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Detailed explanation of the method of judging whether it is a number in java

Release: 2019-12-03 09:54:13
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Detailed explanation of the method of judging whether it is a number in java

Java method to determine whether it is a number: (Recommended: java video tutorial)

1. Use regular expressions

First, import java.util.regex.Pattern and java.util.regex.Matcher

/**
     * 利用正则表达式判断字符串是否是数字
     * @param str
     * @return
     */
    public boolean isNumeric(String str){
           Pattern pattern = Pattern.compile("[0-9]*");
           Matcher isNum = pattern.matcher(str);
           if( !isNum.matches() ){
               return false;
           }
           return true;
    }
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2. Use the functions that come with JAVA

public static boolean isNumeric(String str)
{
  for (int i = 0; i < str.length(); i++)
  {  
    System.out.println(str.charAt(i));
    if (!Character.isDigit(str.charAt(i)))
    {
        return false;
      }
  }
  return true;
}
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3. Use org.apache.commons.lang

org.apache.commons.lang.StringUtils;

boolean isNunicodeDigits=StringUtils.isNumeric("aaa123456789");


http://jakarta.apache.org/commons/lang/api-release/index.html下面的解释:

public static boolean isNumeric(String str)Checks if the String contains only unicode digits. A decimal point is not a unicode digit and returns false.

null will return false. An empty String ("") will return true.

StringUtils.isNumeric(null)   = false

StringUtils.isNumeric("")     = true

StringUtils.isNumeric(" ")   = false

StringUtils.isNumeric("123") = true

StringUtils.isNumeric("12 3") = false

StringUtils.isNumeric("ab2c") = false

StringUtils.isNumeric("12-3") = false

StringUtils.isNumeric("12.3") = false
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4. Determine the ASCII code value

public static boolean isNumeric0(String str)
{  
  for(int i=str.length();--i>=0;)
  {
     int chr=str.charAt(i);
     if(chr<48 || chr>57)
        return false;
  }
  return true;
 }
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5. Determine whether the characters in str are 0-9 one by one

public static boolean isNumeric3(String str)
{
  final String number = "0123456789";
  for(int i = 0;i < number.length; i ++)
  {
   if(number.indexOf(str.charAt(i)) == -1)
     {  
        return false;  
     }  
  }  
  return true;
}
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6. Capture the NumberFormatException exception

public static boolean isNumeric00(String str)
{
  try{
     Integer.parseInt(str);
     return true;
  }catch(NumberFormatException e)
  {
   System.out.println("异常:\"" + str + "\"不是数字/整数...");
   return false;
  }
}
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For more java knowledge, please pay attention to the java basic tutorial column.

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