Question

我的输入为1+1或1-1或2+9或3-12或31-10或11+11，同样明智

我试过

^\d*[\+\-]\d*$

它有效，但我需要的是输入，如+1或+2或+31或1+或7+或-1或-2或-31或1-或7-不匹配。

Answer 1

将*替换为+：

var pat = @"^\d+[+-]\d+$";

查看demo of how this regex works。

+量词将匹配一个或更多数字不允许输入的实例，例如-，+，1+或{{1 }}

此外，您无需在-2字符类中转义+或-（[-+]不必在其开始/结束时进行转义）

这是C# demo：

输出：

var strs = new string[] {"1+1","1-1","2+9","3-12","31-10","11+11","+1","+2","+31","1+","7+","-1","-2","-31","1-","7-"};
foreach (string s in strs) 
{
    var matched = Regex.IsMatch(s, @"^\d+[+-]\d+$");
    Console.WriteLine("{0}: {1}", s, matched);
}

Answer 2

当你想要匹配时，如果1和+之间有一个可选的空格，你可以使用正则表达式

^\d+\s*[+-]\s*\d+$

在https://regex101.com/r/bpCjKM/2

查找示例

解释正则表达式

/
^\d+\s*[\+\-]\s*\d+$
/
g
^ asserts position at start of the string
\d+ matches a digit (equal to [0-9])
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
\s* matches any whitespace character (equal to [\r\n\t\f\v ])
* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
Match a single character present in the list below [\+\-]
\+ matches the character + literally (case sensitive)
\- matches the character - literally (case sensitive)
\s* matches any whitespace character (equal to [\r\n\t\f\v ])
* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
\d+ matches a digit (equal to [0-9])
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
$ asserts position at the end of the string, or before the line terminator right at the end of the string (if any)
Global pattern flags
g modifier: global. All matches (don't return after first match)

Answer 3

对于你的输入尝试这个（即将正则表达式的*替换为+）

^\d+[\+\-]\d+$

*表示零次或多次出现。

+表示一次或多次出现。

正则表达式匹配字符串与+或 - 之间

3 个答案: