问题描述
我要解决的是:使用建议的方法(mapReduce)使用$ in保留ID数组的顺序:MongoDB 的 $in 子句是否保证顺序
What I'm trying to solve is: preserving the order of my array of Ids with $in using this suggested method (mapReduce): Does MongoDB's $in clause guarantee order
我已经完成作业,发现将它们转换为字符串是理想的:比较猫鼬_id和字符串.
I've done my homework, and saw it's ideal to convert them to strings: Comparing mongoose _id and strings.
代码:
var dataIds = [ '57a1152a4d124a4d1ad12d80',
'57a115304d124a4d1ad12d81',
'5795316dabfaa62383341a79',
'5795315aabfaa62383341a76',
'57a114d64d124a4d1ad12d7f',
'57953165abfaa62383341a78' ];
CollectionSchema.statics.all = function() {
var obj = {};
//adds dataIds to obj.scope as inputs , to be accessed in obj.map
obj.scope = {'inputs': dataIds};
obj.map = function() {
//used toString method as suggested in other SO answer, but still get -1 for Id.
var order = inputs.indexOf(this._id.toString());
emit(order, {
doc : this
});
};
obj.reduce = function() {};
obj.out = {inline: 1};
obj.query = {"_id": {"$in": dataIds } };
obj.finalize = function(key, value) {
return value;
};
return Product
.mapReduce(obj)
.then(function(products){
console.log('map products : ', products)
})
};
这就是我一直在找回console.log的产品承诺:
This is what I keep getting back for my console.log in the products promise :
[{ _id: -1, value: null } ]
哪个,使我相信它不能将其中的ObjectId与dataIds的索引进行匹配.但是,如果我只在.find()中使用$ in子句,则将返回正确的乘积,但是顺序不正确.
Which, leads me to believe it's not able to match the ObjectId from this, with an index of dataIds. However, if I just use the $in clause within a .find(), the correct products are returned -- but, in the incorrect order.
由此产生意外行为.
obj.map = function() {
for(var i = 0; i < inputs.length; i++){
if(inputs[i] == this._id.toString()){
}
emit(inputs[i], this);
}
};
发射:
[ { _id: '5795315aabfaa62383341a76', value: null },
{ _id: '57953165abfaa62383341a78', value: null },
{ _id: '5795316dabfaa62383341a79', value: null },
{ _id: '57a114d64d124a4d1ad12d7f', value: null },
{ _id: '57a1152a4d124a4d1ad12d80', value: null },
{ _id: '57a115304d124a4d1ad12d81', value: null } ]
obj.map = function() {
for(var i = 0; i < inputs.length; i++){
if(inputs[i] == this._id.toString()){
var order = i;
}
emit(this._id.toString(), this);
}
};
发射:
[ { _id: 'ObjectId("5795315aabfaa62383341a76")', value: null },
{ _id: 'ObjectId("57953165abfaa62383341a78")', value: null },
{ _id: 'ObjectId("5795316dabfaa62383341a79")', value: null },
{ _id: 'ObjectId("57a114d64d124a4d1ad12d7f")', value: null },
{ _id: 'ObjectId("57a1152a4d124a4d1ad12d80")', value: null },
{ _id: 'ObjectId("57a115304d124a4d1ad12d81")', value: null } ]
现在,如何摆脱ObjectId()包装器?最好是比str.slice()更干净的方法,它会起作用-但是,我觉得必须有一种更"mongo"/更安全的方法来转换ID.
Now, how do I get rid of the ObjectId() wrapper? Preferably, something more clean than str.slice(), which would work -- However, I feel there must be a more "mongo" / safer way of converting the Id.
我签出了文档,但只提到了toString()方法,该方法在地图中似乎无法正常工作: https://docs.mongodb.com/manual/reference/method/ObjectId.toString/
I checked out the docs, but it only mentions the toString() method, which does not seem to be working correctly within map: https://docs.mongodb.com/manual/reference/method/ObjectId.toString/
推荐答案
将其弄清楚了:
obj.map = function() {
for(var i = 0; i < inputs.length; i++){
if(this._id.equals(inputs[i])) {
var order = i;
}
}
emit(order, {doc: this});
};
这篇关于如何在不包含&#39; ObjectId()'的情况下转换mongo ObjectId .toString包装器-只是价值?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!