problems with the values of the promise Promise { value }












0















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










share|improve this question



























    0















    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










    share|improve this question

























      0












      0








      0








      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










      share|improve this question














      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






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      share|improve this question










      asked Nov 22 '18 at 17:21







      user10597291































          2 Answers
          2






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          oldest

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          0














          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

          }
          }





          share|improve this answer































            0














            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);
            });
            });





            share|improve this answer























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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              0














              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

              }
              }





              share|improve this answer




























                0














                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

                }
                }





                share|improve this answer


























                  0












                  0








                  0







                  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

                  }
                  }





                  share|improve this answer













                  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

                  }
                  }






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 22 '18 at 18:08









                  FakeFootballFakeFootball

                  11




                  11

























                      0














                      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);
                      });
                      });





                      share|improve this answer




























                        0














                        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);
                        });
                        });





                        share|improve this answer


























                          0












                          0








                          0







                          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);
                          });
                          });





                          share|improve this answer













                          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);
                          });
                          });






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Nov 22 '18 at 23:02









                          Jake NohJake Noh

                          615




                          615






























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