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;
    }

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;
}

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

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;
 }

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;
}

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;
  }
}

For more java knowledge, please pay attention to the java basic tutorial column.

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