展平mongoDB架构

Question

我有一个现有的深层嵌套的mongoDB架构,我必须弄平,因为我有一个复杂的查询,无法用当前结构有效地进行.这是架构的MWE:

db.test.insert({
    "_id" : ObjectId("58e574a768afb6085ec3a388"),
    "details" : [
            {
                "_id" : ObjectId("58e55f0f68afb6085ec3a2cc"),
                "a" : [
                    {
                        "unit" : "08",
                        "size" : "5",
                        "pos" : "Far",
                        "_id" : ObjectId("58e55f0f68afb6085ec3a2d0")
                    }
                ],
                "b" : [
                    {
                        "unit" : "08",
                        "size" : "5",
                        "pos" : "Far",
                        "_id" : ObjectId("58e55f0f68afb6085ec3a2cd")
                    }
                ],
                "c" : [
                    {
                        "unit" : "08",
                        "size" : "3",
                        "pos" : "Far",
                        "_id" : ObjectId("58e55f0f68afb6085ec3a2ce")
                    }
                ],
                "d" : [
                    {
                        "unit" : "08",
                        "size" : "5",
                        "pos" : "Far",
                        "_id" : ObjectId("58e55f0f68afb6085ec3a2cf")
                    }
                ]
            }
        ]
    })

我想弄平图式.期望的结果是:

"_id" : ObjectId("58e574a768afb6085ec3a388"),
"tests" : [
        {
            "_id" : ObjectId("58e542fb68afb6085ec3a1d2"),
            "aUnit" : "08",
            "aSize" : "5",
            "aPos" : "Far",
            "bPos" : "Far",
            "bSize" : "5",
            "bUnit" : "08",
            "cPos" : "Far",
            "cSize" : "3",
            "cUnit" : "08",
            "dPos" : "Far",
            "dSize" : "5",
            "dUnit" : "08"
                }
            ]

我愿意一次做一个入口类型,我认为我有一个方法可以这样做,但它不起作用.这是我尝试过的:

db.test.find({"tests.$.details.a.unit":{$exists:true}}).forEach(function(doc) {      
    doc.tests = {aUnit:tests.details.a.unit};
    delete tests.details.a.unit;
    db.test.save(doc);
    });

但是,这没有任何改变.如何改进我的查询以展平我的架构？

编辑:我意识到MWE与我打算用它的那个相比有一个小错误.我关闭了每个条目.例如,"a" : [{ ... }],被错误地写为{"a" : [{ ... }]},.但是,它现在已更新.

Answer 1

新的回应

打印数据

db.test.find().forEach(doc => {
  doc.details = doc.details.map( detail => {
    Object.keys(detail).filter( k => k !== "_id" ).forEach( k => {
      detail[k].forEach( item => {
        Object.keys(item).filter(i => i !== "_id" ).forEach( inner => {
          detail[k + inner.charAt(0).toUpperCase() + inner.substr(1)]
            = item[inner];
        })
      });
      delete detail[k];
    });
    return detail;
  });
  printjson(doc);
});

更新数据

db.test.find().forEach(doc => {
  doc.details = doc.details.map( detail => {
    Object.keys(detail).filter( k => k !== "_id" ).forEach( k => {
      detail[k].forEach( item => {
        Object.keys(item).filter(i => i !== "_id" ).forEach( inner => {
          detail[k + inner.charAt(0).toUpperCase() + inner.substr(1)]
            = item[inner];
        })
      });
      delete detail[k];
    });
    return detail;
  });

  ops = [
    ...ops,
    { "updateOne": {
      "filter": { "_id": doc._id },
      "update": { "$set": { "doc.details": doc.details } }
    }}
  ];

  if ( ops.length >= 500 ) {
    db.test.bulkWrite(ops);
    ops = [];
  }
});

if ( ops.length > 0 ) {
  db.test.bulkWrite(ops);
  ops = [];
}

输出表格

{
    "_id" : ObjectId("58e574a768afb6085ec3a388"),
    "details" : [
        {
          "_id" : ObjectId("58e55f0f68afb6085ec3a2cc"),
          "aUnit" : "08",
          "aSize" : "5",
          "aPos" : "Far",
          "bUnit" : "08",
          "bSize" : "5",
          "bPos" : "Far",
          "cUnit" : "08",
          "cSize" : "3",
          "cPos" : "Far",
          "dUnit" : "08",
          "dSize" : "5",
          "dPos" : "Far"
        }
    ]
}

原始数据

{
    "_id" : ObjectId("58e574a768afb6085ec3a388"),
    "tests" : [
      {
        "_id" : ObjectId("58e542fb68afb6085ec3a1d2"),
        "details" : [
          {
            "a" : [
              {
                "unit" : "08",
                "size" : "5",
                "pos" : "Far",
                "_id" : ObjectId("58e542fb68afb6085ec3a1d6")
              }
            ]
          },
          {
            "b" : [
              {
                "pos" : "Drive Side Far",
                "size" : "5",
                "unit" : "08",
                "_id" : ObjectId("58e542fb68afb6085ec3a1d3")
              }
            ]
          },
          {
            "c" : [
              {
                "pos" : "Far",
                "size" : "3",
                "unit" : "08",
                "_id" : ObjectId("58e542fb68afb6085ec3a1d4")
              }
            ]
          },
          {
            "d" : [
              {
                "pos" : "Far",
                "size" : "5",
                "unit" : "08",
                "_id" : ObjectId("58e542fb68afb6085ec3a1d5")
              }
            ]
          }
        ]
      }
    ]
}

原始答案

如果您正在尝试"更新"您的数据,那么它比您尝试的更多.你有几个数组,你需要实际"遍历"数组元素,而不是试图直接访问它们.

这里只是一个"打印""扁平化"数据的示例:

db.test.find().forEach(doc => {
  doc.tests = doc.tests.map( test => {
    test.details.forEach( detail => {
      Object.keys(detail).forEach( key => {
        detail[key].forEach( item => {
          Object.keys(item).forEach( inner => {
            if ( inner !== '_id' ) {
              test[key + inner.charAt(0).toUpperCase() + inner.substr(1)]
                = item[inner];
            }
          });
        });
      });
    });
    delete test.details;
    return test;
  });
  printjson(doc);
})

我相信它给你正在寻找的结构:

{
    "_id" : ObjectId("58e574a768afb6085ec3a388"),
    "tests" : [
        {
            "_id" : ObjectId("58e542fb68afb6085ec3a1d2"),
            "aUnit" : "08",
            "aSize" : "5",
            "aPos" : "Far",
            "bPos" : "Drive Side Far",
            "bSize" : "5",
            "bUnit" : "08",
            "cPos" : "Far",
            "cSize" : "3",
            "cUnit" : "08",
            "dPos" : "Far",
            "dSize" : "5",
            "dUnit" : "08"
        }
    ]

}

现在我没有考虑到你的"details"阵列中任何可能"a"出现多次等密钥的文件的可能性.所以我只是考虑到那里只有一个文件里面有一个"a"或一个"b"等等,并且在将新密钥添加到"details"文档的顶层时总是分配与该密钥匹配的最后找到的值.

如果您的实际情况各不相同,那么您需要修改其中的各种.forEach()循环以使用"index"作为参数并将该索引值包含在键名称的一部分中.即:

"a0Unit": "08",
"a0Size": "05",
"a1Unit": "09",
"a1Size": "06"

但这是一个必要的细节,必要时必须解决,因为这与数据在问题中的呈现方式不同.

但是,如果这非常适合您要更新的内容,那么只需运行具有.bulkWrite()定期执行语句的循环:

let ops = [];

db.test.find().forEach(doc => {
  doc.tests = doc.tests.map( test => {
    test.details.forEach( detail => {
      Object.keys(detail).forEach( key => {
        detail[key].forEach( item => {
          Object.keys(item).forEach( inner => {
            if ( inner !== '_id' ) {
              test[key + inner.charAt(0).toUpperCase() + inner.substr(1)]
                = item[inner];
            }
          });
        });
      });
    });
    delete test.details;
    return test;
  });

  ops = [
    ...ops,
    { "updateOne": {
      "filter": { "_id": doc._id },
      "update": { "$set": { "tests": doc.tests } }
    }}
  ];

  if ( ops.length >= 500 ) {
    db.test.bulkWrite(ops);
    ops = [];
  }
});

if ( ops.length > 0 ) {
  db.test.bulkWrite(ops);
  ops = [];
}

它也出现_id在您使用mongoose的每个数组成员文档中的字段中.所以无论你做什么,都不要尝试使用mongoose本身运行代码.这是您的数据的"一次性"批量更新,应该直接从shell运行.当然,您需要修改架构以适应新结构.

但这就是为什么你应该printjson()先用方法在shell中运行你的数据.