如何让Dapper.NET成为CRUD我的Oracle DB?
我有一个名为PLAYER_LOG
的表,它的标识是由触发器完成的,这里是sql
SELECT SQ_MASTER_SEQUENCE.NEXTVAL INTO tmpVar FROM dual;
:NEW.ID := tmpVar;
我的模特是:
public class PlayerLogStorage : IEntity //-> here is the identity
{
public string Cli { get; set; }
public string PlayerAnswer { get; set; }
public DateTime InsertDate { get; set; }
}
这是我的插入内容:
using (IDbConnection ctx = DbConnectionProvider.Instance.Connection)
{
ctx.Query<PlayerLogStorage>("INSERT INTO PLAYER_LOG (CLI, ANSWER, INSERT_DATE) VALUES (:Cli, :PlayerAnswer, :InsertDate)", new
{
Cli = model.Cli,
PlayerAnswer = model.PlayerAnswer,
InsertDate = model.InsertDate
});
}
这是例外:
ORA-01008: not all variables bound
答案 0 :(得分:11)
我遇到了类似的东西,但是使用了return语句。我发现的技巧是使用DynamicParameters对象。在我使用的系统中,insert语句必须在序列上调用NextVal,它不在触发器中。
var param = new DynamicParameters();
param.Add(name: "Cli", value: model.Cli, direction: ParameterDirection.Input);
param.Add(name: "PlayerAnswer", value: model.PlayerAnswer, direction: ParameterDirection.Input);
param.Add(name: "InsertDate", value: model.InsertDate, direction: ParameterDirection.Input);
param.Add(name: "Id", dbType: DbType.Int32, direction: ParameterDirection.Output);
using (IDbConnection ctx = DbConnectionProvider.Instance.Connection)
{
ctx.Execute("INSERT INTO PLAYER_LOG (CLI, ANSWER, INSERT_DATE) VALUES (:Cli, :PlayerAnswer, :InsertDate) returning Id into :Id", paramList);
}
var Id = param.get<int>("Id");
答案 1 :(得分:6)
除了bwalk2895的答案之外,您还可以将模型对象传递给DynamicParameters的构造函数,然后您只需要添加输出参数。保存几行代码,尤其是对于具有许多属性的对象。例如:
var param = new DynamicParameters(model);
param.Add(name: "Id", dbType: DbType.Int32, direction: ParameterDirection.Output);
using (IDbConnection ctx = DbConnectionProvider.Instance.Connection)
{
ctx.Execute("INSERT INTO PLAYER_LOG (CLI, ANSWER, INSERT_DATE) VALUES (:Cli, :PlayerAnswer, :InsertDate) returning Id into :Id", param);
}
var Id = param.Get<int>("Id");
更新:更正的方法名称