检查字符串是否包含特殊字符

Question

如何检查字符串是否包含特殊字符，例如：

[,],{,},{,),*,|,:,>,

Answer 1

Pattern p = Pattern.compile("[^a-z0-9 ]", Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher("I am a string");
boolean b = m.find();

if (b)
   System.out.println("There is a special character in my string");

Answer 2

You can use the following code to detect special character from string.

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class DetectSpecial{ 
public int getSpecialCharacterCount(String s) {
     if (s == null || s.trim().isEmpty()) {
         System.out.println("Incorrect format of string");
         return 0;
     }
     Pattern p = Pattern.compile("[^A-Za-z0-9]");
     Matcher m = p.matcher(s);
    // boolean b = m.matches();
     boolean b = m.find();
     if (b)
        System.out.println("There is a special character in my string ");
     else
         System.out.println("There is no special char.");
     return 0;
 }
}

Answer 3

If you want to have LETTERS, SPECIAL CHARACTERS and NUMBERS in your password with at least 8 digit, then use this code, it is working perfectly

public static boolean Password_Validation(String password) 
{

    if(password.length()>=8)
    {
        Pattern letter = Pattern.compile("[a-zA-z]");
        Pattern digit = Pattern.compile("[0-9]");
        Pattern special = Pattern.compile ("[!@#$%&*()_+=|<>?{}\\[\\]~-]");
        //Pattern eight = Pattern.compile (".{8}");


           Matcher hasLetter = letter.matcher(password);
           Matcher hasDigit = digit.matcher(password);
           Matcher hasSpecial = special.matcher(password);

           return hasLetter.find() && hasDigit.find() && hasSpecial.find();

    }
    else
        return false;

}

Answer 4

如果它匹配正则表达式[a-zA-Z0-9 ]*则其中没有特殊字符。

Answer 5

What do you exactly call "special character" ? If you mean something like "anything that is not alphanumeric" you can use org.apache.commons.lang.StringUtils class (methods IsAlpha/IsNumeric/IsWhitespace/IsAsciiPrintable).

If it is not so trivial, you can use a regex that defines the exact character list you accept and match the string against it.

Answer 6

All depends on exactly what you mean by "special". In a regex you can specify

• \\W to mean non-alpahnumeric
• \\p{Punct} to mean punctuation characters

I suspect that the latter is what you mean. But if not use a [] list to specify exactly what you want.

Answer 7

Have a look at the java.lang.Character class. It has some test methods and you may find one that fits your needs.

Examples: Character.isSpaceChar(c) or !Character.isJavaLetter(c)

Answer 8

This is tested in android 7.0 up to android 10.0 and it works

Use this code to check if string contains special character and numbers:

  name = firstname.getText().toString(); //name is the variable that holds the string value

  Pattern special= Pattern.compile("[^a-z0-9 ]", Pattern.CASE_INSENSITIVE);
  Pattern number = Pattern.compile("[0-9]", Pattern.CASE_INSENSITIVE);
  Matcher matcher = special.matcher(name);
  Matcher matcherNumber = number.matcher(name);

  boolean constainsSymbols = matcher.find();
  boolean containsNumber = matcherNumber.find();

  if(constainsSymbols == true){
   //string contains special symbol/character
  }
  else if(containsNumber == true){
   //string contains numbers
  }
  else{
   //string doesn't contain special characters or numbers
  }

Answer 9

This worked for me:

String s = "string";
if (Pattern.matches("[a-zA-Z]+", s)) {
 System.out.println("clear");
} else {
 System.out.println("buzz");
}

Answer 10

First you have to exhaustively identify the special characters that you want to check.

Then you can write a regular expression and use

public boolean matches(String regex)

Answer 11

//without using regular expression........

    String specialCharacters=" !#$%&'()*+,-./:;<=>?@[]^_`{|}~0123456789";
    String name="3_ saroj@";
    String str2[]=name.split("");

    for (int i=0;i<str2.length;i++)
    {
    if (specialCharacters.contains(str2[i]))
    {
        System.out.println("true");
        //break;
    }
    else
        System.out.println("false");
    }

Answer 12

//this is updated version of code that i posted /* The isValidName Method will check whether the name passed as argument should not contain- 1.null value or space 2.any special character 3.Digits (0-9) Explanation--- Here str2 is String array variable which stores the the splited string of name that is passed as argument The count variable will count the number of special character occurs The method will return true if it satisfy all the condition */

public boolean isValidName(String name)
{
    String specialCharacters=" !#$%&'()*+,-./:;<=>?@[]^_`{|}~0123456789";
    String str2[]=name.split("");
    int count=0;
    for (int i=0;i<str2.length;i++)
    {
        if (specialCharacters.contains(str2[i]))
        {
            count++;
        }
    }       

    if (name!=null && count==0 )
    {
        return true;
    }
    else
    {
        return false;
    }
}

Answer 13

Pattern p = Pattern.compile("[\\p{Alpha}]*[\\p{Punct}][\\p{Alpha}]*");
        Matcher m = p.matcher("Afsff%esfsf098");
        boolean b = m.matches();

        if (b == true)
           System.out.println("There is a sp. character in my string");
        else
            System.out.println("There is no sp. char.");

Answer 14

Visit each character in the string to see if that character is in a blacklist of special characters; this is O(n*m).

The pseudo-code is:

for each char in string:
  if char in blacklist:
    ...

The complexity can be slightly improved by sorting the blacklist so that you can early-exit each check. However, the string find function is probably native code, so this optimisation - which would be in Java byte-code - could well be slower.

Answer 15

in the line String str2[]=name.split(""); give an extra character in Array... Let me explain by example "Aditya".split("") would return [, A, d,i,t,y,a] You will have a extra character in your Array...
The "Aditya".split("") does not work as expected by saroj routray you will get an extra character in String => [, A, d,i,t,y,a].

I have modified it,see below code it work as expected

 public static boolean isValidName(String inputString) {

    String specialCharacters = " !#$%&'()*+,-./:;<=>?@[]^_`{|}~0123456789";
    String[] strlCharactersArray = new String[inputString.length()];
    for (int i = 0; i < inputString.length(); i++) {
         strlCharactersArray[i] = Character
            .toString(inputString.charAt(i));
    }
    //now  strlCharactersArray[i]=[A, d, i, t, y, a]
    int count = 0;
    for (int i = 0; i <  strlCharactersArray.length; i++) {
        if (specialCharacters.contains( strlCharactersArray[i])) {
            count++;
        }

    }

    if (inputString != null && count == 0) {
        return true;
    } else {
        return false;
    }
}

Answer 16

Convert the string into char array with all the letters in lower case:

char c[] = str.toLowerCase().toCharArray();

Then you can use Character.isLetterOrDigit(c[index]) to find out which index has special characters.

Answer 17

Use java.util.regex.Pattern class's static method matches(regex, String obj)
regex : characters in lower and upper case & digits between 0-9
String obj : String object you want to check either it contain special character or not.

It returns boolean value true if only contain characters and numbers, otherwise returns boolean value false

Example.

String isin = "12GBIU34RT12";<br>
if(Pattern.matches("[a-zA-Z0-9]+", isin)<br>{<br>
   &nbsp; &nbsp; &nbsp; &nbsp;System.out.println("Valid isin");<br>
}else{<br>
   &nbsp; &nbsp; &nbsp; &nbsp;System.out.println("Invalid isin");<br>
}