Javascript mongodb-如何在此查询中隐藏\u id？

如何从这个查询中隐藏_id，我使用express node.js

我有这个查询，我制作了一个API，但我想隐藏这个id

以下是查询：

router.get("/", (req, res) => {
  VerbsDE.find({}, { _id: 0 })
    .limit(1)
    .then(verbs => {
      res.send(verbs);
    });
});

////////////////////////////////////////////////// 这是收藏：

[
      {
        Indicative: {
          Present: [
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            }
          ],
          Perfect: [
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            }
          ],
          Past: [
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            }
          ],
          Pluperfect: [
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            }
          ],
          Future_I: [
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            }
          ],
          Future_II: [
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            }
          ]
        },
        Imperative: {
          Worte: [
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            },
            {
              _id: "5bb9009249efde355376ad29",
              pron: "xxx",
              verb: "xxx xxx xxx"
            }
          ]
        },
        _id: "5bb9009249efde355376ad23",
        verbName: "abbilden",
        __v: 0
      }
    ];

---

我试图隐藏_id，但每次我犯错误时，我都想获取数据，但不想使用id。

有两种主要方法可以做到这一点：

直接从mongo使用排除

对结果使用

.map（）

函数

Mongo排除 与您所做的类似，但您需要正确声明变量，如果集合是动态的（例如每个文档中的“预设”、“过去”等更改），这可能会带来麻烦

您需要以嵌套属性的方式使用“字段”选项，只需更改以下内容：

router.get("/", (req, res) => {
  VerbsDE.find({}, { _id: 0 })
    .limit(1)
    .then(verbs => {
      res.send(verbs);
    });
});

为此：

router.get("/", (req, res) => {
  VerbsDE.find({}, { _id: 0, __v: 0, 'Indicative.Present._id': 0 })
    .limit(1)
    .then(verbs => {
      res.send(verbs);
    });
});

但是，由于文档的当前、过去等分配，这可能需要大量重复。现在让我们试试：

在响应之前使用Map

现在你有：

router.get("/", (req, res) => {
      VerbsDE.find({}, { _id: 0, __v: 0 })
        .limit(1)
        .then(verbs => {
          // We'll use map before sending the response
          res.send(verbs);
        });
    });

因此，映射函数如下所示：

function cleanVerbs(verbs) {
    return verbs.map(doc => {
        // For each doc, make a newDoc
        const newDoc = {};
        for (const mood in doc) {
            // mood will be 'Imperative' 'Indicative', etc.
            if (!newDoc[mood]) {
                // If out newDoc object does not have 'Imperative', etc. property, assign it as object.
                newDoc[mood] = {};
            }
            if (mood === 'verbName') {
                // You have verbName as root property, treat it differently
                newDoc[mood] = doc[mood];
                break; // Avoid further execution on this cycle
            }
            for (const time in doc[mood]) {
                console.log('MOOD & TIME: ', [mood, time]);
                const entries = doc[mood][time];
                const newTimeEntries = entries.map(e => {
                    delete e._id;
                    return e;
                });
                // This will set the newTimeEntries for this Mood's time.
                newDoc[mood][time] = newTimeEntries;
            }
        }
        return newDoc;
    });
}

然后，新的

get

方法如下：

router.get("/", (req, res) => {
      VerbsDE.find({}, { _id: 0, __v: 0 })
        .limit(1)
        .then(verbs => {
          // We'll use map before sending the response
          res.send(cleanVerbs(verbs));
          // Try res.json(cleanVerbs(verbs)) instead :)
        });
    });

编写此路由时，请记住声明

cleanVerbs

函数并将其置于作用域（可在同一文件上访问）中

注意： 我强烈建议将Mongo集合模式更改为：

您必须：

{
    _id: "5bb9009249efde355376ad23",
    verbName: "abbilden",
    grammar: [
      Indicative: {
            Present: [...],
            Past: [...],
      },...
    ],
    __v: 0
}

---

将情绪保持在每个集合的数组中将简化迭代，例如不在

.map（…）

函数上使用

if（情绪==='verbName'）{…}

测试有两种主要方法：

直接从mongo使用排除

对结果使用

.map（）

函数

Mongo排除 与您所做的类似，但您需要正确声明变量，如果集合是动态的（例如每个文档中的“预设”、“过去”等更改），这可能会带来麻烦

您需要以嵌套属性的方式使用“字段”选项，只需更改以下内容：

router.get("/", (req, res) => {
  VerbsDE.find({}, { _id: 0 })
    .limit(1)
    .then(verbs => {
      res.send(verbs);
    });
});

为此：

router.get("/", (req, res) => {
  VerbsDE.find({}, { _id: 0, __v: 0, 'Indicative.Present._id': 0 })
    .limit(1)
    .then(verbs => {
      res.send(verbs);
    });
});

但是，由于文档的当前、过去等分配，这可能需要大量重复。现在让我们试试：

在响应之前使用Map

现在你有：

router.get("/", (req, res) => {
      VerbsDE.find({}, { _id: 0, __v: 0 })
        .limit(1)
        .then(verbs => {
          // We'll use map before sending the response
          res.send(verbs);
        });
    });

因此，映射函数如下所示：

function cleanVerbs(verbs) {
    return verbs.map(doc => {
        // For each doc, make a newDoc
        const newDoc = {};
        for (const mood in doc) {
            // mood will be 'Imperative' 'Indicative', etc.
            if (!newDoc[mood]) {
                // If out newDoc object does not have 'Imperative', etc. property, assign it as object.
                newDoc[mood] = {};
            }
            if (mood === 'verbName') {
                // You have verbName as root property, treat it differently
                newDoc[mood] = doc[mood];
                break; // Avoid further execution on this cycle
            }
            for (const time in doc[mood]) {
                console.log('MOOD & TIME: ', [mood, time]);
                const entries = doc[mood][time];
                const newTimeEntries = entries.map(e => {
                    delete e._id;
                    return e;
                });
                // This will set the newTimeEntries for this Mood's time.
                newDoc[mood][time] = newTimeEntries;
            }
        }
        return newDoc;
    });
}

然后，新的

get

方法如下：

router.get("/", (req, res) => {
      VerbsDE.find({}, { _id: 0, __v: 0 })
        .limit(1)
        .then(verbs => {
          // We'll use map before sending the response
          res.send(cleanVerbs(verbs));
          // Try res.json(cleanVerbs(verbs)) instead :)
        });
    });

编写此路由时，请记住声明

cleanVerbs

函数并将其置于作用域（可在同一文件上访问）中

注意： 我强烈建议将Mongo集合模式更改为：

您必须：

{
    _id: "5bb9009249efde355376ad23",
    verbName: "abbilden",
    grammar: [
      Indicative: {
            Present: [...],
            Past: [...],
      },...
    ],
    __v: 0
}

---

将情绪保留在每个集合的数组中将简化迭代，例如不使用

if（情绪==='verbName'）{…}

测试

.map（…）

函数

您可以尝试使用投影（）：

例如：

router.get("/", (req, res) => {
  VerbsDE.find({}, {projection: { _id: 0 }})
    .limit(1)
    .then(verbs => {
      res.send(verbs);
    });
});

---

您可以尝试使用投影（）：

例如：

router.get("/", (req, res) => {
  VerbsDE.find({}, {projection: { _id: 0 }})
    .limit(1)
    .then(verbs => {
      res.send(verbs);
    });
});

查看此帮助：查看此帮助：