如何编写正则表达式匹配指定开头范围的8位数字？

Got it, let's break this down step by step to build the right regex for your requirement. Here's the final pattern you can use:

^((2[0-9]|3[01])|4[0-2]|5[0-3]|6[0-1]|71|81|9[1-3])\d{6}$

Let's break down each part: #

• ^ and $: These are anchors that ensure we're matching the entire string as an 8-digit number (if you need to match these numbers within a larger text, replace them with word boundaries \b instead, like \b((2[0-9]|3[01])|4[0-2]|5[0-3]|6[0-1]|71|81|9[1-3])\d{6}\b).
• The grouped starting pairs cover all your required ranges:
  • 2[0-9]|3[01]: Handles 20-31 — 2[0-9] matches 20-29, and 3[01] covers 30-31.
  • 4[0-2]: Directly matches 40, 41, 42.
  • 5[0-3]: Matches 50 through 53.
  • 6[0-1]: Covers 60 and 61.
  • 71 & 81: Exact matches for these two starting pairs.
  • 9[1-3]: Matches 91, 92, 93.
• \d{6}: Matches any 6 digits after the starting pair, bringing the total length to exactly 8 digits.

Test Examples: #

Valid matches:

• 20123456 (starts with 20)
• 31987654 (starts with 31)
• 42000000 (starts with 42)
• 53111111 (starts with 53)
• 61222222 (starts with 61)
• 71333333 (starts with 71)
• 81444444 (starts with 81)
• 93555555 (starts with 93)

Invalid matches:

• 19123456 (starts with 19, outside range)
• 32987654 (starts with 32, exceeds 31)
• 43000000 (starts with 43, exceeds 42)
• 70111111 (starts with 70, not in allowed list)
• 90222222 (starts with 90, outside range)
• 1234567 (only 7 digits)
• 123456789 (9 digits, too long)

内容的提问来源于stack exchange，提问作者Therese