CODEWITHSUNDEEP
node.jssqlitesequelize.js
Vegas Rico
Vegas Rico

Reputation: 73

Sequelize: findAll is not a function

I just learning node.js by book.

I follow to instructions in the book and got an error. I spended couple of hours to find a problem, but without success. Hope for your help and understanding.

exports.connectDB = function(){
var SQNote;
var sequlz;

if(SQNote) return SQNote.sync();

return new Promise((resolve, reject) => {
    fs.readFile(process.env.SEQUELIZE_CONNECT, 'utf-8', (err, data) => {
        if(err){
            reject(err);
        }
        else{
            resolve(data);
        }
    });
})
.then(yamltext => {
    return jsyaml.safeLoad(yamltext, 'utf-8');
})
.then(params => {
    sequlz = new Sequelize(params.dbname, 
        params.username, 
        params.password, 
        params.params);

    SQNote = sequlz.define('Note', {
                            notekey: {
                                    type: Sequelize.STRING,
                                    primaryKey: true,
                                    unique: true
                                },
                            title: Sequelize.STRING,
                            body: Sequelize.STRING
                        });
    return SQNote.sync;
});};

Error occurs here:

exports.keylist = function(){
return exports.connectDB()
.then(SQNote => {
    return SQNote.findAll({ attributes: [ 'notekey' ] })
    .then(notes => {
        return notes.map(note => notekey);
    });
});};

Error: error in browser picture

SQNote.findAll is not a function

TypeError: SQNote.findAll is not a function at exports.connectDB.then.SQNote (/home/vegas/NodeJS_course/notes/models/notes-sequelize.js:119:23) at anonymous

Upvotes: 0

Views: 1024

Answers (1)

Vegas Rico
Vegas Rico

Reputation: 73

After spend another few hours I found my mistake. Should be:

return SQNote.sync();

Instead:

return SQNote.sync;

Upvotes: 1

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