MongoDB 查询 - 如果值出现在数组 x 次

Question

{
  "_id": ObjectId("53ab1d2c256072374a5cc63f"),
  "title": "10% Off",
  "endDate": "2015-05-08",
  "limit" : "limited",
  "redemptions": [
    "1f7f5f96be3a",
    "kf40vksk03ps"
  ]
}
{
  "_id": ObjectId("53ab1d2c25607sfdgs74a5cc63f"),
  "title": "20% Off",
  "endDate": "2015-06-07",
  "limit" : "unlimited",
  "redemptions": [
    "1f7f5f96be3a",
    "1f7f5f96be3a",
    "kf40vksk03ps"
  ]
}

故事：一个人可以兑换2次优惠券。 2次后，请勿退货。 如何检查一个值出现的次数少于 2 次？？

希望它像这样简单：

{ 'redemptions' : { $exists: true }, $where : 'this.redemptions.$.1f7f5f96be3a.length < 2' }

如何计算特定值在数组中出现的次数并进行比较？

编辑 所以要增加一些乐趣。我更新了我的模式，所以我需要把它变成一个条件。 if limit = 'unlimited' { return record } if limit = 'limited' { 仅当数组少于 2 个值时才返回 = '1f7f5f96be3a'

Answer 1

你可以用聚合框架做你想做的事：

db.collection.aggregate([
    /* find only documents that have redemption key */
    { $match : { redemptions : { $exists: true }}},
    /* unwind the redemptions array so we can filter documents by coupons */
    { $unwind : "$redemptions" },
    /* filter only coupons you're looking for */
    { $match : { redemptions : "1f7f5f96be3a"}},
    /* group the documents back so we can calculate the count */
    { $group : { _id : "$_id",
                 title : { $first : "$title" },
                 endDate : { $first : "$endDate" },
                 count :  {$sum : 1 }}},
    /* finally, count the documents that have less than 2 coupons */
    { $match : { count : { $lt :2 }}}
]);

编辑：

您只需要更改 $group 和 last $match 阶段：

{ $group : { _id : "$_id",
            title : { $first : "$title" },
            endDate : { $first : "$endDate" },
            limit : { $first : "$limit" },
            count :  {$sum : 1 }}},
/* finally, count the documents that have less than 2 coupons 
   or have limit "unlimited */
{ $match : { $or : [{ count : { $lt :2 }}, {limit : "unlimited"}]}}