java 确定一个字符串中有多少个数字和字母
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Determine how many digits and letters are in a string
提问by user1840435
import java.util.Scanner;
public class digitthingy
{
public static void main(String args[])
{
Scanner s = new Scanner(System.in);
String first="";
int firstnum=0;
System.out.print("Enter a string: ");
first = s.nextLine();
firstnum = first.indexOf("1-100");
System.out.println(firstnum);
}
}
I'd like to know how many numbers there are in a certain string I entered, but I have no idea how to do it. I'm new to programming.
我想知道我输入的某个字符串中有多少个数字，但我不知道该怎么做。我是编程新手。
回答by Peter Wilkinson
This looks like a job for regular expressions!
这看起来像是一份工作regular expressions！
String s = "sdf234sdf234";
System.out.println(s.replaceAll("\D", "").length());
Or perhaps your after each multiple digit number instance?
或者也许你在每个多位数字实例之后？
String s = "sdf234sdf234sdf23";
s = s.replaceAll("^\D+|\D+$", "").replaceAll("\D+", ",");
List<String> numbers = Arrays.asList(s.split(","));
System.out.println(numbers);
回答by ccleve
String myString = "whatever123";
int count = 0;
for (int i = 0; i < myString.length(), i++) {
if (Character.isDigit(myString.charAt(i)) {
count++;
}
}
System.out.println(count);
回答by Delorean
For number of digits:
对于位数：
Pattern digitPattern = Pattern.compile("\d");
Matcher digitMatcher = digitPattern.matcher("asdf 123 qwer");
int digitCount = 0;
while (digitMatcher.find())
digitCount++;
Similar for letters, only use "[a-zA-Z]" regex instead;
与字母类似，仅使用“[a-zA-Z]”正则表达式；
回答by 7er
Count digits in string with java 8:
使用 java 8 计算字符串中的数字：
public long countDigits(String s) {
return s.chars().mapToObj(i -> (char) i).filter(Character::isDigit).count();
}
回答by Lokesh Kumar Gaurav
Try this code I hope it will work
试试这个代码我希望它会起作用
import java.util.Scanner;
public class CountDigitAndCharacter {
public static void main(String args[])
{
Scanner s = new Scanner(System.in);
String first="";
int firstnum=0;
System.out.print("Enter a string: ");
first = s.nextLine();
int numberCount=0,charCount=0;
for(int i=0; i<first.length();i++){
char c=first.charAt(i);
if(c=='0' || c=='1' || c=='2' || c=='3' || c=='4' || c=='5' || c=='6' || c=='7' || c=='8' || c=='9' )
++numberCount;
else
++charCount;
}
System.out.println("numberCount :-"+numberCount +" CharCount :-"+charCount);
}
}
Result:- Enter a string: l0kesh numberCount :-1 CharCount :-5
结果：- 输入字符串：l0kesh numberCount :-1 CharCount :-5
回答by dreamcrash
you can use the Scanner class:
您可以使用 Scanner 类：
Scanner in = new Scanner(System.in);
// Reads a single line from the console
// and stores into name variable
first = s.nextLine();
And then iterate over the String:
然后遍历字符串：
int numbers = 0;
int letters = 0;
for(int i = 0; i < first.length(); i++)
{
if (Character.isDigit (first.charAt(i)) numbers++; // Count the digits
else if (Character.isLetter(first.charAt(i)) letters++; // Count the Letters
else{} // is not a letter and not a digit
}
回答by Srinivas B
try this ...
试试这个 ...
public class digitthingy
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
String s="";
int firstnum=0;
System.out.print("Enter a string: ");
s = sc.nextLine();
int digitCount =0, charCount=0;
for(int i=0;i<s.length();i++){
if(s.charAt(i)==' ') continue;
else if((s.charAt(i) >='a' && s.charAt(i)<='z') || (s.charAt(i)>='A' && s.charAt(i)<='Z'))
charCount++;
else if(s.charAt(i) >= '0' && s.charAt(i)<='9')
digitCount++;
}
System.out.println(digitCount+ " "+ charCount);
}
}
回答by SanjayDVG
public class ArrayDigits
公共类 ArrayDigits
{
{
public static void main(String[] args)
{
String str = new String("Hey123456789234Hey");
String num = new String("0123456789");
int dig_count = 0;
for(int i=0;i<str.length();i++)
{
for(int j=0;j<num.length();j++)
{
if(str.charAt(i)==num.charAt(j))
{
dig_count++;
}
}
}
System.out.println("Total length of String:"+str.length());
System.out.println("Number of digits:"+dig_count);
System.out.println("Number of characters:"+(str.length() - dig_count));
}
}
}
