最近完成了一个nodejs+mongoose的项目，碰到了mongodb的级联查询操作。情形是实现一个排行榜，查看某个公司（organization）下属客户中发表有效文ruan章wen最多的前十人。
Account表：公司的信息单独存在一个account表里。
var AccountSchema = new Schema({ loginname: {type: String}, password: {type: String}, /** * 联系方式 */ //账户公司名 comName: {type: String}, //地址 address: {type: String}, //公司介绍 intro: {type: String} }); mongoose.model('Account', AccountSchema);
Cusomer表：公司的客户群。
var CustomerSchema = new Schema({ /** * 基本信息 */ //密码 password: {type: String}, //归属于哪个Account belongToAccount: {type: ObjectId, ref: 'Account'}, //手机号，登录用 mobile: {type: String}, //真实姓名 realname: {type: String} }); CustomerSchema.index({belongToAccount: 1, mobile: 1}, {unique: true}); mongoose.model('Customer', CustomerSchema);
article表
var articleSchema= new Schema({ belongToAccount: {type: ObjectId, ref: 'Account'}, title: {type: String}, text: {type: String}, createTime: {type: Date, default: Date.now}, author: {type: ObjectId, ref: 'Customer'}, //0,待确认,1 有效 ,-1 无效 status: {type: Number, default: 0} }); articleSchema.index({belongToAccount: 1, createTime:-1,author: 1}, {unique: false}); mongoose.model('article', articleSchema);
这里要做的就是，由accountId→aggregate整理软文并排序→级联author找到作者的姓名及其他信息。
代码如下：
exports.getRankList = function (accountid, callback) { AticleModel.aggregate( {$match: {belongToAccount: mongoose.Types.ObjectId(accountid), status: 1}}, {$group: {_id: {customerId: "$author"}, number: {$sum: 1}}}, {$sort: {number: -1}}).limit(10).exec(function (err, aggregateResult) { if(err){ callback(err); return; } var ep = new EventProxy(); ep.after('got_customer', aggregateResult.length, function (customerList) { callback(null, customerList); }); aggregateResult.forEach(function (item) { Customer.findOne({_id: item._id.customerId}, ep.done(function (customer) { item.customerName = customer.realname; item.customerMobile=cusomer.mobile; // do someting ep.emit('got_customer', item); })); }) }); };
返回的结果格式（这里仅有两条记录，实际为前十）：
[ { _id: { customerId: 559a5b6f51a446602032fs21 }, number: 5, customerName: 'test2', mobile:22 } ， { _id: { customerId: 559a5b6f51a446602041ee6f }, number: 1, customerName: 'test1', mobile: 11 } ]