Question

我必须使用Ruby（不是在Rails上，在Notepad ++上）在我的Mongo数据库中插入文档，许多文档都有重复，但有一些修改。

我想编写一个使用json文件的脚本，读取它，通过检查每个文档是否在数据库中没有任何重复，将其导入MongoDB中，如果有重复，我想组合它，如果它包含任何其他信息：

如：

文件1

{ "Name" : "Lila",
  "Files":
  [
         { "Name": "File3", "Date" : "26-03-2018"}
  ]
}

文件2

 { "Name" : "Lila",
   "Files":
      [
             { "Name": "File1", "Date" : "05-11-2017"},
             { "Name": "File2", "Date" : "26-03-2018"},
             { "Name": "File3", "Date" : "26-03-2018"}
      ]
  }

将它们组合起来：

public void CheckingFiveLetter()
{

    for (int i = 0; i < convertStoredLetters.Length; i++) 
             //convertStoredLetters is consonants and vowels which i get at 
    runtime
    {
        for (int j = 0; j < convertStoredLetters.Length; j++) 
        {
            for (int k = 0; k < convertStoredLetters.Length; k++) 
            {
                for (int l = 0; l < convertStoredLetters.Length; l++)
                {
                    for (int m = 0; m < convertStoredLetters.Length; m++) 
                    {

                        if (i != j && i != k && i != l && i != m  && j != i && j != k && j != l && j != m  && k != i && k != j && k != l && k != m  && l != i && l != j && l != k && l != m  && m != i && m != j && m != k && m != l) 
                        {
                            letterword5.Add (convertStoredLetters.Substring (i, 1) + convertStoredLetters.Substring (j, 1) + convertStoredLetters.Substring (k, 1) + convertStoredLetters.Substring (l, 1) + convertStoredLetters.Substring (m, 1));
                            letterword5.Add (convertStoredLetters.Substring (j, 1) + convertStoredLetters.Substring (k, 1) + convertStoredLetters.Substring (l, 1) + convertStoredLetters.Substring (m, 1) + convertStoredLetters.Substring (i, 1));
                            letterword5.Add (convertStoredLetters.Substring (k, 1) + convertStoredLetters.Substring (l, 1) + convertStoredLetters.Substring (m, 1) + convertStoredLetters.Substring (i, 1) + convertStoredLetters.Substring (j, 1));
                            letterword5.Add (convertStoredLetters.Substring (l, 1) + convertStoredLetters.Substring (m, 1) + convertStoredLetters.Substring (i, 1) + convertStoredLetters.Substring (j, 1) + convertStoredLetters.Substring (k, 1));
                            letterword5.Add (convertStoredLetters.Substring (m, 1) + convertStoredLetters.Substring (i, 1) + convertStoredLetters.Substring (j, 1) + convertStoredLetters.Substring (k, 1) + convertStoredLetters.Substring (l, 1));


                            letterword5.Add (convertStoredLetters.Substring (i, 1) + convertStoredLetters.Substring (m, 1) + convertStoredLetters.Substring (l, 1) + convertStoredLetters.Substring (k, 1) + convertStoredLetters.Substring (j, 1));
                            letterword5.Add (convertStoredLetters.Substring (j, 1) + convertStoredLetters.Substring (i, 1) + convertStoredLetters.Substring (m, 1) + convertStoredLetters.Substring (l, 1) + convertStoredLetters.Substring (k, 1));
                            letterword5.Add (convertStoredLetters.Substring (k, 1) + convertStoredLetters.Substring (j, 1) + convertStoredLetters.Substring (i, 1) + convertStoredLetters.Substring (m, 1) + convertStoredLetters.Substring (l, 1));
                            letterword5.Add (convertStoredLetters.Substring (l, 1) + convertStoredLetters.Substring (k, 1) + convertStoredLetters.Substring (j, 1) + convertStoredLetters.Substring (i, 1) + convertStoredLetters.Substring (m, 1));
                            letterword5.Add (convertStoredLetters.Substring (m, 1) + convertStoredLetters.Substring (l, 1) + convertStoredLetters.Substring (k, 1) + convertStoredLetters.Substring (j, 1) + convertStoredLetters.Substring (i, 1));



                        }   
                    }
                }
            }
        }
    }

    //checking if the combination is present in the word dictionary or not 
    if present add count
    for (int m = 0; m < letterword5.Count; m++)
    {
        bool Checkfi = wordRan.CheckWord (letterword5 [m].ToLower ());
        if (Checkfi == true) 
        {
            fiveLetterWord.Add (letterword5 [m].ToLower ());

        }

    }
    fi = fiveLetterWord.Distinct ().ToList ();

    wordRan.storeFiveWordCount = fi.Count;
    five.text = "Five Possible words : " + fi.Count.ToString ();
    letterword5.Clear ();
    fiveLetterWord.Clear ();

}

我发现在mongo shell中有可能归功于聚合 - 积累$ mergeObjects，但在Ruby中它似乎不存在。

Answer 1

您也可以使用ruby中的所有运算符。您需要首先获取基础集合对象。

require 'mongo' 

db = Mongo::Connection.new.db("mydb")
coll = db.collection('posts')

coll.aggregate([ 
  {"$project" => {"last_name" => 1, "first_name" => 1 }}, 
  {"$match" => {"last_name" => "Jones"}}
])

这是一个示例管道。您可以将mongo shell上为您工作的相同聚合管道提供给aggregate。

有关更多信息，请参阅MongoDB Ruby驱动程序文档： http://www.rubydoc.info/gems/mongo/1.8.2/Mongo%2FCollection%3Aaggregate

是否可以在MongoDB中将两个Json与Ruby结合使用？

1 个答案: