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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 ...
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.
Jan 30, 2011 · String s = String.valueOf(i); String s = Integer.toString(i); Another more concise way is: String s = "" + i; See it working online: ideone. This is particularly useful if the reason you are converting the integer to a string is in order to concatenate it to another string, as it means you can omit the explicit conversion:
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.
Nov 7, 2013 · Here are all the different versions: a) Convert an Integer to a String. Integer one = Integer.valueOf(1); String oneAsString = one.toString(); b) Convert an int to a String. int one = 1; String oneAsString = String.valueOf(one); c) Convert a String to an Integer. String oneAsString = "1";
Nov 5, 2010 · If you say String.valueOf (i), Java converts the integer to a string and returns the result. If you say ""+i, Java creates a StringBuilder object, appends an empty string to it, converts the integer to a string, appends this to the StringBuilder, then converts the StringBuilder to a String. That's a lot of extra steps.
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.
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";
Sep 24, 2012 · try. Integer.parseInt(s); return true; } catch (NumberFormatException ex) return false; If you want to see if something is an Integer object (and hence wraps an int): return o instanceof Integer; Using == to compare doubles (as in your isDoubleInt() method) is dangerous because of loss of precision.
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);
