如何检查两个GUID
是否匹配?如何检查两个GUID是否与.NET(C#)匹配?
使用以下C#代码,如何匹配如果g2
具有相同GUID
作为g1
:
Guid g1 = new Guid("{10349469-c6f7-4061-b2ab-9fb4763aeaed}");
Guid g2 = new Guid("{45DF902E-2ECF-457A-BB0A-E04487F71D63}");
如何检查两个GUID
是否匹配?如何检查两个GUID是否与.NET(C#)匹配?
使用以下C#代码,如何匹配如果g2
具有相同GUID
作为g1
:
Guid g1 = new Guid("{10349469-c6f7-4061-b2ab-9fb4763aeaed}");
Guid g2 = new Guid("{45DF902E-2ECF-457A-BB0A-E04487F71D63}");
ToString()
GUID的方法将导致consisten字符串值,因此它可以用于匹配。
检查this DotNetFiddle进行此测试。
using System;
public class Program
{
public static void Main()
{
// Two distinct GUIDs
Guid g1 = new Guid("{10349469-c6f7-4061-b2ab-9fb4763aeaed}");
Guid g2 = new Guid("{45DF902E-2ECF-457A-BB0A-E04487F71D63}");
// GUID similar to 'g1' but with mixed case
Guid g1a = new Guid("{10349469-c6f7-4061-b2ab-9fb4763AEAED}");
// GUID similear to 'g1' but without braces
Guid g1b = new Guid("10349469-c6f7-4061-b2ab-9fb4763AEAED");
// Show string value of g1,g2 and g3
Console.WriteLine("g1 as string: {0}\n", g1.ToString());
Console.WriteLine("g2 as string: {0}\n", g2.ToString());
Console.WriteLine("g1a as string: {0}\n", g1a.ToString());
Console.WriteLine("g1b as string: {0}\n", g1b.ToString());
// g1 to g1a match result
bool resultA = (g1.ToString() == g1a.ToString());
// g1 to g1b match result
bool resultB = (g1.ToString() == g1b.ToString());
// Show match result
Console.WriteLine("g1 matches to g1a: {0}\n", resultA);
Console.WriteLine("g1 matches to g1b: {0}", resultB);
}
}
输出
G1作为字符串:10349469-c6f7-4061-b2ab-9fb4763aeaed
G2作为字符串:45df902e-2ecf-457A-bb0a-e04487f71d63
G1A作为字符串:10349469-c6f7-4061-b2ab-9fb4763aeaed
g1b as string:10349469-c6f7-4061-b2ab-9fb4763aeaed
G1匹配G1A:真
G1匹配G1B:真
绝对没有必要转换为字符串,你增加了不必要的开销 - 这是一个可怕的想法。只需使用'g1 == g2'或'g1.Equals(g2)' – DavidG
您使用的Guid.Equals
重载。
所以在实用性方面:
Guid g1 = ...
Guid g2 = ...
if (g1.Equals(g2)) { /* guids are equal */ }
注意System.Guid
实现平等的运营商一样,所以下面也将工作:
if (g1 == g2) { /* guids are equal */ }
感谢您指向正确的方向,我认为'Equals'与引用类型相匹配。 –
@AmitKB看一看['public bool Equals(Guid g)'](http://referencesource.microsoft.com/mscorlib/a.html#54bcc19a4028b3f2) –
@MartinBackasch谢谢参考。 –
'Guid'正确实现平等,所以'的Equals '或'=='。 –