c# 已知字符串内容,如果把字符串作为泛型使用
问题描述:
var kind ="PageSkill";
var service =new SkillService<PageSkill>();
如何根据已知变量kind,确定第二行代码的泛型?
答
public static Type GetTypeByName(string typename)
{
Type t = null;
string source = typename;
if (source.IndexOf('<') > 0)
{
List<string> lv = new List<string>();
while (RegexExpand.IsMatch(source, @"<[^<>]+>"))
{
lv.Add(RegexExpand.Match(source, @"(?<=<)[^<>]+(?=>)").Value);
source = RegexExpand.Replace(source, @"<[^<>]+>", "/" + (lv.Count - 1));
}
List<Type[]> args = new List<Type[]>();
for (int i = 0; i < lv.Count; i++)
{
List<Type> arg = new List<Type>();
string[] sp = lv[i].Split(',');
for (int j = 0; j < sp.Length; j++)
{
string s = sp[j].Trim();
if (!string.IsNullOrEmpty(s))
{
if (RegexExpand.IsMatch(s, @"/\d+$"))
{
Match m = RegexExpand.Match(s, @"^([^/\s]+)\s*/(\d+)$");
if (!m.Success)
{
throw new Exception("");
}
Type p = GetTypeByName(m.Groups[1].Value);
Type c = p.MakeGenericType(args[Convert.ToInt32(m.Groups[2].Value)]);
arg.Add(c);
}
else
{
arg.Add(GetTypeByName(s));
}
}
}
args.Add(arg.ToArray());
}
Match f = RegexExpand.Match(source, @"^([^/\s]+)\s*/(\d+)$");
if (!f.Success)
{
throw new Exception("");
}
Type fp = GetTypeByName(f.Groups[1].Value);
Type fc = fp.MakeGenericType(args[Convert.ToInt32(f.Groups[2].Value)]);
return fc;
}
else
{
try
{
t = Type.GetType(source);
if (t != null)
{
return t;
}
Assembly[] assembly = AppDomain.CurrentDomain.GetAssemblies();
foreach (Assembly ass in assembly)
{
t = ass.GetType(source);
if (t != null)
{
return t;
}
Type[] ts = ass.GetTypes();
foreach (Type st in ts)
{
if (RegexExpand.IsMatch(st.FullName, @"\." + RegexExpand.FormatRegExp(source) + @"(`?\d+)?$"))
{
return st;
}
}
}
}
catch (Exception ex)
{
}
}
return t;
}
// 调用的时候很简单
string typename = "Dictionary<Dictionary<string,List<Ajax>>,int>";
Type t = Common.GetTypeByName(typename);
dynamic item = t.Assembly.CreateInstance(t.FullName);
// 代码中出现RegexExpand 可用 Regex 代替,代替后记得加忽略大小写的参数
// FormatRegExp ,用来替换字符串中的特殊符号加转义,比如 .+*[]等
答
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