从字符串中获取多个数字

时间:2021-09-23 11:03:20

I have strings like

我的弦

AS_!SD 2453iur ks@d9304-52kasd

AS_ !SD 2453 iur ks@d9304-52kasd

I need to get the 2 frist numbres of the string:

我需要得到字符串的2个frist numbres:

for that case will be: 2453 and 9304

该病例将为:2453和9304

I don't have any delimiter in the string to try a split, and the length of the numbers and string is variable, I'm working in C# framework 4.0 in a WPF.

我在字符串中没有任何分隔符来尝试分割,而且数字和字符串的长度是可变的,我在WPF中使用c# framework 4.0。

thanks for the help, and sorry for my bad english

谢谢你的帮助,很抱歉我的英语说得不好

3 个解决方案

#1


9  

This solution will take two first numbers, each can have any number of digits

这个解需要两个前数,每个都可以有任意个数

string s =  "AS_!SD 2453iur ks@d9304-52kasd";

MatchCollection matches = Regex.Matches(s, @"\d+");

string[] result = matches.Cast<Match>()
                         .Take(2)
                         .Select(match => match.Value)
                         .ToArray();

Console.WriteLine( string.Join(Environment.NewLine, result) );

will print

将打印

2453
9304

you can parse them to int[] by result.Select(int.Parse).ToArray();

可以通过result.Select(int. parse).ToArray()将它们解析为int[];

#2


0  

You can loop chars of your string parsing them, if you got a exception thats a letter if not is a number them you must to have a list to add this two numbers, and a counter to limitate this.

你可以循环字符串的字符来解析它们,如果你有一个例外那就是一个字母如果不是数字,你必须有一个列表来添加这两个数字,还有一个计数器来限制它。

follow a pseudocode:

遵循一个伪代码:

for char in string:

if counter == 2:
 stop loop

if parse gets exception
 continue

else
 loop again in samestring stating this point
 if parse gets exception
  stop loop
 else add char to list

#3


0  

Alternatively you can use the ASCII encoding:

您也可以使用ASCII编码:

string value = "AS_!SD 2453iur ks@d9304-52kasd";

byte zero = 48; // 0
byte nine = 57; // 9

byte[] asciiBytes = Encoding.ASCII.GetBytes(value);

byte[] asciiNumbers = asciiBytes.Where(b => b >= zero && b <= nine)
                    .ToArray();

char[] numbers = Encoding.ASCII.GetChars(asciiNumbers);

// OR

string numbersString =  Encoding.ASCII.GetString(asciiNumbers);

//First two number from char array
int aNum = Convert.ToInt32(numbers[0]);
int bNum =  Convert.ToInt32(numbers[1]);

//First two number from string
string aString = numbersString.Substring(0,2);

#1


9  

This solution will take two first numbers, each can have any number of digits

这个解需要两个前数,每个都可以有任意个数

string s =  "AS_!SD 2453iur ks@d9304-52kasd";

MatchCollection matches = Regex.Matches(s, @"\d+");

string[] result = matches.Cast<Match>()
                         .Take(2)
                         .Select(match => match.Value)
                         .ToArray();

Console.WriteLine( string.Join(Environment.NewLine, result) );

will print

将打印

2453
9304

you can parse them to int[] by result.Select(int.Parse).ToArray();

可以通过result.Select(int. parse).ToArray()将它们解析为int[];

#2


0  

You can loop chars of your string parsing them, if you got a exception thats a letter if not is a number them you must to have a list to add this two numbers, and a counter to limitate this.

你可以循环字符串的字符来解析它们,如果你有一个例外那就是一个字母如果不是数字,你必须有一个列表来添加这两个数字,还有一个计数器来限制它。

follow a pseudocode:

遵循一个伪代码:

for char in string:

if counter == 2:
 stop loop

if parse gets exception
 continue

else
 loop again in samestring stating this point
 if parse gets exception
  stop loop
 else add char to list

#3


0  

Alternatively you can use the ASCII encoding:

您也可以使用ASCII编码:

string value = "AS_!SD 2453iur ks@d9304-52kasd";

byte zero = 48; // 0
byte nine = 57; // 9

byte[] asciiBytes = Encoding.ASCII.GetBytes(value);

byte[] asciiNumbers = asciiBytes.Where(b => b >= zero && b <= nine)
                    .ToArray();

char[] numbers = Encoding.ASCII.GetChars(asciiNumbers);

// OR

string numbersString =  Encoding.ASCII.GetString(asciiNumbers);

//First two number from char array
int aNum = Convert.ToInt32(numbers[0]);
int bNum =  Convert.ToInt32(numbers[1]);

//First two number from string
string aString = numbersString.Substring(0,2);