Good way to encapsulate Integer.parseInt()

You could return an Integer instead of an int, returning null on parse failure.

It's a shame Java doesn't provide a way of doing this without there being an exception thrown internally though - you can hide the exception (by catching it and returning null), but it could still be a performance issue if you're parsing hundreds of thousands of bits of user-provided data.

EDIT: Code for such a method:

public static Integer tryParse(String text) {
  try {
    return Integer.parseInt(text);
  } catch (NumberFormatException e) {
    return null;
  }
}

Note that I'm not sure off the top of my head what this will do if text is null. You should consider that - if it represents a bug (i.e. your code may well pass an invalid value, but should never pass null) then throwing an exception is appropriate; if it doesn't represent a bug then you should probably just return null as you would for any other invalid value.

Originally this answer used the new Integer(String) constructor; it now uses Integer.parseInt and a boxing operation; in this way small values will end up being boxed to cached Integer objects, making it more efficient in those situations.

What behaviour do you expect when it's not a number?

If, for example, you often have a default value to use when the input is not a number, then a method such as this could be useful:

public static int parseWithDefault(String number, int defaultVal) {
  try {
    return Integer.parseInt(number);
  } catch (NumberFormatException e) {
    return defaultVal;
  }
}

Similar methods can be written for different default behaviour when the input can't be parsed.

In some cases you should handle parsing errors as fail-fast situations, but in others cases, such as application configuration, I prefer to handle missing input with default values using Apache Commons Lang 3 NumberUtils.

int port = NumberUtils.toInt(properties.getProperty("port"), 8080);

To avoid handling exceptions use a regular expression to make sure you have all digits first:

//Checking for Regular expression that matches digits
if(value.matches("\\d+")) {
     Integer.parseInt(value);
}