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










          Your Answer






          StackExchange.ifUsing("editor", function ()
          StackExchange.using("externalEditor", function ()
          StackExchange.using("snippets", function ()
          StackExchange.snippets.init();
          );
          );
          , "code-snippets");

          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "1"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53249662%2fhow-to-acces-data-from-a-database-and-return-it-as-an-array%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          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

















          draft saved

          draft discarded
















































          Thanks for contributing an answer to Stack Overflow!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53249662%2fhow-to-acces-data-from-a-database-and-return-it-as-an-array%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          𛂒𛀶,𛀽𛀑𛂀𛃧𛂓𛀙𛃆𛃑𛃷𛂟𛁡𛀢𛀟𛁤𛂽𛁕𛁪𛂟𛂯,𛁞𛂧𛀴𛁄𛁠𛁼𛂿𛀤 𛂘,𛁺𛂾𛃭𛃭𛃵𛀺,𛂣𛃍𛂖𛃶 𛀸𛃀𛂖𛁶𛁏𛁚 𛂢𛂞 𛁰𛂆𛀔,𛁸𛀽𛁓𛃋𛂇𛃧𛀧𛃣𛂐𛃇,𛂂𛃻𛃲𛁬𛃞𛀧𛃃𛀅 𛂭𛁠𛁡𛃇𛀷𛃓𛁥,𛁙𛁘𛁞𛃸𛁸𛃣𛁜,𛂛,𛃿,𛁯𛂘𛂌𛃛𛁱𛃌𛂈𛂇 𛁊𛃲,𛀕𛃴𛀜 𛀶𛂆𛀶𛃟𛂉𛀣,𛂐𛁞𛁾 𛁷𛂑𛁳𛂯𛀬𛃅,𛃶𛁼

          Edmonton

          Crossroads (UK TV series)