我需要找到并提取字符串中包含的数字。
例如,从这些字符串:
string test = "1 test"
string test1 = " 1 test"
string test2 = "test 99"
我该怎么做呢?
我需要找到并提取字符串中包含的数字。
例如,从这些字符串:
string test = "1 test"
string test1 = " 1 test"
string test2 = "test 99"
我该怎么做呢?
当前回答
使用正则表达式…
Regex re = new Regex(@"\d+");
Match m = re.Match("test 66");
if (m.Success)
{
Console.WriteLine(string.Format("RegEx found " + m.Value + " at position " + m.Index.ToString()));
}
else
{
Console.WriteLine("You didn't enter a string containing a number!");
}
其他回答
获取字符串中包含的所有正数的扩展方法:
public static List<long> Numbers(this string str)
{
var nums = new List<long>();
var start = -1;
for (int i = 0; i < str.Length; i++)
{
if (start < 0 && Char.IsDigit(str[i]))
{
start = i;
}
else if (start >= 0 && !Char.IsDigit(str[i]))
{
nums.Add(long.Parse(str.Substring(start, i - start)));
start = -1;
}
}
if (start >= 0)
nums.Add(long.Parse(str.Substring(start, str.Length - start)));
return nums;
}
如果你也想要负数,只需修改这段代码来处理负号(-)
假设输入如下:
"I was born in 1989, 27 years ago from now (2016)"
得到的数字列表将是:
[1989, 27, 2016]
使用上面的@tim-pietzcker回答,以下将适用于PowerShell。
PS C:\> $str = '1 test'
PS C:\> [regex]::match($str,'\d+').value
1
你也可以试试这个
string.Join(null,System.Text.RegularExpressions.Regex.Split(expr, "[^\\d]"));
string verificationCode ="dmdsnjds5344gfgk65585";
string code = "";
Regex r1 = new Regex("\\d+");
Match m1 = r1.Match(verificationCode);
while (m1.Success)
{
code += m1.Value;
m1 = m1.NextMatch();
}
如果数字有小数点,可以使用下面的方法
using System;
using System.Text.RegularExpressions;
namespace Rextester
{
public class Program
{
public static void Main(string[] args)
{
//Your code goes here
Console.WriteLine(Regex.Match("anything 876.8 anything", @"\d+\.*\d*").Value);
Console.WriteLine(Regex.Match("anything 876 anything", @"\d+\.*\d*").Value);
Console.WriteLine(Regex.Match("$876435", @"\d+\.*\d*").Value);
Console.WriteLine(Regex.Match("$876.435", @"\d+\.*\d*").Value);
}
}
}
结果:
"anything 876.8 anything" ==> 876.8 "anything 876 anything" ==> 876 "$876435" ==> 876435 "$876.435" ==> 876.435
示例:https://dotnetfiddle.net/IrtqVt