如何检查给定的输入值是否可用？在对象中

Question

这是我从console.log(objects) 得到的信息：

{
    "_id" : "5bb20d7556db6915846da55f",
    "gallary" : {
        "profilepic" : [
            "1",
            "2"
        ]

    }
},
{
    "_id" : "5bb20d7556db6915846da55f",
    "gallary" : {
        "profilepic" : [
            "3",
            "4"
        ]

    }
}

我有一个输入数组，例如：let uniqueIDs = ["0","1","10"]。我必须检查gallary.profilepic 给定的输入值是否可用？假设不可用意味着我必须推送一个数组并返回结果。

我的代码：

let uniqueIDs = ["0","1","10"]
db.Gallary.find()
.forEach(function(objects){
    console.log(objects);
    // Here your JS code has to come
})

预期输出：

["0", "10"]

Answer 1

我将首先获取数据库中与您要输入的数据匹配的所有值（否则它将加载您的所有集合条目），然后查看丢失的字符串。如有必要，最后推送一个新条目。

const UPIDs = ['0','1','10'];

const entries = await db.Groups.find({
  'members.regularStudent': {
     $in: UPIDs,
  },
});

// Get the strings that are not in the DB
const missing = UPIDs.filter(x => !entries.some(y => y.members.regularStudent.includes(x)));

// Push the missing str in DB
if (missing.length) {
  const obj = new model({
     members: {
        regularStudent: missing,
     },  
  });

  // ...
}

---

---

const UPIDs = ['0', '1', '10'];

// Thx to @Andy
const entries = [{"_id":"5bb20d7556db6915846da55f","members":{"regularStudent":['1', '2']}}];

/*
const entries = await db.Groups.find({
  'members.regularStudent': {
     $in: UPIDs,
  },
});
*/

// Get the strings that are not in the DB
const missing = UPIDs.filter(x => !entries.some(y => y.members.regularStudent.includes(x)));

// Push the missing str in DB
if (missing.length) {
  console.log('Add new data for values', missing);
}

Answer 2

过滤您的 UPIDs 数组，检查其中哪些项目未包含在您的 members 数据的 regularStudent 属性中：

let UPIDs = ["0","1","10"];
const data = [{
    "_id" : "5bb20d7556db6915846da55f",
    "members" : {
        "regularStudent" : [
            "1",
            "2"
        ]

    }
},
{
    "_id" : "5bb20d7556db6915846da55f",
    "members" : {
        "regularStudent" : [
            "3",
            "4"
        ]

    }
}];

const regularStudents = data.map(d => d.members.regularStudent).flat();
console.log(UPIDs.filter(a => !regularStudents.includes(a)));

Answer 3

您可以将所有 ID 集合在一起（我在示例中使用了 reduce），然后 filter 将 UPID 中未出现在该数组中的所有元素收集起来。

const data = [{"_id":"5bb20d7556db6915846da55f","members":{"regularStudent":["1","2"]}},{"_id":"5bb20d7556db6915846da55f","members":{"regularStudent":["3","4"]}}];
const UPIDs = ["0","1","10"];

// Concatenate the ids in the data together
const ids = data.reduce((acc, c) => acc.concat(c.members.regularStudent), []);

// `filter` out the elements in UPIDs that aren't in the ids array
const out = UPIDs.filter(el => !ids.includes(el));

console.log(out);

更新说明：

这是你的代码和我的代码：

let UPIDs = ['0', '1', '10'];

db.Groups.find().forEach(function(objects){
  const ids = objects.reduce((acc, c) => acc.concat(c.members.regularStudent), []);
  const out = UPIDs.filter(el => !ids.includes(el));  
  console.log(out);
});