如何使用Scala Play2.6从数据库中选择单列 [英] How to select Single column from database using scala Play2.6

问题描述

我有一张表用户详细信息表，它包含在下面的列中

I have one table User details Table and it contains below Columns

用户名，用户名，电子邮件

User id,username,Email

我要选择用户ID为1的电子邮件 而且我希望响应采用正确的JSON格式

I want to select email whose user id is 1 and i want the response as in proper JSON Format

如何在使用Slick的Scala Play 2.6中做到这一点
到目前为止，我已经完成了

How to do that In Scala Play 2.6 with Slick
Till now I have done this

 def getPasswqord(username:String):Future[Seq[(String)]]= {
    val a2 = (sql"""select a.userpassword from user_details_table  a where a.Email=$username or a.Mobile_no=$username""".as[(String)])

    dbConfig.run(a2)

    }

由此我得到的响应格式为"["12345"]".

from this i am getting response in a format "["12345"]".

预期的输出格式为

"[{"Password":"12345"}]"

推荐答案

您应定义自定义

You should define custom Json Writes to format your query result. For example:

import play.api.libs.json.{JsPath, JsValue, Json, Writes}

// Given passwords
val passwords: Seq[(String)] = Seq(("12345"), ("qwerty"))
val writes : Writes[String] = (JsPath \ "password").write[String]

def toJson[T](sequence: Seq[T])(implicit writes: Writes[T]): JsValue = {
  Json.toJson(sequence)
}

toJson(passwords)(writes)

这将输出json对象[{"password":"12345"},{"password":"qwerty"}]

This will output json object [{"password":"12345"},{"password":"qwerty"}]

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