在字符串中找到一个整数。-Regextofindanintegerwithinastring

49  

You're asking for 0 or more digits. You need to ask for 1 or more:

你要求0或更多的数字。你需要要求一个或多个:

"\\d+"

3  

It looks like the other solutions failed to handle +/- and cases like 2e3, which java.lang.Integer.parseInt(String) supports, so I'll take my go at the problem. I'm somewhat inexperienced at regex, so I may have made a few mistakes, used something that Java's regex parser doesn't support, or made it overly complicated, but the statements seemed to work in Kiki 0.5.6.

看起来其他解决方案无法处理+/-和2e3这样的情况，这是java.lang.Integer.parseInt(String)支持的，所以我将着手解决这个问题。我在regex方面有点缺乏经验，所以我可能犯了一些错误，使用了Java的regex解析器不支持的东西，或者使它过于复杂，但是这些语句似乎在Kiki 0.5.6中起作用。

All regular expressions are provided in both an unescaped format for reading, and an escaped format that you can use as a string literal in Java.

所有正则表达式都以一种未转义格式和一种转义格式提供，您可以将这种格式用作Java中的字符串文字。

To get a byte, short, int, or long from a string:

从字符串获取字节、短、整数或长字符串:

unescaped: ([\+-]?\d+)([eE][\+-]?\d+)?
  escaped: ([\\+-]?\\d+)([eE][\\+-]?\\d+)?

...and for bonus points...

…和加分……

To get a double or float from a string:

从字符串中获得双或浮动:

unescaped: ([\+-]?\d(\.\d*)?|\.\d+)([eE][\+-]?(\d(\.\d*)?|\.\d+))?
  escaped: ([\\+-]?\\d(\\.\\d*)?|\\.\d+)([eE][\\+-]?(\\d(\\.\\d*)?|\\.\\d+))?

0  

Heres a handy one I made for C# with generics. It will match based on your regular expression and return the types you need:

这是我为c#用泛型做的一个方便的例子。它将根据您的正则表达式匹配并返回所需的类型:

public T[] GetMatches(string Input, string MatchPattern) where T : IConvertible
    {
        List MatchedValues = new List();
        Regex MatchInt = new Regex(MatchPattern);

        MatchCollection Matches = MatchInt.Matches(Input);
        foreach (Match m in Matches)
            MatchedValues.Add((T)Convert.ChangeType(m.Value, typeof(T)));

        return MatchedValues.ToArray();
    }

then if you wanted to grab only the numbers and return them in an string[] array:

然后，如果您只想获取数字并以字符串[]数组返回它们:

string Test = "22$data44abc";
string[] Matches = this.GetMatches(Test, "\\d+");

Hopefully this is useful to someone...

希望这对某人有用……

0  

In addition to what PiPeep said, if you are trying to match integers within an expression, so that 1 + 2 - 3 will only match 1, 2, and 3, rather than 1, + 2 and - 3, you actually need to use a lookbehind statement, and the part you want will actually be returned by Matcher.group(2) rather than just Matcher.group().

除了PiPeep所说,如果你想在一个表达式匹配整数,所以1 + 2 - 3只会匹配1、2和3,而不是1 + 2 - 3,你需要使用一个向后插入语句,和你想要的一部分会由Matcher.group返回(2)而不是Matcher.group()。

unescaped: ([0-9])?((?(1)(?:[\+-]?\d+)|)(?:[eE][\+-]?\d+)?)
  escaped: ([0-9])?((?(1)(?:[\\+-]?\\d+)|)(?:[eE][\\+-]?\\d+)?)

Also, for things like someNumber - 3, where someNumber is a variable name or something like that, you can use

同样，对于someNumber - 3，其中someNumber是一个变量名或者类似的名称，你可以使用

unescaped: (\w)?((?(1)(?:[\+-]?\d+)|)(?:[eE][\+-]?\d+)?)
  escaped: (\\w)?((?(1)(?:[\\+-]?\\d+)|)(?:[eE][\\+-]?\\d+)?)

Although of course that wont work if you are parsing a string like The net change to blahblah was +4

当然，如果您正在解析一个字符串，比如对blahblah的净更改为+4，那么这是行不通的

0  

the java spec actually gives this monster of a regex for parsing doubles. however it is considered bad practice, just trying to parse with the intended type, and catching the error, tends to be slightly more readable.

java规范实际上给了这个巨大的regex用于解析加倍。然而，这被认为是不好的做法，只是试图用预期的类型进行解析，并捕获错误，这样做的可读性会稍微高一些。

DOUBLE_PATTERN = Pattern
        .compile("[\\x00-\\x20]*[+-]?(NaN|Infinity|((((\\p{Digit}+)(\\.)?((\\p{Digit}+)?)"
                + "([eE][+-]?(\\p{Digit}+))?)|(\\.((\\p{Digit}+))([eE][+-]?(\\p{Digit}+))?)|"
                + "(((0[xX](\\p{XDigit}+)(\\.)?)|(0[xX](\\p{XDigit}+)?(\\.)(\\p{XDigit}+)))"
                + "[pP][+-]?(\\p{Digit}+)))[fFdD]?))[\\x00-\\x20]*");