Как в символьной строке найти число

есть string строка, например

string str ="ddadsadwd 1337 dsdggf";

нужно найти в ней число и выписать в переменную int

задан 20 ноя 2016 в 14:07

#include <iostream>
#include <string>
#include <cstdlib>

int main(void)
{
    std::string s ="ddadsadwd 1337 dsdggf";
    size_t digits = s.find_first_of( "1234567890+-" );
    if( digits <= s.size() ) {
        std::cout << "Number found: " << atoi( s.c_str() + digits ) << std::endl;
    }
    else {
        std::cout << "Number is not found" << std::endl;
    }
    return 0;
}

ответ дан 20 ноя 2016 в 17:16

Вариант с regex:

#include <iostream>
#include <string>
#include <regex>
using namespace std;

int main() {
    string s = "ddadsadwd 1337 dsdggf";
    regex p("\d+");
    regex_token_iterator<string::iterator> numbers( s.begin(), s.end(), p );
    string t = *numbers;
    int first_number = atoi( t.c_str() );
    cout << first_number << endl;
    return 0;
}

P.S. Почему не работает (*numbers).c_str()?

ответ дан 20 ноя 2016 в 15:41

for(i=0; i<n; i++)//n — длинна строки, чтобы её найти можно использовать, например, strlen()
{
if(str==z)//str — стока, z — искомое число
{
y=z;
}
}

switch(cBuf)
    {
        case ‘.’:{iCile=1;iDrobBuf=0.1;}break;
        case ‘0’:ret=ret*iCile+0*iDrob;break;
        case ‘1’:ret=ret*iCile+1*iDrob;break;
        case ‘2’:ret=ret*iCile+2*iDrob;break;
        case ‘3’:ret=ret*iCile+3*iDrob;break;
        case ‘4’:ret=ret*iCile+4*iDrob;break;
        case ‘5’:ret=ret*iCile+5*iDrob;break;
        case ‘6’:ret=ret*iCile+6*iDrob;break;
        case ‘7’:ret=ret*iCile+7*iDrob;break;
        case ‘8’:ret=ret*iCile+8*iDrob;break;
        case ‘9’:ret=ret*iCile+9*iDrob;break;

   
    }

I have a requirement to find and extract a number contained within a string.

For example, from these strings:

string test = "1 test"
string test1 = " 1 test"
string test2 = "test 99"

How can I do this?

Here is another Linq approach which extracts the first number out of a string.

string input = "123 foo 456";
int result = 0;
bool success = int.TryParse(new string(input
                     .SkipWhile(x => !char.IsDigit(x))
                     .TakeWhile(x => char.IsDigit(x))
                     .ToArray()), out result);

Examples:

string input = "123 foo 456"; // 123
string input = "foo 456";     // 456
string input = "123 foo";     // 123

answered Sep 28, 2016 at 6:35

Just use a RegEx to match the string, then convert:

Match match = Regex.Match(test , @"(d+)");
if (match.Success) {
   return int.Parse(match.Groups[1].Value);
}

answered Jan 19, 2011 at 10:29

You can do this using String property like below:

 return new String(input.Where(Char.IsDigit).ToArray()); 

which gives only number from string.

For those who want decimal number from a string with Regex in TWO line:

decimal result = 0;
decimal.TryParse(Regex.Match(s, @"d+").Value, out result);

Same thing applys to float, long, etc…

answered Oct 5, 2016 at 3:19

Here is another simple solution using Linq which extracts only the numeric values from a string.

var numbers = string.Concat(stringInput.Where(char.IsNumber));

Example:

var numbers = string.Concat("(787) 763-6511".Where(char.IsNumber));

Gives: «7877636511»

answered Apr 11, 2022 at 14:08

var match=Regex.Match(@"a99b",@"d+");
if(match.Success)
{
    int val;
    if(int.TryParse(match.Value,out val))
    {
        //val is set
    }
}

answered Jan 19, 2011 at 10:29

The question doesn’t explicitly state that you just want the characters 0 to 9 but it wouldn’t be a stretch to believe that is true from your example set and comments. So here is the code that does that.

        string digitsOnly = String.Empty;
        foreach (char c in s)
        {
            // Do not use IsDigit as it will include more than the characters 0 through to 9
            if (c >= '0' && c <= '9') digitsOnly += c;
        }

Why you don’t want to use Char.IsDigit() — Numbers include characters such as fractions, subscripts, superscripts, Roman numerals, currency numerators, encircled numbers, and script-specific digits.

answered Jan 18, 2015 at 18:26

var outputString = String.Join("", inputString.Where(Char.IsDigit));

Get all numbers in the string.
So if you use for examaple ‘1 plus 2′ it will get ’12’.

answered Sep 16, 2015 at 14:21

Extension method to get all positive numbers contained in a string:

    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;
    }

If you want negative numbers as well simply modify this code to handle the minus sign (-)

Given this input:

"I was born in 1989, 27 years ago from now (2016)"

The resulting numbers list will be:

[1989, 27, 2016]

answered May 24, 2016 at 19:08

An interesting approach is provided here by Ahmad Mageed, uses Regex and StringBuilder to extract the integers in the order in which they appear in the string.

An example using Regex.Split based on the post by Ahmad Mageed is as follows:

var dateText = "MARCH-14-Tue";
string splitPattern = @"[^d]";
string[] result = Regex.Split(dateText, splitPattern);
var finalresult = string.Join("", result.Where(e => !String.IsNullOrEmpty(e)));
int DayDateInt = 0;

int.TryParse(finalresult, out DayDateInt);

Here is my Algorithm

    //Fast, C Language friendly
    public static int GetNumber(string Text)
    {
        int val = 0;
        for(int i = 0; i < Text.Length; i++)
        {
            char c = Text[i];
            if (c >= '0' && c <= '9')
            {
                val *= 10;
                //(ASCII code reference)
                val += c - 48;
            }
        }
        return val;
    }

answered Oct 11, 2018 at 9:25

here is my solution

string var = "Hello345wor705Ld";
string alpha = string.Empty;
string numer = string.Empty;
foreach (char str in var)
{
    if (char.IsDigit(str))
        numer += str.ToString();
    else
        alpha += str.ToString();
}
Console.WriteLine("String is: " + alpha);
Console.WriteLine("Numeric character is: " + numer);
Console.Read();

answered Jan 18, 2014 at 10:39

You will have to use Regex as d+

d matches digits in the given string.

answered Jan 19, 2011 at 10:28

string s = "kg g L000145.50rn";
        char theCharacter = '.';
        var getNumbers = (from t in s
                          where char.IsDigit(t) || t.Equals(theCharacter)
                          select t).ToArray();
        var _str = string.Empty;
        foreach (var item in getNumbers)
        {
            _str += item.ToString();
        }
        double _dou = Convert.ToDouble(_str);
        MessageBox.Show(_dou.ToString("#,##0.00"));

answered Jan 19, 2016 at 7:25

Using @tim-pietzcker answer from above, the following will work for PowerShell.

PS C:> $str = '1 test'
PS C:> [regex]::match($str,'d+').value
1

answered Feb 19, 2019 at 20:45