ViewModelProvider.AndroidViewModelFactory.getInstance(getApplication())). get(xxxViewModel .class) 发现还是创建失败 然后检查发现xxxViewModel 的构造函数有一行因为别的地方修改后报错,但是没捕获异常,修复之后就好了。
In this tutorial, we shall learn to fix Kotlin: Cannot create an instance of an abstract class. Following is a sample of the error you might get during compilation. Error:(2, 27) Kotlin: Cannot create an instance of an abstract class Solution In Kotlin, we cannot create an instance of ...
VC.net 创建 ref class 后 C#调用报错 Cannot create an instance of the abstract class or interface CS0144 同样的错误:http://computer-programming-forum.com/7-vc.net/aa44def3c57690b4.htm 发现如上面链接所说,他是vc.net 继承了一个.net的abstract Class , 但是没有override 所有应该override的方法 ...
'<typename>' cannot be used as an attribute because it is not a class '<typename>' cannot inherit from <type> '' because it expands the access of the base <type> outside the assembly '<typename>' cannot shadow a 'MustOverride' method implicitly declared for property '<propertyname>' ...
Cannot create an instance of ... because Type.ContainsGenericParameters is true. Cannot create folder because a file or directory with the same name already exists Cannot create the instance of Abstract or interface 'syste..data.common.dbconnection Cannot delete mdf file after it has been access...
throw new ExecutorException("Do not know how to create an instance of " + resultType); } 这个方法是根据结果集返回值的类型创建出相应的bean字段对象 1、当实体使用无参构造器时 mybatis会调用createResultObject方法中 代码语言:txt AI代码解释
In this scenario, you may receive an error message that resembles the following: Cannot create an instance of OLE DB provider "provider_name" for linked server "linked_server_name" Cause This issue can occur if...
Could not create an instance of type QTY. Type is an interface or abstract class and cannot be instantiated. The class hierarchy is the following : [System.Xml.Serialization.XmlIncludeAttribute(typeof(IVXB_TS))] public abstract partial class ANY : object, System.ComponentModel.INotifyPropertyChange...
create(resultType); } else if (shouldApplyAutomaticMappings(resultMap, false)) { return createByConstructorSignature(rsw, resultType, constructorArgTypes, constructorArgs); } throw new ExecutorException("Do not know how to create an instance of " + resultType); } 这个方法是根据结果集返回值的...
Cannot create an instance of OLE DB provider "OraOLEDB.Oracle" for linked server "xxxxxxx". 在SQL SERVER 2008 R2下用Windows 身份认证的登录名创建了一个访问ORACLE数据库的链接服务器xxxxx,测试成功,木有问题,但是其它登录名使用该链接服务器时,报如下错误: 消息 7302,级别 16,状态 1,第 1 行 Cannot ...