Reputation: 1159
What is the proper way to get the result of this MySQL Query out of GetAllFarms and into a variable called err and farms? Sorry, doing a quick code try and coming from a different language.
var err, farms = GetAllFarms()
console.log("GetAllFarms:")
console.log(farms)
console.log(err)
function GetAllFarms(callback) {
query = db.query("SELECT * FROM farms ", function (err, result) {
console.log("DEBUG:QUERY//");
console.log(query.sql);
// console.log(result)
if (err) {
// console.log(err)
return callback(err, null)
} else {
// console.log(result)
return callback(null, result)
}
});
// db.end()
console.log("query")
console.log(query.result)
return query
}
Any help is much appreciated. Thanks
Upvotes: 0
Views: 34
Reputation: 1713
You have to decide wether you want to provide result via callback or with return. Don't mix them, it's confusable.
Callback approach
var err, farms = GetAllFarms()
console.log("GetAllFarms:")
console.log(farms)
console.log(err)
function GetAllFarms(callback) {
query = db.query("SELECT * FROM farms ", function (err, result) {
console.log("DEBUG:QUERY//");
console.log(query.sql);
// console.log(result)
if (err) {
// console.log(err)
return callback(err, null)
} else {
// console.log(result)
return callback(null, result)
}
});
// db.end()
console.log("query")
console.log(query.result)
}
// usage
GetAllFarms((error, result) => {
if (error) {
// handle error
}
// process result
})
Promise approach
var err, farms = GetAllFarms()
console.log("GetAllFarms:")
console.log(farms)
console.log(err)
function GetAllFarms() {
return new Promise((resolve, rejct) => {
db.query("SELECT * FROM farms ", function (err, result) {
console.log("DEBUG:QUERY//");
console.log(query.sql);
if (err) {
return reject(err)
} else {
return resolve(result)
}
});
});
}
// usage
(async () => {
const res = await GetAllFarms();
// or
GetAllFarms().then(/* ... */).catch(/* ... */);
})
Upvotes: 2