試圖使 function 返回 mysql 'SELECT * FROM' 結果，最終返回 undefined

Question

I'm trying to make a function that returns an array of stats from SQL 'SELECT * FROM', when I don't use a function to return it, it works, but I can't use that array outside of 'function(錯誤，結果）'

con.query(`SELECT * FROM MemberPeniz WHERE memberId = '${memberID}'`, function(err, results) {
            if(err) {
                throw err;
            }
            console.log(results)
        })

我試過這個：

function checkTimeoutMEMBERid(memberID) {
        con.query(`SELECT * FROM MemberPeniz WHERE memberId = '${memberID}'`, function(err, results) {
            if(err) {
                throw err;
            }
            return results
        })

    }
    const result = checkTimeoutMEMBERid(message.author.id);
    console.log(result)

但正如我所說，無論我傳入什么 id，它都會返回 undefined

Answer 1

這是因為，您將結果返回給回調，而回調不會返回給調用者 function。 您可以將 function 更新為，

function checkTimeoutMEMBERid(memberId) {
return new Promise((resolve, reject) =>{
    con.query(`SELECT * FROM MemberPeniz WHERE memberId = ?`,[memberId], (err, results) => {
        if(err) {
            return reject(err)
        }
        resolve(results)
    })
  })
}
checkTimeoutMEMBERid(message.author.id)
.then(result =>
{
console.log(result)
})