忽略已知的一组键，并假设每个键仅出现一次:

Ignoring the known set of keys, and assuming each key appears only once:

string str = "V1,B=V1,C=V1,V2,V3,D=V1,V2,A=V1,=V2,V3";

var splitByEqual = new[] {'='};

var values = Regex.Split(str, @",(?=\w+=)")
    .Select(token => token.Split(splitByEqual, 2))
    .ToDictionary(pair => pair.Length == 1 ? "" : pair.First(),
                  pair => pair.Last());

• 正则表达式为非常简单:用逗号分隔，后跟一个密钥(任何字母数字)和等号. (如果我们允许A=V1,V2=V3，则将无效)
• 现在我们有了集合{V1，B=V1，C=V1,V2,V3，D=V1,V2，A=V1,=V2,V3}.我们将其除以=，但不超过一次.
• 接下来，我们创建一个字典.这条线有点难看，但并不是太重要-我们已经有了所需的数据.我还使用了空字符串而不是null.
• The regex is pretty simple: split by commas that are followed by a key (any alphanumeric) and an equal sign. (If we allow A=V1,V2=V3 this wouldn't work)
• Now we have the collection {V1,B=V1,C=V1,V2,V3,D=V1,V2,A=V1,=V2,V3}. We split that by =, but not more than once.
• Next we create a dictionary. This line is a little ugly, but isn't too important - we already have the data we need. I'm also using an empty string instead of null.

如果我们确实想使用已知的键列表，可以将模式更改为:

If we do want to use the known list of keys, we can change the pattern to:

var splitPattern = @",(?=(?:" + String.Join("|", keys.Select(Regex.Escape))) + ")=)";

并使用Regex.Split(str, splitPattern).