如何筛选不在request与friendList数组中的用户并解决重复问题

我正在开发一个简单API，需要筛选出不在request和friendList对象数组中的用户。用map方法实现时，出现了对象被重复推入数组的问题。

相关数据 #

API 返回数据 #

[{"_id":"6304e42231ef2e7a4dec924d","email":"devrajstha@gmail.com","name":"Deepa Shrestha","phone":9819339011,"profile":"https://eduzim.co.zw/news/wp-content/uploads/2022/09/6486/pictures-former-rhythm-city-actress-pearl-petronella-tshuma-set-to-star-in-good-men-meet-the-whole-cast.jpg","dob":"2001-03-26T08:00:00.000Z","join_at":"2022-08-23T14:27:43.249Z","password":"Devraj 123@","__v":0,"friendList":[],"request":[{"_id":"63cd4c1a675ce5566d3ee6df","sender":"dstha221@gmail.com","receiver":"devrajstha@gmail.com","sendAt":"2023-01-22T14:45:20.147Z","status":"NA","__v":0}]},...]

现有JS代码 #

router.get("/people_y_m_n", (req, res) => { 
  let users = []; 
  let new_array=[]; 
  const myEmail = 'dstha22123@gmail.com'; 
  userModel.aggregate([
    { $lookup: { 
        from: "friend_lists", 
        localField: "email", 
        foreignField: "friend", 
        as: "friendList", 
      } 
    },
    { $lookup: { 
        from: "frieend_requests", 
        localField: "email", 
        foreignField: "receiver", 
        as: "request", 
      } 
    }
  ]).exec((err, docs) => { 
    console.log(docs.length); 
    docs.map((doc) => { 
      console.log(myEmail==doc.request.sender); 
      doc.request.map((val)=>{ 
        if(doc.email!=val.sender||doc.email!=val.receiver && myEmail!=val.receiver || myEmail!=val.sender ){ 
          new_array.push({ email:doc.email,name:doc.name,profile:doc.profile,_id:doc._id }) 
        } 
      }); 
      doc.friendList.map((val)=>{ 
        if(doc.email!=val.friend||doc.email!=val.me && myEmail!=val.friend || myEmail!=val.me ){ 
          new_array.push({ email:doc.email,name:doc.name,profile:doc.profile,_id:doc._id }) 
        } 
      }); 
      new_array.map((val)=>{ 
        if(val.email!=doc.email){ 
          users.push({ email:doc.email,name:doc.name,profile:doc.profile,_id:doc._id }) 
        } 
      }); 
    }); 
    res.send(users); 
  }); 
});

当前API输出（含重复） #

[{"email":"devrajstha88@gmail.com","name":"Devraj Shrestha","profile":"https://i.etsystatic.com/5713076/r/il/db3028/660348889/il_1588xN.660348889_byu0.jpg","_id":"6304e43b31ef2e7a4dec9251"},...]

new_array输出 #

[{"email":"devrajstha@gmail.com","name":"Deepa Shrestha","profile":"https://eduzim.co.zw/news/wp-content/uploads/2022/09/6486/pictures-former-rhythm-city-actress-pearl-petronella-tshuma-set-to-star-in-good-men-meet-the-whole-cast.jpg","_id":"6304e42231ef2e7a4dec924d"},...]

问题分析与修复 #

核心问题 #

1. 逻辑判断错误：if条件的运算符优先级和逻辑关系混乱，导致不符合条件的用户被多次推入数组。
2. 滥用map方法：map用于数组元素转换，不应用来做遍历判断，应该用forEach或直接条件判断。
3. 重复添加逻辑混乱：遍历request和friendList时，每满足一次条件就推一次用户；后续对new_array的遍历逻辑完全错误，进一步加剧重复。

修复后的代码 #

router.get("/people_y_m_n", (req, res) => { 
  const myEmail = 'dstha22123@gmail.com'; 
  userModel.aggregate([
    { $lookup: { 
        from: "friend_lists", 
        localField: "email", 
        foreignField: "friend", 
        as: "friendList", 
      } 
    },
    { $lookup: { 
        from: "frieend_requests", 
        localField: "email", 
        foreignField: "receiver", 
        as: "request", 
      } 
    }
  ]).exec((err, docs) => { 
    if (err) {
      return res.status(500).send(err);
    }
    // 过滤出符合条件的用户：非好友、无相关好友请求、排除自己
    const filteredUsers = docs.filter(doc => {
      if (doc.email === myEmail) return false;
      
      // 检查是否为好友
      const isFriend = doc.friendList.some(val => 
        (val.me === myEmail && val.friend === doc.email) || 
        (val.me === doc.email && val.friend === myEmail)
      );
      if (isFriend) return false;
      
      // 检查是否存在好友请求
      const hasRequest = doc.request.some(val => 
        (val.sender === myEmail && val.receiver === doc.email) || 
        (val.sender === doc.email && val.receiver === myEmail)
      );
      if (hasRequest) return false;
      
      return true;
    }).map(doc => ({
      email: doc.email,
      name: doc.name,
      profile: doc.profile,
      _id: doc._id
    }));
    
    res.send(filteredUsers); 
  }); 
});

修复要点 #

1. 用filter做筛选：直接对docs数组过滤，只保留符合条件的用户，从根源避免重复。
2. 明确逻辑判断：
   • 先排除当前用户自己
   • 用some方法检查是否存在好友关系，只要有一条匹配就排除
   • 用some方法检查是否存在好友请求，只要有一条匹配就排除
3. 统一转换结构：过滤后用map一次性转换为需要的输出字段，避免多次操作。
4. 添加错误处理：捕获数据库查询错误，返回500状态码友好处理。

内容的提问来源于stack exchange，提问作者Debaraj Stha