Add "where" query in Sequalize based on flag - angularjs

Question

I want to add "where" query in Sequalize based on flag . Let me know how to get this.
For example if the flag is true then I want to add this "where" class otherwise I don't want to add the "where" clause.
models.Users.findAll({
where: {
email: req.query.email
}
}).then(function (list) {
res.send(list);
}).catch(function (error) {
res.status(500).json(error);
});
models.Users.findAll({
}).then(function (list) {
res.send(list);
}).catch(function (error) {
res.status(500).json(error);
});

Answer 1

You can create JSON first which you are providing in where clause and then pass it in where clause.. something like
var jsonObj = null ;
if (req.query.email){
jsonObj = {"email": req.query.email};
} else {
jsonObj = {"email": {$ne:null}}; //otherwise it will return all rows, if email is not null in your database
}
models.Users.findAll({
where: jsonObj
}).then(function (list) {
res.send(list);
}).catch(function (error) {
res.status(500).json(error);
});
Hope this will help

Answer 2

You can start with an empty query object, and add a where property to it when appropriate:
let query = {};
if (someCondition) {
query.where = { email: req.query.email };
}
models.Users.findAll(query).then(...);

Answer 3

You can use simple one-line if statement inside the findAll method call. Let's assume that variable is your boolean flag
models.Users.findAll(variable ? { where: { email: req.query.email } } : {}).then(list => {
res.send(list);
}).catch(error => {
res.status(500).json(error);
});
Or you can assign the options object to a variable and use the same if statement, but outside the findAll call
let options = variable ? { where: { email: req.query.email } } : {};
models.Users.findAll(options).then(list => {
// ...
});