正则表达式以防止尾随空格和额外空格 [英] Regex to prevent trailing spaces and extra spaces
问题描述
现在我有一个正则表达式可以防止用户输入任何特殊字符.唯一允许的字符是 A 到 Z、0 到 9 或空格.
Right now I have a regex that prevents the user from typing any special characters. The only allowed characters are A through Z, 0 through 9 or spaces.
我想改进这个正则表达式以防止以下情况:
I want to improve this regex to prevent the following:
- 没有前导/训练空格 - 如果用户在条目之前或之后输入一个或多个空格，则不允许.
- 禁止双空格 - 如果用户多次键入空格键，则不允许.
我现在用来防止特殊字符的正则表达式如下所示，看起来工作得很好，即:
The Regex I have right now to prevent special characters is as follows and appears to work just fine, which is:
^[a-zA-Z0-9 ]+$
根据其他一些想法，我尝试了所有这些选项，但没有奏效:
Following some other ideas, I tried all these options but they did not work:
^\A\s+[a-zA-Z0-9 ]+$\A\s+
/s*^[a-zA-Z0-9 ]+$/s*
我可以帮忙处理这个代码吗?同样，我只想要字母 A-Z、数字 0-9，并且没有前导或尾随空格.
Could I get a helping hand with this code? Again, I just want letters A-Z, numbers 0-9, and no leading or trailing spaces.
谢谢.
推荐答案
您可以使用以下正则表达式:
You can use the following regex:
^[a-zA-Z0-9]+(?: [a-zA-Z0-9]+)*$
参见正则表达式演示.
正则表达式将匹配开头的字母数字(1 个或多个)，然后匹配单个空格的零个或多个块，后跟一个或多个字母数字.
The regex will match alphanumerics at the start (1 or more) and then zero or more chunks of a single space followed with one or more alphanumerics.
作为替代，这里有一个基于前瞻的正则表达式(但效率较低):
As an alternative, here is a regex based on lookaheads (but is thus less efficient):
^(?!.* {2})(?=\S)(?=.*\S$)[a-zA-Z0-9 ]+$
查看正则表达式演示
(?!.* {2}) 不允许连续空格，(?=.*\S$) 要求末尾有一个非空格字符串和 (?=\S) 在开始时需要它.
The (?!.* {2}) disallows consecutive spaces and (?=.*\S$) requires a non-whitespace to be at the end of the string and (?=\S) requires it at the start.
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