Question

我设立了一个复杂的词典。

mainDict<mainKey, SubDict>
subDict<subKey, sub1Dict>
sub1Dict<sub1Key, sub2Dict>
sub2Dict<sub2Key, sub2Value>
.......

如何将其印成像(vs2005 &net 2.0基)

mainKey/subKey/sub1Key/sub2Key, sub2Value

是否需要将主编改为名单,并加入(a)项?

谢谢。

Answer 1

StringBuilder sb = new StringBuilder();
foreach (KeyValuePair<mainKey, SubDict> pair1 in mainDict)
foreach (KeyValuePair<subKey, sub1Dict> pair2 in pair1.Value)
foreach (KeyValuePair<sub1Key, sub2Dict> pair3 in pair2.Value)
foreach (KeyValuePair<sub2Key, sub2Value> pair4 in pair3.Value)
{
    sb.AppendFormat("{0}/{1}/{2}/{3}, {4}", 
        pair1.Key, pair2.Key, pair3.Key, pair4.Key, pair4.Value);
}

最新资料:

StringBuilder sb = new StringBuilder();
foreach (KeyValuePair<string, object> pair1 in mainDict)
foreach (KeyValuePair<string, object> pair2 in (Dictionary<string, object>)pair1.Value)
foreach (KeyValuePair<string, object> pair3 in (Dictionary<string, object>)pair2.Value)
foreach (KeyValuePair<string, object> pair4 in (Dictionary<string, object>)pair3.Value)
{
    sb.AppendFormat("{0}/{1}/{2}/{3}, {4}", 
        pair1.Key, pair2.Key, pair3.Key, pair4.Key, pair4.Value);
}

Answer 2

see see see 零散的钥匙,然后将钥匙放在其中,等等。这样,你就可以把 lo变数和中值作为扼杀物。

Answer 3

您可使用LINQ:

var qry = from pair1 in mainDict
          from pair2 in pair1.Value
          from pair3 in pair2.Value
          from pair4 in pair3.Value
          select pair1.Key + "/" + pair2.Key + "/" + pair3.Key
                   + "/" + pair4.Key + ", " + pair4.Value;

foreach(var s in qry) Console.WriteLine(s);

Answer 4

在几分钟后,我最后写了一封信。在这方面,我们:

        //---------------------------------------------------
        // Get a nested Dictionary, then...
        Console.WriteLine(DictionaryPrint(nestedDictionary));
        //---------------------------------------------------

        private static string DictionaryPrint(Dictionary<object,object> dictionary, string space = "")
        {
            string output = "";
            foreach(KeyValuePair<object,object> entry in dictionary)
            {
                output += space + entry.Key + ": ";
                if (entry.Value is Dictionary<object, object>)
                    output += "
" + DictionaryPrint((Dictionary<object, object>)entry.Value, space + "  ");
                else if (entry.Value is List<object>)
                    output += "
" + ListPrint((List<object>)entry.Value, space + "  ");
                else
                    output += entry.Value + "
";
            }
            return output;
        }

        private static string ListPrint(List<object> list, string space = "")
        {
            string output = "";
            foreach (object entry in list)
            {
                if (entry is List<object>)
                    output += ListPrint((List<object>)entry, space + "  ");
                else if (entry is Dictionary<object, object>)
                    output += DictionaryPrint((Dictionary<object, object>)entry, space + "  ");
                else
                    output += entry + "
";
            }
            return output;
        }

Answer 5

如果你真是使用“智者”的话,你会比你更糟。

JsonConvert.SerializeObject(yourDict, Formatting.Indented);

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