Why I still get the 200 response status code instead of 404?-CodePudding

I just making delete query to mysql with sequelize.

const model = require('../../../config/model/index');
const controller = {};

controller.drop =  async function(req, res) {
  try {
    await model.activitys.destroy({
      where: {
        id: req.params.id
      }
    })
    check_id = model.activitys.findAll({
      where: {
        id: req.params.id
      }
    })
    if(check_id!=null){
      res.status(200).json({
        status: "Success",
        message: "Success",
        data: {}
      })
    }else{    
      res.status(404).json({
        status: "Not Found",
        message: `Activity with ID ${id} Not Found`,
        data: {}
      })    
    }
  } catch (error) {
    res.status(404).json({
      status: "Error",
      message: error.message
    })
  }
}

module.exports = controller;

I want to delete the data on DB from id parameters, it's work for deleting the data. But when I try to delete by id that not exist in my DB, it's still get the 200 status code.

How to make that will return 404 status code if there's no data exists in DB ?

CodePudding user response：

If you want to check if a single record exists in DB then use findOne or findByPk instead of findAll. findAll always returns an array (either empty or not) and that means it's always not equal to null:

check_id = model.activitys.findOne({
      where: {
        id: req.params.id
      }
    })