javascript - 正则表达式字符串最多包含 1 个 '#'

Question

需要正则表达式通过模式匹配字符串，排除字符串中“#”超过 1 次的情况

var srg = new RegExp(/([a-z0-9#]+[\s])/g);

字符串:

fdsffmsd,fmsd 
qwjswkds03sj 
ewew 
rfekwjkr#jfkdlsf 
wiru0ksd#erjk#jkls 
# 
casdw##kfdl 

结果

fdsffmsd，fmsd

qwjswkds03sj

哎呀

rfekwjkr#jfkdlsf

wiru0ksd#erjk#jkls

#

casdw##kfdl

Fiddle

谢谢

更新:

请原谅我的任务解释不太好

我更新了新的 fiddle New fiddle

var srg = new RegExp(/((\-)[a-z0-9#]+[\s])/g);

我们做什么:

• 搜索“-”字符

• 如果找到 - 匹配 a-z0-9# 但不满足任何空格

• 需要添加条件，仅在模式捕获中匹配 # 0 或 1 次

可能类似于 [a-z0-9#{0,1}] 表示 # 0 次或一次...

Answer 1

描述

这个表达式将:

• 要求字符串以 - 字符开头
• 允许#在字符串中最多出现1次
• 防止字符串中出现空格
• 允许使用一个或多个 a-z0-9 字符
• 允许#字符出现在字符串中的任何位置，包括开头和结尾

^(?!(?:.*?#){2,})-[a-z0-9#]+$

NODE                     EXPLANATION
--------------------------------------------------------------------------------
  ^                        the beginning of the string
--------------------------------------------------------------------------------
  (?!                      look ahead to see if there is not:
--------------------------------------------------------------------------------
    (?:                      group, but do not capture (at least 2
                             times (matching the most amount
                             possible)):
--------------------------------------------------------------------------------
      .*?                      any character except \n (0 or more
                               times (matching the least amount
                               possible))
--------------------------------------------------------------------------------
      #                        '#'
--------------------------------------------------------------------------------
    ){2,}                    end of grouping
--------------------------------------------------------------------------------
  )                        end of look-ahead
--------------------------------------------------------------------------------
  -                        '-'
--------------------------------------------------------------------------------
  [a-z0-9#]+               any character of: 'a' to 'z', '0' to '9',
                           '#' (1 or more times (matching the most
                           amount possible))
--------------------------------------------------------------------------------
  $                        before an optional \n, and the end of the
                           string

示例

Live Demo

示例

-abcdefghijklmnopqrstuvwxyz1234567890     = good
-abcdefghijklmnopq#rstuvwxyz1234567890    = good
-abcdefghijklmnopq##rstuvwxyz1234567890   = bad
-#abcdefghijklmnopqrstuvwxyz1234567890    = good
-##abcdefghijklmnopqrstuvwxyz1234567890   = bad
-#abcdefghijklmnopqrstuvwxyz1234567890    = good
-abcdefghijklmnopqrstuvwxyz1234567890#    = good
-#abcdefghijklmnopqrstuvwxyz1234567890#   = bad
#-abcdefghijklmnopqrstuvwxyz1234567890    = bad