使用 mongoose 和 node express 查找一个集合中的所有文档，更改数据并保存到另一个集合中

Question

我正在尝试从 1 个集合中获取 select 数据，删除我不需要的字段，并添加一些新字段，并传递在每个文档上为新字段复制的所需数据，并将它们保存到不同的集合中.

我是接近还是远离？ 任何帮助，将不胜感激。

router.post('', (req, res, next) => {
    const taskDay = req.body.taskDay;
    switch (taskDay) {
        case 'taskSunday':
            MasterTask.find({taskSunday: true}).forEach(tempData => {
                const taskSheet = new Tasksheet({
                    taskDate: req.body.taskDate,
                    taskTicket: req.body.taskTicket,
                    taskTime: tempData.taskTime,
                    taskCompany: tempData.taskCompany,
                    taskDescription: tempData.taskDescription,
                    taskCompleted: req.body.taskCompleted,
                    taskComments: req.body.taskComments,
                    taskCompletedBy: req.body.taskCompletedBy,
                    taskTimeStamp: req.body.taskTimeStamp,
                    taskKnowledgeTitle: tempData.taskKnowledgeTitle,
                    taskKnowledgeUrl: tempData.taskKnowledgeUrl,
                    taskType: tempData.taskType
                }); 
                Tasksheet.insertMany(taskSheet).then(result => {
                    console.log(result);
                    res.status(200).json({message: 'task sheet created'})
                });
            });
            break;

在 find 中提取的示例数据：

    _id: "5f0acb7209e5981d10bb765a",
    taskTime: "00:00",
    taskCompany: "CompanyName",
    taskDescription: "Some Needed Daily Task",
    taskSunday: true,
    taskMonday: true,
    taskTuesday: true,
    taskWednesday: true,
    taskThursday: true,
    taskFriday: true,
    taskSaturday: true,
    taskDisable: false,
    taskEndOfMonth: null,
    taskKnowledgeTitle: "Some Title",
    taskKnowledgeUrl: "Some URL",
    taskForSpecificDays: null,
    taskType: "Core",
    __v: 0
  }

我试图保存到新集合的数据转换：

  {
    id: "mongooseID",
    taskTicket: "",
    taskDate: null,
    taskTime: "00:00",
    taskCompany: "somename",
    taskDescription: "This is an example description of a task.  This is going to be an extra long line of text to test.",
    taskCompleted: false,
    taskComments: "",
    taskCompletedBy: "",
    taskTimeStamp: "",
    taskKnowledgeTitle: "Test",
    taskKnowledgeUrl: "http://www.google.com",
    taskType: "Core"
  }

Answer 1

尝试这个。

MasterTask.find(
{
    taskSunday: true
},
function(err, tempData) {
    let taskSheet = [];
    tempData.forEach(t => {
        taskSheet.push({
            taskDate: req.body.taskDate,
            taskTicket: req.body.taskTicket,
            taskTime: t.taskTime,
            taskCompany: t.taskCompany,
            taskDescription: t.taskDescription,
            taskCompleted: req.body.taskCompleted,
            taskComments: req.body.taskComments,
            taskCompletedBy: req.body.taskCompletedBy,
            taskTimeStamp: req.body.taskTimeStamp,
            taskKnowledgeTitle: t.taskKnowledgeTitle,
            taskKnowledgeUrl: t.taskKnowledgeUrl,
            taskType: t.taskType
        })
    })

    Tasksheet.insertMany(taskSheet, (err, t) => {
        console.log(err)
        console.log(t)
    })
})

Answer 2

好吧，你做的有点不对。 如果您想简单地重新导出您的集合值（不修改它），您最好使用诸如mongodump或mongoexport之类的实用程序（取决于场景），然后通过mongorestore或mongoimport导入它

如果你想用JS代码处理你的问题，那很好，但是上面的代码有点错误，这是为什么呢？ 因为如果你在一个集合中处理大量的文档，这段代码永远不会完成。

所以最好使用带有 cursor 的查找查询，如下所示：

async function t () {
   try {
        await collection_name.find({taskSunday: true}).lean().cursor({batchSize: 10}).eachAsync(async (document) => {
                /**
                * Your could write any logic here related with your document
                */
                const taskSheet = new Tasksheet({
                    taskDate: document.taskDate,
                    taskTicket: document.taskTicket,
                    taskTime: document.taskTime,
                    taskCompany: document.taskCompany,
                    taskDescription: document.taskDescription,
                    taskCompleted: document.taskCompleted,
                    taskComments: document.taskComments,
                    taskCompletedBy: document.taskCompletedBy,
                    taskTimeStamp: document.taskTimeStamp,
                    taskKnowledgeTitle: document.taskKnowledgeTitle,
                    taskKnowledgeUrl: document.taskKnowledgeUrl,
                    taskType: document.taskType
                }); 
                taskSheet.save()
        }, { parallel: 10})
   } catch (e) {
       console.error(e)
   }
}

此代码将保证完成，因为它从您的原始collection_name中获取每 10 个文档（它取决于{batchSize: 10}和{ parallel: 10}值）并在 10 个流中一个接一个地插入。 将允许您不加载 RAM 中的所有集合。

并且还允许您即时修改必要的数据