检查字符串是否只包含字母

Question

The idea is to have a String read and to verify that it does not contain any numeric characters. So something like "smith23" would not be acceptable.

Answer 1

What do you want? Speed or simplicity? For speed, go for a loop based approach. For simplicity, go for a one liner RegEx based approach.

Speed

public boolean isAlpha(String name) {
    char[] chars = name.toCharArray();

    for (char c : chars) {
        if(!Character.isLetter(c)) {
            return false;
        }
    }

    return true;
}

Simplicity

public boolean isAlpha(String name) {
    return name.matches("[a-zA-Z]+");
}

Answer 2

Java 8 lambda expressions. Both fast and simple.

boolean allLetters = someString.chars().allMatch(Character::isLetter);

Answer 3

或者，如果您使用的是 Apache Commons，请使用[StringUtils.isAlpha()] 。

Answer 4

First import Pattern :

import java.util.regex.Pattern;

Then use this simple code:

String s = "smith23";
if (Pattern.matches("[a-zA-Z]+",s)) { 
  // Do something
  System.out.println("Yes, string contains letters only");
}else{
  System.out.println("Nope, Other characters detected");    
}

This will output:

Nope, Other characters detected

Answer 5

I used this regex expression (".*[a-zA-Z]+.*") . With if not statement it will avoid all expressions that have a letter before, at the end or between any type of other character.

String strWithLetters = "123AZ456";
if(! Pattern.matches(".*[a-zA-Z]+.*", str1))
 return true;
else return false

Answer 6

A quick way to do it is by:

public boolean isStringAlpha(String aString) {
    int charCount = 0;
    String alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

    if (aString.length() == 0) {
        return false; //zero length string ain't alpha
    }

    for (int i = 0; i < aString.length(); i++) {
        for (int j = 0; j < alphabet.length(); j++) {
            if (aString.substring(i, i + 1).equals(alphabet.substring(j, j + 1))
                    || aString.substring(i, i + 1).equals(alphabet.substring(j, j + 1).toLowerCase())) {
                charCount++;
            }
        }

        if (charCount != (i + 1)) {
            System.out.println("\n**Invalid input! Enter alpha values**\n");
            return false;
        }
    }

    return true;
}

Because you don't have to run the whole aString to check if it isn't an alpha String .

Answer 7

private boolean isOnlyLetters(String s){
    char c=' ';
    boolean isGood=false, safe=isGood;
    int failCount=0;
    for(int i=0;i<s.length();i++){
        c = s.charAt(i);
        if(Character.isLetter(c))
            isGood=true;
        else{
            isGood=false;
            failCount+=1;
        }
    }
    if(failCount==0 && s.length()>0)
        safe=true;
    else
        safe=false;
    return safe;
}

I know it's a bit crowded. I was using it with my program and felt the desire to share it with people. It can tell if any character in a string is not a letter or not. Use it if you want something easy to clarify and look back on.

Answer 8

Faster way is below. Considering letters are only az,AZ.

public static void main( String[] args ){ 
        System.out.println(bestWay("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
        System.out.println(isAlpha("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));

        System.out.println(bestWay("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa1aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
        System.out.println(isAlpha("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa1aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
    }

    public static boolean bettertWay(String name) {
        char[] chars = name.toCharArray();
        long startTimeOne = System.nanoTime();
        for(char c : chars){
            if(!(c>=65 && c<=90)&&!(c>=97 && c<=122) ){
                System.out.println(System.nanoTime() - startTimeOne);
                    return false;
            }
        }
        System.out.println(System.nanoTime() - startTimeOne);
        return true;
    }


    public static boolean isAlpha(String name) {
        char[] chars = name.toCharArray();
        long startTimeOne = System.nanoTime();
        for (char c : chars) {
            if(!Character.isLetter(c)) {
                System.out.println(System.nanoTime() - startTimeOne);
                return false;
            }
        }
        System.out.println(System.nanoTime() - startTimeOne);
        return true;
    }

Runtime is calculated in nano seconds. It may vary system to system.

5748//bettertWay without numbers
true
89493 //isAlpha without  numbers
true
3284 //bettertWay with numbers
false
22989 //isAlpha with numbers
false

Answer 9

Check this,i guess this is help you because it's work in my project so once you check this code

if(! Pattern.matches(".*[a-zA-Z]+.*[a-zA-Z]", str1))
 {
   String not contain only character;
 }
else 
{
  String contain only character;
}

Answer 10

        String expression = "^[a-zA-Z]*$";
        CharSequence inputStr = str;
        Pattern pattern = Pattern.compile(expression);
        Matcher matcher = pattern.matcher(inputStr);
        if(matcher.matches())
        {
              //if pattern matches 
        }
        else
        {
             //if pattern does not matches
        }

Answer 11

尝试使用正则表达式： String.matches

Answer 12

public boolean isAlpha(String name)
{
    String s=name.toLowerCase();
    for(int i=0; i<s.length();i++)
    {
        if((s.charAt(i)>='a' && s.charAt(i)<='z'))
        {
            continue;
        }
        else
        {
           return false;
        }
    }
    return true;
}

Answer 13

While there are many ways to skin this cat, I prefer to wrap such code into reusable extension methods that make it trivial to do going forward. When using extension methods, you can also avoid RegEx as it is slower than a direct character check. I like using the extensions in the Extensions.cs NuGet package. It makes this check as simple as:

1. Add the https://www.nuget.org/packages/Extensions.cs package to your project.
2. Add " using Extensions; " to the top of your code.
3. "smith23".IsAlphabetic() will return False whereas "john smith".IsAlphabetic() will return True. By default the .IsAlphabetic() method ignores spaces, but it can also be overridden such that "john smith".IsAlphabetic(false) will return False since the space is not considered part of the alphabet.
4. Every other check in the rest of the code is simply MyString.IsAlphabetic() .

Answer 14

Feels as if our need is to find whether the character are only alphabets. Here's how you can solve it-

Character.isAlphabet(int(c)) 

helps to check if the characters of the string are alphabets or not. where c is

s.charAt(elementIndex)

Answer 15

To allow only ASCII letters, the character class \\p{Alpha} can be used. (This is equivalent to [\\p{Lower}\\p{Upper}] or [a-zA-Z] .)

boolean allLettersASCII = str.matches("\\p{Alpha}*");

For allowing all Unicode letters, use the character class \\p{L} (or equivalently, \\p{IsL} ).

boolean allLettersUnicode = str.matches("\\p{L}*");

See the Pattern documentation .

Answer 16

I found an easy of way of checking a string whether all its digit is letter or not.

public static boolean isStringLetter(String input) {

    boolean b = false;
    for (int id = 0; id < input.length(); id++) {
        if ('a' <= input.charAt(id) && input.charAt(id) <= 'z') {
            b = true;
        } else if ('A' <= input.charAt(id) && input.charAt(id) <= 'Z') {
            b = true;
        } else {
            b = false;
        }
    }
    return b;
}

I hope it could help anyone who is looking for such method.

Answer 17

使用 StringUtils.isAlpha() 方法，它会让你的生活变得简单。

Answer 18

Solution: I prefer to use a simple loop.

public static boolean isAlpha(String s) {
    if (s == null || s.length() == 0) {
        return false;
    }

    char[] chars = s.toCharArray();
    for (int index = 0; index < chars.length; index++) {
        int codePoint = Character.codePointAt(chars, index);
        if (!Character.isLetter(codePoint)) {
            return false;
        }
    }

    return true;
}