How to acces data from a database and return it as an array?










0















I'm new to Javascript and I have to solve a task for my Homework.
I have to get specific data (album titles and the album IDs) from a Database in one class, return this Data to another class and format it with template literals there.



getAlbums: function (db) 
sql = `SELECT DISTINCT Title, AlbumId FROM albums`;
let title = , id = ;
db.all(sql, , (err, rows) =>
if (err) throw err;
rows.forEach((row)=>
title.push(row.Title);
id.push(row.AlbumId);
);
);
return [title, id];



this is the function where I get the data from the database. When I put a console.log(title) inside the db.all function, the titles array is filled correctly. But it does not return this array to the other class, I get an error:




TypeError: Cannot read property 'forEach' of undefined.




If I put a console.log(title) outside the db.all function, it returns an empty array and I don't get an error. The output is just undefined.



The method call in my other class:



showAlbum: function (db) 
let albumID = model.getAlbums(db);
//let album = "";
//let albId = "";

albumID.forEach((element) =>
//album = element.title;
//albId = element.id;
console.log(`Albumtitle: $element.title AlbumID: $element.idn`);
);



How can I transfer the arrays correctly?










share|improve this question




























    0















    I'm new to Javascript and I have to solve a task for my Homework.
    I have to get specific data (album titles and the album IDs) from a Database in one class, return this Data to another class and format it with template literals there.



    getAlbums: function (db) 
    sql = `SELECT DISTINCT Title, AlbumId FROM albums`;
    let title = , id = ;
    db.all(sql, , (err, rows) =>
    if (err) throw err;
    rows.forEach((row)=>
    title.push(row.Title);
    id.push(row.AlbumId);
    );
    );
    return [title, id];



    this is the function where I get the data from the database. When I put a console.log(title) inside the db.all function, the titles array is filled correctly. But it does not return this array to the other class, I get an error:




    TypeError: Cannot read property 'forEach' of undefined.




    If I put a console.log(title) outside the db.all function, it returns an empty array and I don't get an error. The output is just undefined.



    The method call in my other class:



    showAlbum: function (db) 
    let albumID = model.getAlbums(db);
    //let album = "";
    //let albId = "";

    albumID.forEach((element) =>
    //album = element.title;
    //albId = element.id;
    console.log(`Albumtitle: $element.title AlbumID: $element.idn`);
    );



    How can I transfer the arrays correctly?










    share|improve this question


























      0












      0








      0








      I'm new to Javascript and I have to solve a task for my Homework.
      I have to get specific data (album titles and the album IDs) from a Database in one class, return this Data to another class and format it with template literals there.



      getAlbums: function (db) 
      sql = `SELECT DISTINCT Title, AlbumId FROM albums`;
      let title = , id = ;
      db.all(sql, , (err, rows) =>
      if (err) throw err;
      rows.forEach((row)=>
      title.push(row.Title);
      id.push(row.AlbumId);
      );
      );
      return [title, id];



      this is the function where I get the data from the database. When I put a console.log(title) inside the db.all function, the titles array is filled correctly. But it does not return this array to the other class, I get an error:




      TypeError: Cannot read property 'forEach' of undefined.




      If I put a console.log(title) outside the db.all function, it returns an empty array and I don't get an error. The output is just undefined.



      The method call in my other class:



      showAlbum: function (db) 
      let albumID = model.getAlbums(db);
      //let album = "";
      //let albId = "";

      albumID.forEach((element) =>
      //album = element.title;
      //albId = element.id;
      console.log(`Albumtitle: $element.title AlbumID: $element.idn`);
      );



      How can I transfer the arrays correctly?










      share|improve this question
















      I'm new to Javascript and I have to solve a task for my Homework.
      I have to get specific data (album titles and the album IDs) from a Database in one class, return this Data to another class and format it with template literals there.



      getAlbums: function (db) 
      sql = `SELECT DISTINCT Title, AlbumId FROM albums`;
      let title = , id = ;
      db.all(sql, , (err, rows) =>
      if (err) throw err;
      rows.forEach((row)=>
      title.push(row.Title);
      id.push(row.AlbumId);
      );
      );
      return [title, id];



      this is the function where I get the data from the database. When I put a console.log(title) inside the db.all function, the titles array is filled correctly. But it does not return this array to the other class, I get an error:




      TypeError: Cannot read property 'forEach' of undefined.




      If I put a console.log(title) outside the db.all function, it returns an empty array and I don't get an error. The output is just undefined.



      The method call in my other class:



      showAlbum: function (db) 
      let albumID = model.getAlbums(db);
      //let album = "";
      //let albId = "";

      albumID.forEach((element) =>
      //album = element.title;
      //albId = element.id;
      console.log(`Albumtitle: $element.title AlbumID: $element.idn`);
      );



      How can I transfer the arrays correctly?







      javascript arrays foreach sqlite3






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 11 '18 at 14:31







      anja-96

















      asked Nov 11 '18 at 14:24









      anja-96anja-96

      13




      13






















          1 Answer
          1






          active

          oldest

          votes


















          0














          Try changing your JS code to the following.



          getAlbums: function (data) 
          sql = `SELECT DISTINCT Title, AlbumId FROM albums`;
          let title = , id = ;
          data.data.all(sql, , (err, rows) =>
          if (err) throw err;
          rows.forEach((row)=>
          title.push(row.Title);
          id.push(row.AlbumId);
          );
          );
          return [title, id];



          then



          showAlbum: function (data) 
          let albumID = model.getAlbums(data.data);
          //let album = "";
          //let albId = "";

          albumID.forEach((element) =>
          //album = element.title;
          //albId = element.id;
          console.log(`Albumtitle: $element.title AlbumID: $element.idn`);
          );






          share|improve this answer

























          • unfortunately, I now get the following Error: TypeError: Cannot read property 'all' of undefined

            – anja-96
            Nov 11 '18 at 14:56












          • Change your db.all to data.db.all and check if it's working?

            – Yaseen Hussain
            Nov 11 '18 at 14:58












          • Wherever you are using your db argument, try changing it to data.db

            – Yaseen Hussain
            Nov 11 '18 at 15:00











          • Still the same, but now it is "TypeError: Cannot read property 'db' of undefined"

            – anja-96
            Nov 11 '18 at 15:07











          • Please check the edited comment with update. Lemme know if it works.

            – Yaseen Hussain
            Nov 11 '18 at 15:13










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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0














          Try changing your JS code to the following.



          getAlbums: function (data) 
          sql = `SELECT DISTINCT Title, AlbumId FROM albums`;
          let title = , id = ;
          data.data.all(sql, , (err, rows) =>
          if (err) throw err;
          rows.forEach((row)=>
          title.push(row.Title);
          id.push(row.AlbumId);
          );
          );
          return [title, id];



          then



          showAlbum: function (data) 
          let albumID = model.getAlbums(data.data);
          //let album = "";
          //let albId = "";

          albumID.forEach((element) =>
          //album = element.title;
          //albId = element.id;
          console.log(`Albumtitle: $element.title AlbumID: $element.idn`);
          );






          share|improve this answer

























          • unfortunately, I now get the following Error: TypeError: Cannot read property 'all' of undefined

            – anja-96
            Nov 11 '18 at 14:56












          • Change your db.all to data.db.all and check if it's working?

            – Yaseen Hussain
            Nov 11 '18 at 14:58












          • Wherever you are using your db argument, try changing it to data.db

            – Yaseen Hussain
            Nov 11 '18 at 15:00











          • Still the same, but now it is "TypeError: Cannot read property 'db' of undefined"

            – anja-96
            Nov 11 '18 at 15:07











          • Please check the edited comment with update. Lemme know if it works.

            – Yaseen Hussain
            Nov 11 '18 at 15:13















          0














          Try changing your JS code to the following.



          getAlbums: function (data) 
          sql = `SELECT DISTINCT Title, AlbumId FROM albums`;
          let title = , id = ;
          data.data.all(sql, , (err, rows) =>
          if (err) throw err;
          rows.forEach((row)=>
          title.push(row.Title);
          id.push(row.AlbumId);
          );
          );
          return [title, id];



          then



          showAlbum: function (data) 
          let albumID = model.getAlbums(data.data);
          //let album = "";
          //let albId = "";

          albumID.forEach((element) =>
          //album = element.title;
          //albId = element.id;
          console.log(`Albumtitle: $element.title AlbumID: $element.idn`);
          );






          share|improve this answer

























          • unfortunately, I now get the following Error: TypeError: Cannot read property 'all' of undefined

            – anja-96
            Nov 11 '18 at 14:56












          • Change your db.all to data.db.all and check if it's working?

            – Yaseen Hussain
            Nov 11 '18 at 14:58












          • Wherever you are using your db argument, try changing it to data.db

            – Yaseen Hussain
            Nov 11 '18 at 15:00











          • Still the same, but now it is "TypeError: Cannot read property 'db' of undefined"

            – anja-96
            Nov 11 '18 at 15:07











          • Please check the edited comment with update. Lemme know if it works.

            – Yaseen Hussain
            Nov 11 '18 at 15:13













          0












          0








          0







          Try changing your JS code to the following.



          getAlbums: function (data) 
          sql = `SELECT DISTINCT Title, AlbumId FROM albums`;
          let title = , id = ;
          data.data.all(sql, , (err, rows) =>
          if (err) throw err;
          rows.forEach((row)=>
          title.push(row.Title);
          id.push(row.AlbumId);
          );
          );
          return [title, id];



          then



          showAlbum: function (data) 
          let albumID = model.getAlbums(data.data);
          //let album = "";
          //let albId = "";

          albumID.forEach((element) =>
          //album = element.title;
          //albId = element.id;
          console.log(`Albumtitle: $element.title AlbumID: $element.idn`);
          );






          share|improve this answer















          Try changing your JS code to the following.



          getAlbums: function (data) 
          sql = `SELECT DISTINCT Title, AlbumId FROM albums`;
          let title = , id = ;
          data.data.all(sql, , (err, rows) =>
          if (err) throw err;
          rows.forEach((row)=>
          title.push(row.Title);
          id.push(row.AlbumId);
          );
          );
          return [title, id];



          then



          showAlbum: function (data) 
          let albumID = model.getAlbums(data.data);
          //let album = "";
          //let albId = "";

          albumID.forEach((element) =>
          //album = element.title;
          //albId = element.id;
          console.log(`Albumtitle: $element.title AlbumID: $element.idn`);
          );







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 11 '18 at 15:13

























          answered Nov 11 '18 at 14:50









          Yaseen HussainYaseen Hussain

          357




          357












          • unfortunately, I now get the following Error: TypeError: Cannot read property 'all' of undefined

            – anja-96
            Nov 11 '18 at 14:56












          • Change your db.all to data.db.all and check if it's working?

            – Yaseen Hussain
            Nov 11 '18 at 14:58












          • Wherever you are using your db argument, try changing it to data.db

            – Yaseen Hussain
            Nov 11 '18 at 15:00











          • Still the same, but now it is "TypeError: Cannot read property 'db' of undefined"

            – anja-96
            Nov 11 '18 at 15:07











          • Please check the edited comment with update. Lemme know if it works.

            – Yaseen Hussain
            Nov 11 '18 at 15:13

















          • unfortunately, I now get the following Error: TypeError: Cannot read property 'all' of undefined

            – anja-96
            Nov 11 '18 at 14:56












          • Change your db.all to data.db.all and check if it's working?

            – Yaseen Hussain
            Nov 11 '18 at 14:58












          • Wherever you are using your db argument, try changing it to data.db

            – Yaseen Hussain
            Nov 11 '18 at 15:00











          • Still the same, but now it is "TypeError: Cannot read property 'db' of undefined"

            – anja-96
            Nov 11 '18 at 15:07











          • Please check the edited comment with update. Lemme know if it works.

            – Yaseen Hussain
            Nov 11 '18 at 15:13
















          unfortunately, I now get the following Error: TypeError: Cannot read property 'all' of undefined

          – anja-96
          Nov 11 '18 at 14:56






          unfortunately, I now get the following Error: TypeError: Cannot read property 'all' of undefined

          – anja-96
          Nov 11 '18 at 14:56














          Change your db.all to data.db.all and check if it's working?

          – Yaseen Hussain
          Nov 11 '18 at 14:58






          Change your db.all to data.db.all and check if it's working?

          – Yaseen Hussain
          Nov 11 '18 at 14:58














          Wherever you are using your db argument, try changing it to data.db

          – Yaseen Hussain
          Nov 11 '18 at 15:00





          Wherever you are using your db argument, try changing it to data.db

          – Yaseen Hussain
          Nov 11 '18 at 15:00













          Still the same, but now it is "TypeError: Cannot read property 'db' of undefined"

          – anja-96
          Nov 11 '18 at 15:07





          Still the same, but now it is "TypeError: Cannot read property 'db' of undefined"

          – anja-96
          Nov 11 '18 at 15:07













          Please check the edited comment with update. Lemme know if it works.

          – Yaseen Hussain
          Nov 11 '18 at 15:13





          Please check the edited comment with update. Lemme know if it works.

          – Yaseen Hussain
          Nov 11 '18 at 15:13

















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