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The two main ways to do this are using the method valueOf() and method parseInt() of the Integer class. Suppose you are given a String like this. String numberInString = "999"; Then you can convert it into integer by using. int numberInInteger = Integer.parseInt(numberInString); And alternatively, you can use.
Jul 9, 2009 · Or even another alternative is to use Java's built-in java.text.NumberFormat object to see if, after parsing the string the parser position is at the end of the string. If it is, we can assume the entire string is numeric: public static boolean isNumeric(String str) {. ParsePosition pos = new ParsePosition(0); NumberFormat.getInstance().parse ...
Integer class has static method toString() - you can use it: int i = 1234; String str = Integer.toString(i); Returns a String object representing the specified integer. The argument is converted to signed decimal representation and returned as a string, exactly as if the argument and radix 10 were given as arguments to the toString(int, int ...
Mar 26, 2011 · Only use Integer.parserInt() after you ensured, that the String really contains an integer value. If doing so and Integer.parseInt still throws an exception, then you know that you have a problem in your code which you should fix (or, for the sake of completeness, that Integer.parseInt itself is buggy, but this option is quite unlikely) –
ByException - integer data: 45 ByException - non-integer data: 465 ByRegex - integer data: 272 ByRegex - non-integer data: 131 ByCompiledRegex - integer data: 45 ByCompiledRegex - non-integer data: 26 ByJonas - integer data: 8 ByJonas - non-integer data: 2
Jun 25, 2009 · Your best bet is to use Integer.parseInt. This will return an int, but this can be auto-boxed to an Integer. This is slightly faster than valueOf, as when your numbers are between -128 and 127 it will use the Integer cache and not create new objects. The slowest is the Apache method. private String data = "99";
Dec 8, 2010 · Cannot convert String to Integer in Java-1. Find average of numbers in text file. 0.
Apr 20, 2010 · Actually, int result = a * 10000 + b * 1000 + c * 100 + d * 10 + e; String s = Integer.toString(result); will work. Note: this will only work when a is greater than 0 and all of b, c, d and e are in [0, 9]. For example, if b is 15, Michael's method will get you the result you probably want. edited Apr 20, 2010 at 11:54.
Instead of using String.format (**). it's good if you use DecimalFormat java API which builds for this type of purposes.Let me explain with code. String pattern = "000"; double value = 12; //can be 536 or any. DecimalFormat formatter = new DecimalFormat(pattern); String formattedNumber = formatter.format(value);
Integer.parseInt("4") == Integer.parseInt("04") That is it. You can convert a numeric string into integer using Integer.parseInt(String) method, which returns an int type. And then comparison is same as 4 == 4. answered Apr 2, 2013 at 12:36. Rohit Jain. 213k 45 414 532.
