node.js 仅从 Sequelize ORM 获取数据值
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Get only dataValues from Sequelize ORM
提问by Gijo Varghese
I'm using the sequelize ORM to fetch data from a PSQL DB. However, when I retrieve something, a whole bunch of data is given. The only data I want is inside 'dataValues'. Of course, I can use object.dataValues. But, is there any other good solutions?
我正在使用 sequelize ORM 从 PSQL DB 获取数据。但是，当我检索某些东西时，会给出一大堆数据。我想要的唯一数据是在“dataValues”里面。当然，我可以使用 object.dataValues。但是，还有其他好的解决方案吗？
I'm using Sequelize 4.10
我正在使用 Sequelize 4.10
采纳答案by Gijo Varghese
The problem occurs only when I log it using:
仅当我使用以下方法记录时才会出现问题：
console.log(Model.findAll());
console.log(Model.findAll());
If I save it to a variable, I can directly access objects inside without using "dataValues"
如果我把它保存到一个变量中，我可以直接访问里面的对象，而不用使用“ dataValues”
回答by Shivam
Yes you can
是的你可以
Model.findAll({
raw: true,
//Other parameters
});
would return just the data and not the model instance
将只返回数据而不是模型实例
回答by C Deuter
Sequelize wraps all it's return values in a virtual object that contains meta data. If you have an object and you just want the undecorated data values, you can unwrap them like so:
Sequelize 将所有返回值包装在包含元数据的虚拟对象中。如果你有一个对象并且你只想要未修饰的数据值，你可以像这样解开它们：
Model.findById(1).then(data => {
console.log(data.get({ plain: true }));
});
Additionally if you just want to print out the object you can use the .toJSONmethod.
此外，如果您只想打印出对象，您可以使用该.toJSON方法。
Model.findById(1).then(data => {
console.log(data.toJSON());
});
回答by Masoud Tavakkoli
Finally I found answer after searching a lot. you should do something like this
经过大量搜索，我终于找到了答案。你应该做这样的事情
const users = await db.users.findAll({})
.map(el => el.get({ plain: true })) // add this line to code
source: github issue
来源：github问题
回答by defraggled
To clarify Masoud Tavakkoli's answer (which is not immediately clear on that github answer): using element.get({ plain: true })returns an array of objectswith each attribute key:value pairs.
为了澄清 Masoud Tavakkoli 的答案（在那个 github 答案上不是很清楚）： usingelement.get({ plain: true })返回一个具有每个属性键：值对的对象数组。
If you just want an array of one specific attribute's values (eg user ids) instead of objects, you can use something like this:
如果您只想要一个包含特定属性值（例如用户 ID）而不是对象的数组，您可以使用以下内容：
const users = await User.findAll({
attributes: ["id"],
where: {} // Your filters here
}).map(u => u.get("id")) // [1,2,3]
Nika Kasradze's answer actually achieves the same outcome as a middle-step; using the JSON stringifier generates the same array output. It's possible this is faster than mapping, but I'm not sure.
Nika Kasradze 的回答实际上达到了与中间步骤相同的结果；使用 JSON 字符串生成器生成相同的数组输出。这可能比映射更快，但我不确定。
const users = await User.findAll({
attributes: ["id"],
where: {} // Your filters here
})
const userIds = JSON.stringify(users)) // [1,2,3]
回答by Nika Kasradze
For nested entities this stupid workaround works for me:
对于嵌套实体，这个愚蠢的解决方法对我有用：
let row = await Model.findOne({
include: [ /*your includes here*/ ]
});
row = JSON.parse( JSON.stringify(row, null, 4) );
Somehow JSON.stringify(row, null, 4)removes all the extra stuff and pretends the dataValuesproperty of the object is the object itself. Then the JSON.parse(...)puts the object back together.
以某种方式JSON.stringify(row, null, 4)删除所有额外的东西并假装dataValues对象的属性是对象本身。然后JSON.parse(...)将对象放回原处。
EDIT:
编辑：
so apparently I'm new to typescript so I didn't need this at all. the console.log(row)printing huge object got me confused. I miss good ol' c# times :'(
所以显然我是打字稿的新手，所以我根本不需要这个。在console.log(row)打印巨物让我感到困惑。我想念好 ol' c# 时代 :'(
