獲取完整的對象圖形字符串C＃

Question

我通過各種藏品循環，如果特定錯誤條件滿足，那麼我需要完整的對象圖，即這指數有問題。

示例代碼：

foreach (var sale in allSales) { 
    foreach (var discount in sale.orders.detail.discounts) { 
     if (errorConditionMet) { 
      // print full object graph. For example, perhaps it's the second Sale (index one), but first discount object (index zero): 
      // We have "discount" object, but want to print: 
      // allSales[1].sale.orders.detail.discounts[0] 
     } 

這可能只是保持計數器（而且可能是更好的性能）：

string.Format("allSales[{0}].sale.orders.detail.discounts[{1}]", saleCount, discountCount); 
    // prints: allSales[1].sale.orders.detail.discounts[0] 

但我不知道這是否可能用C＃反射？我需要在這多個類，所以這將是偉大的一個對象傳遞給方法，並返回對象圖，完全動態：

var resultOne = GetViaReflection(discount); 
// returns: allSales[1].sale.orders.detail.discounts[0] 
var resultTwo = GetViaReflection(currentAnimal); 
// returns: animals[3].animal.types[2] 

Answer 1

使用普通的循環？ 我不知道類型的allSales和sale.orders.detail.discounts但我認爲它是安全的假設他們至少IEnumerable<T>

而且List<T>會得到我們一些更多的功能從IEnumerable<T>

//If I wrap the IEnumerable in a list I get access to an indexer and count 
var allSalesList = new List<T>(allSales); 
for (int i = 0; i < allSalesList.Count; i++) { 
    var sale = allSales[i]; 

    //If I wrap the IEnumerable in a list I get access to an indexer and count 
    var discounts = new List<T>(sale.orders.detail.discounts); 


    for (int j = 0; i < discounts.Count; j++) { 
     var discount = discounts[j]; 
     if (errorConditionMet) { 
      // print full object graph. For example, perhaps it's the second Sale (index one), but first discount object (index zero): 
      // We have "discount" object, but want to print: 
      // allSales[1].sale.orders.detail.discounts[0] 


      //Your formatting here. I'm using C# 6 string interpolation, but its basically a string format. 
      Console.WriteLine($"allSales[{i}].sale.orders.detail.discounts[{j}]") 
     } 
    } 
} 

Answer 2

一想到這一點，但我認爲這不可能像你想的那樣。

我有一個SO question similar to this而回是在尋找方法名。這裏的保存優雅是我可以回去調用堆棧，但我不相信有任何東西像對象層次結構的調用堆棧。對象根本不知道任何有關它們的引用的其他對象。例如：

public class A { 
    public List<B> Objects { get; set; } 
} 

public class B { 
    public B(int i) { } 
    //blah 
} 

public static void Main(string[] args) 
{ 
    //Establish a simple object Heiarchy 
    //Root: A 
    /* 

    A 
    |-B1 
    |-B2 
    |-B3 

    */ 
    var alpha = new A() 
    { 
     Objects = new List<B>() 
     { 
      new B(1), 
      new B(2), 
      new B(3) 
     } 
    } 

    //MagicMethod<T>(object objectToTrace) is the mythical method that we're looking for 

    //What you're looking for is something like this: 
    MagicMethod(alpha.Objects[1]); //Should output "alpha.Objects[1]" 

    //But what if we make another reference to the same object? 

    var beta = alpha.Objects[1]; 

    //Now what would MagicMethod() produce? 
    MagicMethod(beta); //would have to produce "beta" 

    //How is it possible that when We've called MagicMethod() with 
    //fundamentally the same argument, we get two different outputs? 
} 

正如你可以看到，我們的MagicMethod()着可能知道哪些引用應該打印。因此，即使如果一個對象有一個關於所有引用自身的地方的記錄，它不可能選擇正確的一個。

我希望我能夠轉達我想給你的線。我會在這裏說：我不知道這是否屬實，但我無法想象一種可能是真實的方式。

Answer 3

如果我們拋開了明顯的問題是allSales能改變的，賺了指數無用的第二....

var salesWithErrors = allSales.Select((sale,saleIdx => 
    new { Sale = sale, // not really needed for the particular example 
     Index = saleIdx, 
     DiscountsWithErrors = sale.orders.detail.discounts 
      .Select((d,i)=>new { 
       Discount = d, 
       Index = i, 
      }) 
      .Where(di=>isErrorConditionMet(d.Discount)) 
    }) 
    .Where(saleDiscountErrors => saleDiscountErrors.DiscountsWithErrors.Any()) 

var results = string.Join(Environment.NewLine, 
    salesWithErrors.SelectMany(sde=>sde.DiscountsWithErrors.Select(d=>new{ 
    SaleId = sde.Sale.Id, 
    SaleIndex = sde.Index, 
    DiscountId = d.Discount.Id 
    DiscountIndex = d.Index 
    }) 
    .Select(sdi=>$"allSales[{sdi.SaleIndex}].sale.orders.detail.discounts[{sdi.DiscountIndex}]")); 

相反，短暫的集合中，輸出指標的你可以（應該），而不是輸出的ID這是更持久，讓你發現他們在你的數據庫對象的

... 
.Select(sdi=>$"allSales[{sdi.SaleId}].sale.orders.detail.discounts[{sdi.DiscountId }]"