需要的正則表達式來驗證用戶名

Question

需要一個正則表達式來驗證用戶名，其中：

1. 應該允許尾隨空格，但不是空格字符之間
2. 必須至少包含一個字母，可以包含字母和數字
3. 7-15字元（字母數字）
4. 不能包含特殊字符
5. 下劃線允許

不知道如何做到這一點。任何幫助表示讚賞。謝謝。

這是我用的是什麼，但它可以讓人物之間的空間

"(?=.*[a-zA-Z])[a-zA-Z0-9_]{1}[_a-zA-Z0-9\\s]{6,14}" 

示例：用戶名 不允許有空格的用戶名

Answer 1

試試這個：

foundMatch = Regex.IsMatch(subjectString, @"^(?=.*[a-z])\w{7,15}\s*$", RegexOptions.IgnoreCase); 

是也允許使用_，因爲您在嘗試時允許這樣做。

所以基本上我使用三條規則。一個用於檢查是否存在至少一個字母。 另一個檢查字符串是否只包含alpha和_，最後我接受尾隨空格，並且至少有7個最大值爲15個alpha。你的軌道很好。堅持下去，你會在這裏也可以回答問題:)

擊穿：

" 
^   # Assert position at the beginning of the string 
(?=   # Assert that the regex below can be matched, starting at this position (positive lookahead) 
    .  # Match any single character that is not a line break character 
     *  # Between zero and unlimited times, as many times as possible, giving back as needed (greedy) 
    [a-z] # Match a single character in the range between 「a」 and 「z」 
) 
\w   # Match a single character that is a 「word character」 (letters, digits, etc.) 
    {7,15} # Between 7 and 15 times, as many times as possible, giving back as needed (greedy) 
\s   # Match a single character that is a 「whitespace character」 (spaces, tabs, line breaks, etc.) 
    *  # Between zero and unlimited times, as many times as possible, giving back as needed (greedy) 
$   # Assert position at the end of the string (or before the line break at the end of the string, if any) 
"