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?
Regex.Split can extract numbers from strings. You get all the numbers that are found in a string.
string input = "There are 4 numbers in this string: 40, 30, and 10.";
// Split on one or more non-digit characters.
string[] numbers = Regex.Split(input, @"\D+");
foreach (string value in numbers)
{
if (!string.IsNullOrEmpty(value))
{
int i = int.Parse(value);
Console.WriteLine("Number: {0}", i);
}
}
Output:
Number: 4 Number: 40 Number: 30 Number: 10
Another simple solution using Regex You should need to use this
using System.Text.RegularExpressions;
and the code is
string var = "Hello3453232wor705Ld";
string mystr = Regex.Replace(var, @"\d", "");
string mynumber = Regex.Replace(var, @"\D", "");
Console.WriteLine(mystr);
Console.WriteLine(mynumber);
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
if the number has a decimal points, you can use below
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);
}
}
}
results :
"anything 876.8 anything" ==> 876.8
"anything 876 anything" ==> 876
"$876435" ==> 876435
"$876.435" ==> 876.435
Sample : https://dotnetfiddle.net/IrtqVt
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.
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]
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);
Did the reverse of one of the answers to this question: How to remove numbers from string using Regex.Replace?
// Pull out only the numbers from the string using LINQ
var numbersFromString = new String(input.Where(x => x >= '0' && x <= '9').ToArray());
var numericVal = Int32.Parse(numbersFromString);
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();
string s = "kg g L000145.50\r\n";
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"));
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
Based on the last sample I created a method:
private string GetNumberFromString(string sLongString, int iLimitNumbers)
{
string sReturn = "NA";
int iNumbersCounter = 0;
int iCharCounter = 0;
string sAlphaChars = string.Empty;
string sNumbers = string.Empty;
foreach (char str in sLongString)
{
if (char.IsDigit(str))
{
sNumbers += str.ToString();
iNumbersCounter++;
if (iNumbersCounter == iLimitNumbers)
{
return sReturn = sNumbers;
}
}
else
{
sAlphaChars += str.ToString();
iCharCounter++;
// reset the counter
iNumbersCounter = 0;
}
}
return sReturn;
}
1.5
? Exponential notation like1.5E45
? – Tim Pietzcker