如何测试只有字母或撇号的字符串?
问题描述:
我需要检查一个字符串的几件事情:如何测试只有字母或撇号的字符串?
- 的第一个字母必须大写字母
- 只有
- 的2个省略号一个最多可以使用字母或单引号(')
下面的代码总是返回字符串不符合,除非我只输入1个大写字母。我如何纠正它以满足我的要求?
Private Function setName(ByVal pName As String)
Dim letters As Integer()
Dim aphCount As Integer = 0
Dim isvalid As Boolean = True
For i As Integer = 0 To pName.Length - 1 Step 1
ReDim letters(i)
letters(i) = Asc(pName.Substring(i, 1))
Next
If Not letters(0) >= 65 And letters(0) <= 90 Then
isvalid = False
End If
For i As Integer = 1 To pName.Length - 1 Step 1
If letters(i) >= 39 And letters(i) <= 122 Then
If letters(i) = 39 Then
aphCount += 1
If aphCount > 2 Then
isvalid = False
End If
ElseIf letters(i) >= 65 And letters(i) <= 90 Then
ElseIf letters(i) >= 97 And letters(i) <= 122 Then
Else isvalid = False
End If
Else isvalid = False
End If
Next
If isvalid = False Then
If MsgBox("you put in an invalid name", MsgBoxStyle.RetryCancel, "name error") = MsgBoxResult.Cancel Then
pName = "Hero" & heroCount
Else
pName = inputName()
pName = setName(pName)
End If
End If
Return pName
End Function
编辑:感谢大家的帮助,我一定要学习正则表达式,并与一些作品(使用Java来了,我切换到Java,因为这是一个项目,我学习编码,Java有更多的报价):
public void checkName(String name) throws IllegalArgumentException{
String noSpaceName = name.replaceAll("\\s+","");
String pattern = "^[A-Z][A-Za-z]*'?[A-Za-z]*'?[A-Za-z]*";
Pattern re = Pattern.compile(pattern);
Matcher m = re.matcher(noSpaceName);
if (m.matches()){
name.replaceAll("\\s+"," ");
super.setName(name);
}else throw new IllegalArgumentException ("Exception: Name is invalid");
}
答
您可以使用正则表达式在两行中执行此操作。
Private Function IsValidName(name As String) As Boolean
Dim match = Regex.Match(name, "^[A-Z]((')|[A-Za-z])*$")
Return match.Success AndAlso match.Groups(2).Captures.Count <= 2
End Function
答
您的代码对于您要做的事情而言非常复杂。它很长,它包含很多magic numbers。您可以使用正则表达式或通过定期比较字符串中的每个元素来解决此问题。这是第二种方式的示例:
Private Function IsValidName(name As String) As Boolean
If String.IsNullOrWhiteSpace(name) Then Return False
Return Char.IsUpper(name.AsEnumerable().First()) AndAlso
name.AsEnumerable().All(Function(c) Char.IsLetter(c) OrElse c.Equals("'"c)) AndAlso
name.AsEnumerable().Count(Function(c) c.Equals("'"c)) <= 2
End Function
请重新格式化代码以获得更好的可读性 - 很难读取您的函数的签名和主体。 –