How to check if a String is numeric in Java

Question

How would you check if a String was a number before parsing it?

Answer 1

if you are on android, then you should use:

android.text.TextUtils.isDigitsOnly(CharSequence str)

documentation can be found here

keep it simple. mostly everybody can "re-program" (the same thing).

Answer 2

Parallel checking for very long strings using IntStream

In Java 8, the following tests if all characters of the given string are within '0' to '9'. Mind that the empty string is accepted:

string.chars().unordered().parallel().allMatch( i -> '0' <= i && '9' >= i )

Answer 3

This is the fastest way i know to check if String is Number or not:

public static boolean isNumber(String str){
  int i=0, len=str.length();
  boolean a=false,b=false,c=false, d=false;
  if(i<len && (str.charAt(i)=='+' || str.charAt(i)=='-')) i++;
  while( i<len && isDigit(str.charAt(i)) ){ i++; a=true; }
  if(i<len && (str.charAt(i)=='.')) i++;
  while( i<len && isDigit(str.charAt(i)) ){ i++; b=true; }
  if(i<len && (str.charAt(i)=='e' || str.charAt(i)=='E') && (a || b)){ i++; c=true; }
  if(i<len && (str.charAt(i)=='+' || str.charAt(i)=='-') && c) i++;
  while( i<len && isDigit(str.charAt(i)) ){ i++; d=true;}
  return i==len && (a||b) && (!c || (c && d));
}
static boolean isDigit(char c){
  return c=='0' || c=='1' || c=='2' || c=='3' || c=='4' || c=='5' || c=='6' || c=='7' || c=='8' || c=='9';
}

Answer 4

Java 8 lambda expressions.

String someString = "123123";
boolean isNumeric = someString.chars().allMatch( Character::isDigit );

Answer 5

Parse it (i.e. with Integer#parseInt ) and simply catch the exception. =)

To clarify: The parseInt function checks if it can parse the number in any case (obviously) and if you want to parse it anyway, you are not going to take any performance hit by actually doing the parsing.

If you would not want to parse it (or parse it very, very rarely) you might wish to do it differently of course.

Answer 6

public static boolean isNumeric(String str)
{
    return str.matches("-?\\d+(.\\d+)?");
}

CraigTP's regular expression (shown above) produces some false positives. E.g. "23y4" will be counted as a number because '.' matches any character not the decimal point.

Also it will reject any number with a leading '+'

An alternative which avoids these two minor problems is

public static boolean isNumeric(String str)
{
    return str.matches("[+-]?\\d*(\\.\\d+)?");
}