让我们从单个标识符开始:
- 它的第一个字符必须是字母或下划线
- 它可以包含字母、下划线和数字
所以标识符的正则表达式是
[A-Za-z_][A-Za-z0-9_]*
接下来,我们应该将标识符与.
(不要忘记转义.
)一个标识符链接起来,后跟零个或多个.
+ 标识符:
^[A-Za-z_][A-Za-z0-9_]*(?:\.[A-Za-z_][A-Za-z0-9_]*)*$
如果它必须恰好是两个标识符(而不是,比如说abc.def.hi
- 三个)
^[A-Za-z_][A-Za-z0-9_]*\.[A-Za-z_][A-Za-z0-9_]*$
测试:
string[] tests = new string[] {
"Personnel.FirstName1", // the only string that should be matched
"2Personnel.FirstName1",
"Personnel.33FirstName",
"Personnel..FirstName",
"Personnel.;FirstName",
"Personnel.FirstName.",
"Personnel.FirstName ",
" Personnel.FirstName",
" Personnel. FirstName",
" 23Personnel.3FirstName",
} ;
string pattern = @"^[A-Za-z_][A-Za-z0-9_]*(\.[A-Za-z_][A-Za-z0-9_]*)*$";
var results = tests
.Select(test =>
$"{"\"" + test + "\"",-25} : {(Regex.IsMatch(test, pattern) ? "matched" : "failed")}"");
Console.WriteLine(String.Join(Environment.NewLine, results));
结果:
"Personnel.FirstName1" : matched
"2Personnel.FirstName1" : failed
"Personnel.33FirstName" : failed
"Personnel..FirstName" : failed
"Personnel.;FirstName" : failed
"Personnel.FirstName." : failed
"Personnel.FirstName " : failed
" Personnel.FirstName" : failed
" Personnel. FirstName" : failed
" 23Personnel.3FirstName" : failed
编辑:如果应该接受文化特定名称(如)(请参阅 Rand Random 的评论),则范围应更改为-任何字母。异国情调的可能性 - 文化特定的数字(例如波斯数字 - )可以通过更改为来解决äöü.FirstName
[A-Za-z]
\p{L}
۰۱۲۳۴۵۶۷۸۹
0-9
\d
// culture specific letters, but not digits
string pattern = @"^[\p{L}_][\p{L}0-9_]*(?:\.[\p{L}_][\p{L}0-9_]*)*$";
如果每个标识符不应该超过一定的长度(比如,16
),我们应该重新设计初始标识符模式:强制字母或下划线后跟[0..16-1] == {0,15}
字母、数字或下划线
[A-Za-z_][A-Za-z0-9_]{0,15}
我们有
string pattern = @"^[A-Za-z_][A-Za-z0-9_]{0,15}(?:\.[A-Za-z_][A-Za-z0-9_]{0,15})*$";