problems with the values of the promise Promise { value }
I'm exporting a function where it returns the value of a query in mysql, when I do the path of the Json and the values that I assign from the function appear as Promise {value} (example Promise {61}), and I do not know how to fix the value of the promise.
query.js
const mysql = require('mysql2/promise');
const vertodo = async function(req){
const connection = await mysql.createConnection({
host: '192.168.0.222',
port: 3306,
user: 'root',
database: 'project',
password : '123456789'
});
return [rows, fields] = await connection.query('Select COUNT(*) total from table1 where id=' +req.id);
connection.end();
}
module.exports.vertodo = vertodo;
index.js
const data_mysql = require("./query")
// value jsonObj
// [{ id: 1,
// name: 'XTRA',
// proceso: 0
// },
// { id: 2,
// name: 'Maq',
// proceso: 0
// }]
for (var i = 0; i < jsonObj.length; i++) {
//jsonObj[i].proceso = 0;
jsonObj[i].procesa = data_mysql.vertodo(jsonObj[i]).then( rows =>{
return rows[0][0].total
})
// jsonObj[i].proceso = Promise { 61 } , Promise { 33 };
}
setTimeout(function(){
for (var i = 0; i < jsonObj.length; i++) {
console.log(jsonObj[i].procesa); // Promise { 61}, Promise { 33 }
}
}, 3000);
I do not know how I can solve it, thanks
javascript node.js promise mysql2
add a comment |
I'm exporting a function where it returns the value of a query in mysql, when I do the path of the Json and the values that I assign from the function appear as Promise {value} (example Promise {61}), and I do not know how to fix the value of the promise.
query.js
const mysql = require('mysql2/promise');
const vertodo = async function(req){
const connection = await mysql.createConnection({
host: '192.168.0.222',
port: 3306,
user: 'root',
database: 'project',
password : '123456789'
});
return [rows, fields] = await connection.query('Select COUNT(*) total from table1 where id=' +req.id);
connection.end();
}
module.exports.vertodo = vertodo;
index.js
const data_mysql = require("./query")
// value jsonObj
// [{ id: 1,
// name: 'XTRA',
// proceso: 0
// },
// { id: 2,
// name: 'Maq',
// proceso: 0
// }]
for (var i = 0; i < jsonObj.length; i++) {
//jsonObj[i].proceso = 0;
jsonObj[i].procesa = data_mysql.vertodo(jsonObj[i]).then( rows =>{
return rows[0][0].total
})
// jsonObj[i].proceso = Promise { 61 } , Promise { 33 };
}
setTimeout(function(){
for (var i = 0; i < jsonObj.length; i++) {
console.log(jsonObj[i].procesa); // Promise { 61}, Promise { 33 }
}
}, 3000);
I do not know how I can solve it, thanks
javascript node.js promise mysql2
add a comment |
I'm exporting a function where it returns the value of a query in mysql, when I do the path of the Json and the values that I assign from the function appear as Promise {value} (example Promise {61}), and I do not know how to fix the value of the promise.
query.js
const mysql = require('mysql2/promise');
const vertodo = async function(req){
const connection = await mysql.createConnection({
host: '192.168.0.222',
port: 3306,
user: 'root',
database: 'project',
password : '123456789'
});
return [rows, fields] = await connection.query('Select COUNT(*) total from table1 where id=' +req.id);
connection.end();
}
module.exports.vertodo = vertodo;
index.js
const data_mysql = require("./query")
// value jsonObj
// [{ id: 1,
// name: 'XTRA',
// proceso: 0
// },
// { id: 2,
// name: 'Maq',
// proceso: 0
// }]
for (var i = 0; i < jsonObj.length; i++) {
//jsonObj[i].proceso = 0;
jsonObj[i].procesa = data_mysql.vertodo(jsonObj[i]).then( rows =>{
return rows[0][0].total
})
// jsonObj[i].proceso = Promise { 61 } , Promise { 33 };
}
setTimeout(function(){
for (var i = 0; i < jsonObj.length; i++) {
console.log(jsonObj[i].procesa); // Promise { 61}, Promise { 33 }
}
}, 3000);
I do not know how I can solve it, thanks
javascript node.js promise mysql2
I'm exporting a function where it returns the value of a query in mysql, when I do the path of the Json and the values that I assign from the function appear as Promise {value} (example Promise {61}), and I do not know how to fix the value of the promise.
query.js
const mysql = require('mysql2/promise');
const vertodo = async function(req){
const connection = await mysql.createConnection({
host: '192.168.0.222',
port: 3306,
user: 'root',
database: 'project',
password : '123456789'
});
return [rows, fields] = await connection.query('Select COUNT(*) total from table1 where id=' +req.id);
connection.end();
}
module.exports.vertodo = vertodo;
index.js
const data_mysql = require("./query")
// value jsonObj
// [{ id: 1,
// name: 'XTRA',
// proceso: 0
// },
// { id: 2,
// name: 'Maq',
// proceso: 0
// }]
for (var i = 0; i < jsonObj.length; i++) {
//jsonObj[i].proceso = 0;
jsonObj[i].procesa = data_mysql.vertodo(jsonObj[i]).then( rows =>{
return rows[0][0].total
})
// jsonObj[i].proceso = Promise { 61 } , Promise { 33 };
}
setTimeout(function(){
for (var i = 0; i < jsonObj.length; i++) {
console.log(jsonObj[i].procesa); // Promise { 61}, Promise { 33 }
}
}, 3000);
I do not know how I can solve it, thanks
javascript node.js promise mysql2
javascript node.js promise mysql2
asked Nov 22 '18 at 17:21
user10597291
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Your function "vertodo" returns a promise. You need to wait for the promise to resolve with the query response data. The easiest way to do this is to use the async/await syntax. See below.
index.js
const data_mysql = require("./query")
runQuery()
async function runQuery() {
for (var i = 0; i < jsonObj.length; i++) {
//jsonObj[i].proceso = 0;
queryResponse = await data_mysql.vertodo(jsonObj[i])
// process queryResponse
}
}
answered Nov 22 '18 at 18:08
FakeFootballFakeFootball
11
add a comment |
I don't fully understand what you want to do, but I assume that you want to run those queries in loop and wait until all finish and then log results.
If those queries need to be run in sequence.
async function runQueries() {
for (const item of jsonObj) {
item.procesa = await data_mysql.vertodo(item);
}
}
runQueries.then(function () {
jsonObj.forEach(function (item) {
console.log(item.procesa);
});
});
answered Nov 22 '18 at 23:02
Jake NohJake Noh
615
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your function "vertodo" returns a promise. You need to wait for the promise to resolve with the query response data. The easiest way to do this is to use the async/await syntax. See below.
index.js
const data_mysql = require("./query")
runQuery()
async function runQuery() {
for (var i = 0; i < jsonObj.length; i++) {
//jsonObj[i].proceso = 0;
queryResponse = await data_mysql.vertodo(jsonObj[i])
// process queryResponse
}
}
answered Nov 22 '18 at 18:08
FakeFootballFakeFootball
11
add a comment |
Your function "vertodo" returns a promise. You need to wait for the promise to resolve with the query response data. The easiest way to do this is to use the async/await syntax. See below.
index.js
const data_mysql = require("./query")
runQuery()
async function runQuery() {
for (var i = 0; i < jsonObj.length; i++) {
//jsonObj[i].proceso = 0;
queryResponse = await data_mysql.vertodo(jsonObj[i])
// process queryResponse
}
}
answered Nov 22 '18 at 18:08
FakeFootballFakeFootball
11
add a comment |
Your function "vertodo" returns a promise. You need to wait for the promise to resolve with the query response data. The easiest way to do this is to use the async/await syntax. See below.
index.js
const data_mysql = require("./query")
runQuery()
async function runQuery() {
for (var i = 0; i < jsonObj.length; i++) {
//jsonObj[i].proceso = 0;
queryResponse = await data_mysql.vertodo(jsonObj[i])
// process queryResponse
}
}
answered Nov 22 '18 at 18:08
FakeFootballFakeFootball
11
Your function "vertodo" returns a promise. You need to wait for the promise to resolve with the query response data. The easiest way to do this is to use the async/await syntax. See below.
index.js
const data_mysql = require("./query")
runQuery()
async function runQuery() {
for (var i = 0; i < jsonObj.length; i++) {
//jsonObj[i].proceso = 0;
queryResponse = await data_mysql.vertodo(jsonObj[i])
// process queryResponse
}
}
answered Nov 22 '18 at 18:08
FakeFootballFakeFootball
11
answered Nov 22 '18 at 18:08
FakeFootballFakeFootball
11
answered Nov 22 '18 at 18:08
FakeFootballFakeFootball
11
answered Nov 22 '18 at 18:08
FakeFootballFakeFootball
11
11
add a comment |
add a comment |
I don't fully understand what you want to do, but I assume that you want to run those queries in loop and wait until all finish and then log results.
If those queries need to be run in sequence.
async function runQueries() {
for (const item of jsonObj) {
item.procesa = await data_mysql.vertodo(item);
}
}
runQueries.then(function () {
jsonObj.forEach(function (item) {
console.log(item.procesa);
});
});
answered Nov 22 '18 at 23:02
Jake NohJake Noh
615
add a comment |
I don't fully understand what you want to do, but I assume that you want to run those queries in loop and wait until all finish and then log results.
If those queries need to be run in sequence.
async function runQueries() {
for (const item of jsonObj) {
item.procesa = await data_mysql.vertodo(item);
}
}
runQueries.then(function () {
jsonObj.forEach(function (item) {
console.log(item.procesa);
});
});
answered Nov 22 '18 at 23:02
Jake NohJake Noh
615
add a comment |
I don't fully understand what you want to do, but I assume that you want to run those queries in loop and wait until all finish and then log results.
If those queries need to be run in sequence.
async function runQueries() {
for (const item of jsonObj) {
item.procesa = await data_mysql.vertodo(item);
}
}
runQueries.then(function () {
jsonObj.forEach(function (item) {
console.log(item.procesa);
});
});
answered Nov 22 '18 at 23:02
Jake NohJake Noh
615
I don't fully understand what you want to do, but I assume that you want to run those queries in loop and wait until all finish and then log results.
If those queries need to be run in sequence.
async function runQueries() {
for (const item of jsonObj) {
item.procesa = await data_mysql.vertodo(item);
}
}
runQueries.then(function () {
jsonObj.forEach(function (item) {
console.log(item.procesa);
});
});
answered Nov 22 '18 at 23:02
Jake NohJake Noh
615
answered Nov 22 '18 at 23:02
Jake NohJake Noh
615
answered Nov 22 '18 at 23:02
Jake NohJake Noh
615
answered Nov 22 '18 at 23:02
Jake NohJake Noh
615
615
add a comment |
add a comment |
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