sequelize获取查询的所有数据,而无需执行我的条件“ where”

时间:2018-09-27 03:30:42

标签: mysql node.js sequelize.js

im尝试使用以下序列化代码从数据库mysql中获取一些值:

/**
 * signInGoogle
 */

private fun signInGoogle() = launch{

    //init google sigin
    val gso = GoogleSignInOptions.Builder(GoogleSignInOptions.DEFAULT_SIGN_IN)
                .requestIdToken(GOOGLE_WEBCLIENT_ID)
                .requestEmail()
                .build()



    val gApiClient = GoogleApiClient.Builder(activity)
            .addApi(Auth.GOOGLE_SIGN_IN_API,gso)
            .build()

    val c = gApiClient.blockingConnect()

     if(c.isSuccess && gApiClient.isConnected){
         gApiClient.clearDefaultAccountAndReconnect().await()
     }

    val mGoogleSignInClient = GoogleSignIn.getClient(activity, gso)
    val signInIntent = mGoogleSignInClient.signInIntent

    //start activit
    startActivityForResult(signInIntent, GOOGLE_SIGNIN)

}//end request google login

问题在于,sequelize不仅获取表中所有条件的数据,而且还获取表中的所有值。

有什么主意吗?

我们最近更改了AWS中的服务器项目(将elasticbeanstalk更改为其他实例eb)

1 个答案:

答案 0 :(得分:2)

我只能找到的问题是 Where 应该是小写,例如 where

subSector.findAll({
    Where: {
        Enterprise_Evaluation_ID: 2
    }
})

将其更改为此:

subSector.findAll({
    where: { // <--------- HERE
        Enterprise_Evaluation_ID: 2 // <--------- Also make sure the field name is proper
    }
})