node.js - Sequelize - NodeJS - 返回 o 函数未定义

Question

我试图了解如何修复我的代码，因为我的函数返回未定义。我正在用 Sequelize 和 NodeJS 编写我的代码。

exports.getCreditsStudent = function(sgecode,collections,grade,year){
mssql.query('SELECT total FROM [dbo].[creditosAvulsos] WHERE codsge = $sgecode1 and nivelensino_idnivelensino = $collections1 and produto_idproduto = $grade1 and ano = $year1',
{ bind:{sgecode1: sgecode, collections1: collections, grade1: grade, year1: year}, type: mssql.QueryTypes.SELECT})
.then(total => {
    return total
  })

}

我在我的 Service.js 上调用这个函数，像这样:

examples = db.getCreditsStudent(sgecode,collections,grade,year);

问题出在哪里？如何解决？

谢谢。

Answer 1

函数 getCreditsStudent 实际上不返回任何内容。相反，它只是定义了一个异步进程。您需要返回您定义的 promise ，然后使用 async/await 或 promise 链来使用结果。

exports.getCreditsStudent = function(sgecode,collections,grade,year){
  return mssql.query('SELECT total FROM [dbo].[creditosAvulsos] WHERE codsge = $sgecode1 and 
  nivelensino_idnivelensino = $collections1 and produto_idproduto = $grade1 and ano =  $year1',
  { bind:{sgecode1: sgecode, collections1: collections, grade1: grade, year1: year},
  type: mssql.QueryTypes.SELECT})
  .then(total => {
      return total
    })
}

然后在异步函数中你可以做

examples = await db.getCreditsStudent(sgecode,collections,grade,year);

或者你可以使用

db.getCreditsStudent(sgecode, collections, grade, year)
.then(value => {
   examples = value;
   // whatever you want to do with examples...
});