sequelize gets all the data of a query without executing my condition 'where'

Question

im trying to get some values from the database mysql with this sequelize code:

subSector .sync() .then(() => subSector.findAll({ Where: { Enterprise_Evaluation_ID: 2 } })) .then(result => { console.log(result) }) .catch(error => { console.log(error) }) }

the problem is that sequelize get all values of the table and not only the data of the condition where.

any idea?

we recently change the project of server in AWS (change of elasticbeanstalk to other instance eb)

The syntax looks ok and I don't think it is caused due to server change. Try other where conditions for testing. — Ankur Jyoti Phukan
– Ankur Jyoti Phukan, Commented Sep 27, 2018 at 3:54

Vivek Doshi · Accepted Answer · 2018-09-27 04:27:53Z

Only issue I can find is , Where should be lower case like where

subSector.findAll({ Where: { Enterprise_Evaluation_ID: 2 } })

Change it to this :

subSector.findAll({ where: { // <--------- HERE Enterprise_Evaluation_ID: 2 // <--------- Also make sure the field name is proper } })

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