简体   繁体   English

试图使 function 返回 mysql 'SELECT * FROM' 结果,最终返回 undefined

[英]Trying to make function that returns mysql 'SELECT * FROM' results, ended up returning undefined

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(err, results)' 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(错误,结果)'

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

I tried this:我试过这个:

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)

but as I say it returns undefined no matter what id I pass in但正如我所说,无论我传入什么 id,它都会返回 undefined

This is because, you are returning results to callback which will not return to caller function.这是因为,您将结果返回给回调,而回调不会返回给调用者 function。 You can update function as,您可以将 function 更新为,

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)
})

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM