如何使用参数执行 sql 查询？

Question

val resultList: List[List[String]] =
  sql"""
       select * from exchange_transaction
     """
    .map(
      rs =>
        List(
          rs.string("transaction_id")
        )
    )
    .list()
    .apply()

执行上面的代码，就可以了但是这个 sql "select * from exchange_transaction" 不是固定的。我想要这个 sql 像一个参数，就像下面的（shorthand 代码.....）

def findMemberList(segmentExecuteSql: String
    val resultList: List[List[String]] =
      sql"""
           ＄segmentExecuteSql
         """
        .map(
          rs =>
            List(
              rs.string("transaction_id")
            )
        )
        .list()
        .apply()

但是执行后报错消息是

[error] s.StatementExecutor$$anon - SQL execution failed (Reason: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ''select * from  exchange_transaction'' at line 1
Query is : 
           'select * from  exchange_transaction'
         ):

   'select * from  exchange_transaction'

我该怎么办，如何解决异常？

Answer 1

你可以试试把resultList变成一个函数：

import scalikejdbc.SQL

def resultList(sqlQuery: String): List[List[String]] =
  SQL(sqlQuery)
    .map(
      rs =>
        List(
          rs.string("transaction_id")
        )
    )
    .list()
    .apply()

Answer 2

val resultList: List[List[String]] =
  SQL(segmentExecuteSql)
    .map(
      rs =>
        List(
          rs.string("transaction_id")
        )
    )
    .list()
    .apply()

如何使用参数执行 sql 查询？

How can i execute the sql query with parameter?

mysql

scala

playframework

scalikejdbc