我正在尝试为包含虚拟字段的REST API生成精简记录。Keystone.js/mongoose虚拟字段精简记录
如何实现虚拟域的猫鼬的官方文档:
http://mongoosejs.com/docs/guide.html
我的模型:
var keystone = require('keystone')
, Types = keystone.Field.Types
, list = new keystone.List('Vendors');
list.add({
name : {
first: {type : Types.Text}
, last: {type : Types.Text}
}
});
list.schema.virtual('name.full').get(function() {
return this.name.first + ' ' + this.name.last;
});
list.register();
现在,让我们来查询模型:
var keystone = require('keystone'),
vendors = keystone.list('Vendors');
vendors.model.find()
.exec(function(err, doc){
console.log(doc)
});
虚拟字段名.full不在这里:
[ { _id: 563acf280f2b2dfd4f59bcf3,
__v: 0,
name: { first: 'Walter', last: 'White' } }]
但是,如果我们这样做:
vendors.model.find()
.exec(function(err, doc){
console.log(doc.name.full); // "Walter White"
});
那么虚拟所示。
我想原因是,当我做一个console.log(文档)的Mongoose document.toString()方法被调用,它不包括虚拟默认情况下。很公平。这是可以理解的。
要包含虚函数在任何的转换方法,你必须去:
doc.toString({virtuals: true})
doc.toObject({virtuals: true})
doc.toJSON({virtuals: true})
然而,这包括键我不希望我的REST API泵出我的用户:
{ _id: 563acf280f2b2dfd4f59bcf3,
__v: 0,
name: { first: 'Walter', last: 'White', full: 'Walter White' },
_: { name: { last: [Object], first: [Object] } },
list:
List {
options:
{ schema: [Object],
noedit: false,
nocreate: false,
nodelete: false,
autocreate: false,
sortable: false,
hidden: false,
track: false,
inherits: false,
searchFields: '__name__',
defaultSort: '__default__',
defaultColumns: '__name__',
label: 'Vendors' },
key: 'Vendors',
path: 'vendors',
schema:
Schema {
paths: [Object],
subpaths: {},
virtuals: [Object],
nested: [Object],
inherits: {},
callQueue: [],
_indexes: [],
methods: [Object],
statics: {},
tree: [Object],
_requiredpaths: [],
discriminatorMapping: undefined,
_indexedpaths: undefined,
options: [Object] },
schemaFields: [ [Object] ],
uiElements: [ [Object], [Object] ],
underscoreMethods: { name: [Object] },
fields: { 'name.first': [Object], 'name.last': [Object] },
fieldTypes: { text: true },
relationships: {},
mappings:
{ name: null,
createdBy: null,
createdOn: null,
modifiedBy: null,
modifiedOn: null },
model:
{ [Function: model]
base: [Object],
modelName: 'Vendors',
model: [Function: model],
db: [Object],
discriminators: undefined,
schema: [Object],
options: undefined,
collection: [Object] } },
id: '563acf280f2b2dfd4f59bcf3' }
我可以随时当然只是删除不需要的钥匙,但是这似乎并不完全正确:
vendors.model.findOne()
.exec(function(err, doc){
var c = doc.toObject({virtuals: true});
delete c.list;
delete c._;
console.log(c)
});
这产生了我需要的东西:
{ _id: 563acf280f2b2dfd4f59bcf3,
__v: 0,
name: { first: 'Walter', last: 'White', full: 'Walter White' },
id: '563acf280f2b2dfd4f59bcf3' }
是否没有更好的方式获得精益记录?
我认为你是对的,而我不想要的字段是属于Keystone js的内部字段。如果我运行本地Mongo DB或Mongoose,则不会在结果集中包含.list和._字段。 – ChrisRich