Use Long.parseLong()
Copy Long.parseLong("0", 10) // returns 0L
Long.parseLong("473", 10) // returns 473L
Long.parseLong("-0", 10) // returns 0L
Long.parseLong("-FF", 16) // returns -255L
Long.parseLong("1100110", 2) // returns 102L
Long.parseLong("99", 8) // throws a NumberFormatException
Long.parseLong("Hazelnut", 10) // throws a NumberFormatException
Long.parseLong("Hazelnut", 36) // returns 1356099454469L
Long.parseLong("999") // returns 999L
Answer from Mike Christensen on Stack OverflowVideos
Use Long.parseLong()
Copy Long.parseLong("0", 10) // returns 0L
Long.parseLong("473", 10) // returns 473L
Long.parseLong("-0", 10) // returns 0L
Long.parseLong("-FF", 16) // returns -255L
Long.parseLong("1100110", 2) // returns 102L
Long.parseLong("99", 8) // throws a NumberFormatException
Long.parseLong("Hazelnut", 10) // throws a NumberFormatException
Long.parseLong("Hazelnut", 36) // returns 1356099454469L
Long.parseLong("999") // returns 999L
To convert a String to a Long (object), use Long.valueOf(String s).longValue();
See link
See the reference documentation for the String class: String s = String.valueOf(date);
If your Long might be null and you don't want to get a 4-letter "null" string, you might use Objects.toString, like: String s = Objects.toString(date, null);
EDIT:
You reverse it using Long l = Long.valueOf(s); but in this direction you need to catch NumberFormatException
String strLong = Long.toString(longNumber);
Simple and works fine :-)
This is a common way to do it:
long l = Long.parseLong(str);
There is also this method: Long.valueOf(str); Difference is that parseLong returns a primitive long while valueOf returns a new Long() object.
The method for converting a string to a long is Long.parseLong. Modifying your example:
String s = "1333073704000";
long l = Long.parseLong(s);
// Now l = 1333073704000