Question

我想只接受％符号（可选）作为带数字的字符串变量，如输入中的百分比。 E.G，1000％或100％或2000％或普通字符串输入为1000或2000。

我们可以借助正则表达式来完成吗？或者其他一些验证。

请帮忙

Answer 1

^[0-9]+%?$

假设至少需要一个数字，并且只允许正数（或0）。

如果你还想允许小数（你可以在你的问题中提到过），试试

^[0-9]+(?:\.[0-9]+)?%?$

这允许任何正整数或小数（但不是缩写形式，如.1或1.）。当然不支持指数表示法。

Answer 2

您可以尝试使用数字和％作为可选字符。

^\d+(\.\d+)?%?$

Answer 3

绝对。听起来你也想在这里消毒，所以“[0-9] {1，}％”或“[0-9] *。？[0-9] {1，}％”确保你匹配小数百分比。

通常我可能会说，jfgi但这太多了（好玩。编辑）

http://www.regular-expressions.info/reference.html

你可以使用正则表达式之外的东西来做，但这只是一个循环，断言输入的每个字符都是十进制数字，句点或百分号，并且它们是以适当的顺序。

希望这有帮助！

Answer 4

尝试以下正则表达式

^[+-]?\d+(\.\d+)?%?$

应该使用小数

Answer 5

基于此评论：

它似乎与1000％，100但现在的模式完美配合小数值的问题，如101.50或199.50％

^[-+]?\d*\.?\d+\b%?$

说明：

"
^        # Assert position at the beginning of the string
[-+]     # Match a single character present in the list “-+”
   ?        # Between zero and one times, as many times as possible, giving back as needed (greedy)
[0-9]    # Match a single character in the range between “0” and “9”
   *        # Between zero and unlimited times, as many times as possible, giving back as needed (greedy)
\.       # Match the character “.” literally
   ?        # Between zero and one times, as many times as possible, giving back as needed (greedy)
[0-9]    # Match a single character in the range between “0” and “9”
   +        # Between one and unlimited times, as many times as possible, giving back as needed (greedy)
\b       # Assert position at a word boundary
%        # Match the character “%” literally
   ?        # Between zero and one 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)
"

作为旁注：你应该真的能够从@Tim的回答中考虑可选部分。看看其他答案提出的教程。

需要仅接受％符号作为输入字符串

5 个答案: