Javascript regexp - 仅当第一个字符不是星号时
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Javascript regexp - only if first character is not an asterisk
提问by Splynx
I am using a javascript validator which will let me build custom validation based on regexp
我正在使用一个 javascript 验证器，它可以让我基于正则表达式构建自定义验证
From their website: regexp=^[A-Za-z]{1,20}$allow up to 20 alphabetic characters.
从他们的网站：regexp=^[A-Za-z]{1,20}$最多允许 20 个字母字符。
This will return an error if the entered data in the input field is outside this scope.
如果输入字段中输入的数据超出此范围，这将返回错误。
What I need is the string that will trigger an error for the inputfield if the value has an asterix as the first character.
如果值的第一个字符是星号，我需要的是将触发输入字段错误的字符串。
I can make it trigger the opposite (an error if the first character is NOT an asterix) with:
我可以让它触发相反的（如果第一个字符不是星号则错误）：
regexp=[\u002A]
Heeeeelp please :-D
请嘿嘿嘿：-D
回答by Cameron
How about:
怎么样：
^[^\*]
Which matches any input that does notstart with an asterisk; judging from the example regex, any input which does not match the regex will be cause a validation error, so with the double negative you should get the behaviour you want :-)
匹配任何不以星号开头的输入；从示例正则表达式来看，任何与正则表达式不匹配的输入都将导致验证错误，因此使用双重否定，您应该得到您想要的行为:-)
Explanation of my regex:
我的正则表达式的解释：
- The first
^means "at the start of the string" - The
[...]construct is a character class, which matches a single character among the ones enclosed within the brackets - The
^in the beginning of the character class means "negate the character class", i.e. match any character that's notone of the ones listed - The
\*means a literal*;*has a special meaning in regular expressions, so I've escaped it with a backslash. As Rob has pointed out in the comments, it's not strictly necessary to escape (most) special characters within a character class
- 第一个
^意思是“在字符串的开头” - 该
[...]结构是一个字符类，不匹配的括号中的那些中的单个字符 - 将
^在字符类手段的开始“否定字符类”，即匹配的任何字符未列出的一个 - 所述
\*指文字*;*在正则表达式中具有特殊含义，所以我用反斜杠对其进行了转义。正如 Rob 在评论中指出的那样，在字符类中转义（大多数）特殊字符并不是绝对必要的
回答by DuckMaestro
How about ^[^\*].+.
怎么样^[^\*].+。
Broken down:
分解：
^= start of string.[^\*]= any one character not the '*'..+= any other character at least once.
^= 字符串的开始。[^\*]= '*' 以外的任何一个字符。.+= 任何其他字符至少一次。
回答by Kamil Szot
You can invert character class by using ^ after [
您可以在 [ 之后使用 ^ 来反转字符类
regexp=[^\u002A]
regexp=[^\u002A]
