猫鼬查找功能搜索不返回所需的结果

Question

>我在Mongodb中有一个用户列表，需要根据一些过滤器进行搜索，如下图所示：只有性别是强制性的，用户可能有也可能没有其他详细信息

    User.find({
     "gender": userEntry.gender,
     "dob": { $gte: convert.getDobFromAge(userEntry.ageHigherLimit), $lte: convert.getDobFromAge(userEntry.ageLowerLimit), $exist: false },
     "details.chest": { $gte: userEntry.chestLowerLimit, $lte: userEntry.chestHigherLimit, $exist: false },
     "details.waist": { $gte: userEntry.waistLowerLimit, $lte: userEntry.waistHigherLimit, $exist: false },
     "details.height": { $gte: userEntry.heightLowerLimit, $lte: userEntry.heightHigherLimit, $exist: false },
     "details.weight": { $gte: userEntry.weightLowerLimit, $lte: userEntry.weightHigherLimit, $exist: false }
   },   function (err, users) {
          return res.render('client/search.html', { users: users });
   });

上面是要搜索的猫鼬查询，userEntry 看起来像这样

 userEntry={
  "gender":2,
  "ageLowerLimit":28,"ageHigherLimit":40,
  "chestLowerLimit":"","chestHigherLimit":"",
  "heightLowerLimit":"","heightHigherLimit":"",
  "waistLowerLimit":"","waistHigherLimit":"",
  "weightLowerLimit":"","weightHigherLimit":"",
  "state":"","city":"",
  "country":"","skin_color":"",
  "profession_type":"","experience":"",
  "hair_type":""
}

我的问题是 find 函数，它应该搜索性别为"2"且年龄>=28 和年龄=<40 的所有记录（从上面的查询中给我空数组，即使一条记录满足它），给出满足上述条件的所有结果，无论其他字段是否为空或不存在。任何帮助将不胜感激。

正如正确建议的那样，将查询更改为但仍获取了 0 条记录

var query = {
  details: {}
};
if (userEntry.gender) {
  query.gender = userEntry.gender;
}
if(userEntry.ageLowerLimit && userEntry.ageHigherLimit ) {
  query.dbo = { $gte: convert.getDobFromAge(userEntry.ageHigherLimit), $lte: convert.getDobFromAge(userEntry.ageLowerLimit)};
}
console.log(query);
User.find(query, function (err, users) {
  if(!err) {
    console.log(users);
    return res.render('client/search.html', { users: users });
  }
  console.log(err);
});

}）;

尝试获取的记录之一

{ "_id" : ObjectId("59c3f47e6388613b94556b78"), "name" : "tanzeel", "email" : "[e-mail]", "password" : "$2a$10$kvachEZL0vEPPJiS7bIAMeGMXiZ.MRaZmrBECXB207jme1I4JEn6i", "created_at" : ISODate("2017-09-21T17:18:54.822Z"), "role" : 1, "following" : [ ], "dp" : "/dp/default.jpg", "gender" : 2, "__v" : 0, "dob" : ISODate("1994-11-29T00:00:00Z"), "details" : {  "height" : 160, "weight" : 65, "profession_type" : "Actor", "skin_color" : "Tan", "eye_color" : "Black", "waist" : 32, "chest" : 35 } }

Answer 1

您从MongoDB获得了正确的值，但不是预期的值，因为您的查询构建不正确。此外$exist无效运算符应该$exists并且您与某些字段（如 details.waist = ""）的空字符串进行比较userEntry.weightLowerLimit因为 是空的。但是，您应该正确构建查询以获得预期的结果。可以这样尝试...

var query = {
  details: {}
};
if (userEntry.gender) {
  query.gender = userEntry.gender;
}
if(userEntry.ageLowerLimit && userEntry.ageHigherLimit ) {
  query.dbo = { $gte: convert.getDobFromAge(userEntry.ageHigherLimit), $lte: convert.getDobFromAge(userEntry.ageLowerLimit)};
}
if(userEntry.chestLowerLimit &&  userEntry.chestHigherLimit) {
  query['details.chest'] = { $gte: userEntry.chestLowerLimit, $lte: userEntry.chestHigherLimit };
}
//... for others conditions
User.find(query, function (err, users) {
  if(!err) {
    return res.render('client/search.html', { users: users });
  }
  console.log(err);
});