Javascript 来自 Sequelize 的 findAll() 没有得到
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findAll() from Sequelize doesn't get
提问by German Gonzalez
I'm using Sequelize with MySQL.
我在 MySQL 中使用 Sequelize。
When I run this code:
当我运行此代码时：
usuarioService.getAll = function () {
Usuario.findAll().then(function (users) {
//return users;
console.dir(users);
});
}
Instead of get the user, I get:
我没有得到用户，而是得到：
http://i.stack.imgur.com/uLhmN.png
http://i.stack.imgur.com/uLhmN.png
Help me, please! I'm going crazy!
请帮帮我！我要疯了！
Thanks
谢谢
回答by hargasinski
Sequelize is returning an array of instanceobjects in users. An instanceobject has a number of convenience methods attached to it that allow you to act on it.
Sequelize 正在返回instance用户中的对象数组。一个instance对象附有许多方便的方法，允许您对其进行操作。
If you want to get just the data with your fields as keys, use get({plain: true}). For example, for the first object in the array users[0].get({plain: true}). If you want to keep using the instances, you can just use get with the name of your field. For example, users[0].get('nombre').
如果您只想获取以字段为键的数据，请使用get({plain: true}). 例如，对于数组中的第一个对象users[0].get({plain: true})。如果您想继续使用这些实例，您可以使用带有您的字段名称的 get。例如，users[0].get('nombre')。
You should be also able to access the properties directly on the object, even if they're not being logged, such as users[0].nombre.
您还应该能够直接访问对象上的属性，即使它们没有被记录，例如users[0].nombre.
Edit
编辑
This is not related to the original question, but your comment on another answer. Make sure you are doing things asynchronously. The code should be:
这与原始问题无关，而是您对另一个答案的评论。确保您正在异步执行操作。代码应该是：
usuarioService.getAll = function (cb) {
Usuario.findAll().then(function (users) {
return cb(null, users);
}).catch(function(err) {
return cb(err);
});
}
Then when calling this method you would do something like:
然后在调用此方法时，您将执行以下操作：
router.get('your_path', function(req, res, next) {
serv.getAll(function(err, users) {
if (err) {
// your err handling code
}
// users is now a valid js array
// could send it in res.json(users)
});
});
or
或者
Since Sequelize uses promises, doing this using promises would be the best way.
由于 Sequelize 使用 Promise，因此使用 Promise 将是最好的方法。
usuarioService.getAll = function () {
return Usuario.findAll({ raw: true });
}
Then when calling this method you would do something like:
然后在调用此方法时，您将执行以下操作：
router.get('your_path', function(req, res, next) {
serv.getAll().then(function(users) {
res.render('usuarios/index',{
users: users
})
}).catch(function(err) {
// your error handling code here
});
});
回答by LT-
You are returning a user.
您正在返回用户。
The first bit you see is the SQL query that Sequelize is executing for you.
您首先看到的是 Sequelize 为您执行的 SQL 查询。
The bit that says
说的那一点
dataValues:
{ usuario_id: 1,
...
}
is your user. findAll()should give you an array with all of your users.
是你的用户。findAll()应该给你一个包含所有用户的数组。
If you just want the dataValues returned you can just pass in raw: true.
如果您只想返回 dataValues，则只需传入raw: true.
usuarioService.getAll = function () {
Usuario.findAll({ raw: true }).then(function (users) {
//return users;
console.dir(users);
});
}
