2017-10-09 80 views
0
let userModel = sequelize.define('userModel',{ 
//Schema 
},{ 
//options 
    instanceMethods:{ 
     checkPassword(){} 
    } 
}) 
......... 
async Login(){ 
    let ins = await userModel.findOne('xxx') 
    ins.checkPassword // <= is not a function 
} 

看來該實例方法沒有加載到userModel.prototype上。而我的console.log(ins),checkPassword存在於屬性instanceMethods中。無法獲取續集實例方法

如何調用instanceMethods?

sequelize版本:4.11.1

+0

它工作時,我寫了''userModel.prototype.checkPassword''。那麼選項''instanceMethods''如何工作? – seeLuck

回答