How to determine whether it is a number in java:
1. Use regular expressions
/** * 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 JAVA’s own function
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. Catching NumberFormatException
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
java basic tutorial.
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