試圖從兩個列表

Question

下面創建列表是我的課：

public class Test 
    { 
     public int TestId { get; set; } 
     public List<VariantsRank> VariantsRanks { get; set; } 
    } 

    public class VariantsRank 
    { 
     public string Name { get; set; } 
     public int Rank { get; set; } 
    } 

以下2種方法：

testList的

var testList = CreateDataFromList();//return List<Test> 
Var testListfromDb = GetDatafromDB();//return List<Test> 

記錄：

testListfromDb的

[0] : 100 
     List of VariantRanks 

[1]: 101 
     List of VariantRanks 

記錄：

[0] : 100 
     List of VariantRanks // Order of variants will be different here and that is important to me 

[1]: 101 
    List of VariantRanks // Order of variants will be different here and that is important to me 

下面是我的變量，我想最後的輸出：

var final = new List<Test>(); 

方案：

）

testList = 100 and 101 
testListfromDb = 101 

Expected Output for 1st scenario : final = 100 and 101(2 records and this 101 record will be from testListfromDb) 

）

testList = 100 and 101 
    testListfromDb = 100 

    Expected Output for 2nd scenario : final = 100 and 101(2 records and this 100 record will be from testListfromDb) 

）

testList = 100 and 101 
    testListfromDb = 100 and 101 

    Expected Output for 3rd scenario : final = 100 and 101 (2 records and both will be from testListfromDb) 

我已經做了3次的情況，但我沒有得到如何解決第一和第二方案。

代碼：

var final = new List<Test>(); 
var testList = CreateDataFromList(list); //return List<Test> 
Var testListfromDb = GetDatafromDB();//return List<Test> 
    if (testListfromDb == null) 
      final = testList; 
    else 
     { 
      if (testList.Count == testListfromDb.Count) // 3rd scenario 
       final = testListfromDb; 

     } 

更新：

public void Method() 
     { 
      var testList = new List<Test>(); 
      testList.Add(new Test { TestId = 100 }); 
      testList.Add(new Test { TestId = 101 }); 
      var testListfromDb = new List<Test>(); 
      var final = new List<Test>(); 

      //scenario 1 
      testListfromDb.Add(new Test { TestId = 101 }); 
      //expected output in final variable 
      final[0] = testList[0]; 
      final[1] = testListfromDb[0]; 

      //scenario 2 
      testListfromDb.Add(new Test { TestId = 100 }); 
      //expected output in final variable 
      final[0] = testListfromDb[0]; 
      final[1] = testList[1]; 

      //scenario 3 
      testListfromDb.Add(new Test { TestId = 100 }); 
      testListfromDb.Add(new Test { TestId = 101 }); 
      //expected output in final variable 
      final[0] = testListfromDb[0]; 
      final[1] = testListfromDb[1]; 
     } 

Answer 1

我認爲解決您的問題：

public List<Test> Process(List<Test> testList, List<Test> testListfromDb) 
{ 
    return 
     testListfromDb 
      .Concat(testList) 
      .GroupBy(x => x.TestId) 
      .SelectMany(x => x.Take(1)) 
      .ToList(); 
} 

如果更改Test這個定義：

public class Test 
{ 
    public int TestId { get; set; } 
    public string Source { get; set; } 
} 

那麼這個代碼可以測試的結果：

var scenario1 = Process(new List<Test>() 
{ 
    new Test { TestId = 100, Source = "Test" }, 
    new Test { TestId = 101, Source = "Test" } 
}, new List<Test>() 
{ 
    new Test { TestId = 101, Source = "DB" } 
}); 

var scenario2 = Process(new List<Test>() 
{ 
    new Test { TestId = 100, Source = "Test" }, 
    new Test { TestId = 101, Source = "Test" } 
}, new List<Test>() 
{ 
    new Test { TestId = 100, Source = "DB" } 
}); 

var scenario3 = Process(new List<Test>() 
{ 
    new Test { TestId = 100, Source = "Test" }, 
    new Test { TestId = 101, Source = "Test" } 
}, new List<Test>() 
{ 
    new Test { TestId = 100, Source = "DB" }, 
    new Test { TestId = 101, Source = "DB" } 
});