How to Do an Integer.Parseint() for a Decimal Number

How to do an Integer.parseInt() for a decimal number?

0.01 is not an integer (whole number), so you of course can't parse it as one. Use Double.parseDouble or Float.parseFloat instead.

Integer.parseInt and string format with decimal number

From your description I understood that you want to print an int. You can code it like this:

    public static void main(final String[] args) throws ParseException {
            NumberFormat formatter = NumberFormat.getInstance();
            formatter.setParseIntegerOnly(true);
            System.out.println("Test");
            final String nombre = "3.0";
            int entier;
            entier=formatter.parse(nombre).intValue();
            System.out.println("result :" + entier);
        }

NumberFormat will do the job

How to convert String to decimal using Integer.parseInt(String, int)

First argumet of Integer.parseint function is a string which express a number, and second argumet is a base (or radix).

"0x16" is made of two parts, "0x" shows that the base is 16 and "16" shows the number in hexadecimal.

If you want to get integer value of "0x16", the code will be

int aValue;
aValue = Integer.parseint("16",16);

and then, you can show the value in decimal by following code

printf("%d" , aValue);

Points are
1. Binary, decimal and hexadecimal are formats to express integer values by strings.
2. "0x16" and "0b10" are combination of 'base' and 'value'(in string).
3. To convert combination string into integer, use Inter.parseint with 'string' as first argument and 'base' as second argument. (0x means 'base is 16', 0b means 'base is 2' etc.)
Ex:
"0x16" : parseint("16",16);
"0b10" : parseint("10",2);

How can I convert a decimal string value to int in java

In kotlin you can do this as

val str = "3.0"
val result = str.toDouble().toInt()

Problems with JavaScript parseInt() decimal string

If the input string is "14,82" and you want the value 14.82, you'll need to convert the , to a . and then use parseFloat:

var total_amount_int = parseFloat(
        total_amount_string.replace(/,/g, ".")
    ).toFixed(2);

parseInt will only parse the leading part of the string that defines a whole number ("int" = "integer" = "whole number"), so it stops at the ,. parseFloat will parse a decimal number, but only understands . as the decimal point, not ,, even in locales where , is the decimal point.

Integer.parseInt converts everything to zero

You pass invalid input

As the comments indicate, you add a newline to the text being parsed as an integer, thereby adulterating the input:

txt = txt + "\n";  

So you are violating the contract of Integer.parse method. To quote the Javadoc:

… The characters in the string must all be decimal digits…

Your attempt at parsing is throwing an NumberFormatException because you passed faulty input to that method.

int good = Integer.parseInt( "42" ) ;
int bad = Integer.parseInt( "666" + "\n" ) ; // Throws a `NumberFormatException` because of the appended newline.

See that code run live at IdeOne.com.

You asked:

Tell me how to solve this problem.

Pass valid input to the method, as documented: Text with only digits, and optional -/+ in front.

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