return user; } public void setUser(User user) { this.user = user; } @JSON(name="user2") public User getUser2() { return user2; } public void setUser2(User user2) { this.user2 = user2; } } 

目前无论我打电话给哪种方法，我仍然得到以下结果：

 {"user":{"firstName":"John","lastName":"Smith"},"user2":{"firstName":"Smith","lastName":"John"}} 

可能吗？

更新：

我修改了代码：

 public class JsonAction extends ActionSupport{ private User user = new User("John","Smith"); private User user2 = new User("Smith","John"); public String populate(){ return "populate"; } @Action(value="/getJson", results = { @Result(name="success", type="json",params = { "includeProperties", "user"})}) public String test(){ return "success"; } @Action(value="/getJson2", results = { @Result(name="success", type="json",params = { "includeProperties", "user2"})}) public String test2(){ return "success"; } @JSON(name="user") public User getUser() { return user; } public void setUser(User user) { this.user = user; } @JSON(name="user2") public User getUser2() { return user2; } public void setUser2(User user2) { this.user2 = user2; } } 

现在我来了

 {"user":{}} 

和

 {"user2":{}} 

Answer 1

是的，有可能，解决方案需要使用包含/排除参数。

以下是一个例子。

方法getJson1和getJson2显示includeParameters，而getJson3显示excludeParameters。

注意：虽然该示例使用字符串作为include / exclude参数的参数，但该字符串被解释为正则表达式。 所以我可以将action3上的“string1，string2”替换为“string *”。

有关更多信息，请参阅： https ： //cwiki.apache.org/confluence/display/WW/JSON%20Plugin

 package struts2; import com.opensymphony.xwork2.ActionSupport; import org.apache.struts2.convention.annotation.Action; import org.apache.struts2.convention.annotation.ParentPackage; import org.apache.struts2.convention.annotation.Result; @ParentPackage("json-default") public class Test2 extends ActionSupport { private String string1 = "One"; private String string2 = "Two"; private String other = "Other"; public String getString1() { return this.string1; } public String getString2() { return this.string2; } public String getOther() { return this.other; } @Action(value="/getJson1", results = { @Result(type = "json", params = { "includeProperties", "string1" })}) public String action1() { return ActionSupport.SUCCESS; } @Action(value="/getJson2", results = { @Result(type = "json", params = { "includeProperties", "string2" })}) public String action2() { return ActionSupport.SUCCESS; } @Action(value="/getJson3", results = { @Result(type = "json", params = { "excludeProperties", "string1, string2" })}) public String action3() { return ActionSupport.SUCCESS; } } 

… / getJson1返回{“string1”：“One”}

… / getJson2返回{“string2”：“Two”}

… / getJson3返回{“other”：“Other”}

Answer 2

您必须包含要序列化的所有属性。 这包括User类的属性，例如：

 @Action(value="/getJson", results = { @Result(name="success", type="json",params = { "includeProperties", "user\.firstName, user\.lastName"})}) 

但是，另一种获得此工作的forms可能是使用正则表达式：

 @Action(value="/getJson", results = { @Result(name="success", type="json",params = { "includeProperties", "user\..*"})}) 

问候。

Answer 3

此操作提供两个属性：user和user2。

如果/ getJson和/ getJson2都映射到此操作类，则它们都将使用可用属性进行响应：user和user2。