如何将String转换为int?

"1234"  →  1234

当前回答

对于Android开发者来说,以下是Kotlin的各种解决方案:

// Throws exception if number has bad form
val result1 = "1234".toInt()
// Will be null if number has bad form
val result2 = "1234"
    .runCatching(String::toInt)
    .getOrNull()
// Will be the given default if number has bad form
val result3 = "1234"
    .runCatching(String::toInt)
    .getOrDefault(0)
// Will be return of the else block if number has bad form
val result4 = "1234"
    .runCatching(String::toInt)
    .getOrElse {
        // some code
        // return an Int
    }

其他回答

正如我在GitHub上写的:

public class StringToInteger {
    public static void main(String[] args) {
        assert parseInt("123") == Integer.parseInt("123");
        assert parseInt("-123") == Integer.parseInt("-123");
        assert parseInt("0123") == Integer.parseInt("0123");
        assert parseInt("+123") == Integer.parseInt("+123");
    }

    /**
     * Parse a string to integer
     *
     * @param s the string
     * @return the integer value represented by the argument in decimal.
     * @throws NumberFormatException if the {@code string} does not contain a parsable integer.
     */
    public static int parseInt(String s) {
        if (s == null) {
            throw new NumberFormatException("null");
        }
        boolean isNegative = s.charAt(0) == '-';
        boolean isPositive = s.charAt(0) == '+';
        int number = 0;
        for (int i = isNegative ? 1 : isPositive ? 1 : 0, length = s.length(); i < length; ++i) {
            if (!Character.isDigit(s.charAt(i))) {
                throw new NumberFormatException("s=" + s);
            }
            number = number * 10 + s.charAt(i) - '0';
        }
        return isNegative ? -number : number;
    }
}

使用Integer.parseInt()并将其放入try。。。catch块来处理任何错误,例如,

private void ConvertToInt(){
    String string = txtString.getText();
    try{
        int integerValue=Integer.parseInt(string);
        System.out.println(integerValue);
    }
    catch(Exception e){
       JOptionPane.showMessageDialog(
         "Error converting string to integer\n" + e.toString,
         "Error",
         JOptionPane.ERROR_MESSAGE);
    }
 }

这是一个完整的程序,所有条件都是正的和负的,不使用库

import java.util.Scanner;


public class StringToInt {

    public static void main(String args[]) {
        String inputString;
        Scanner s = new Scanner(System.in);
        inputString = s.nextLine();

        if (!inputString.matches("([+-]?([0-9]*[.])?[0-9]+)")) {
            System.out.println("Not a Number");
        }
        else {
            Double result2 = getNumber(inputString);
            System.out.println("result = " + result2);
        }
    }


    public static Double getNumber(String number) {
        Double result = 0.0;
        Double beforeDecimal = 0.0;
        Double afterDecimal = 0.0;
        Double afterDecimalCount = 0.0;
        int signBit = 1;
        boolean flag = false;

        int count = number.length();
        if (number.charAt(0) == '-') {
            signBit = -1;
            flag = true;
        }
        else if (number.charAt(0) == '+') {
            flag = true;
        }
        for (int i = 0; i < count; i++) {
            if (flag && i == 0) {
                continue;
            }
            if (afterDecimalCount == 0.0) {
                if (number.charAt(i) - '.' == 0) {
                    afterDecimalCount++;
                }
                else {
                    beforeDecimal = beforeDecimal * 10 + (number.charAt(i) - '0');
                }
            }
            else {
                afterDecimal = afterDecimal * 10 + number.charAt(i) - ('0');
                afterDecimalCount = afterDecimalCount * 10;
            }
        }
        if (afterDecimalCount != 0.0) {
            afterDecimal = afterDecimal / afterDecimalCount;
            result = beforeDecimal + afterDecimal;
        }
        else {
            result = beforeDecimal;
        }
        return result * signBit;
    }
}

也可以从删除所有非数字字符开始,然后解析整数:

String mystr = mystr.replaceAll("[^\\d]", "");
int number = Integer.parseInt(mystr);

但请注意,这只适用于非负数。

int foo = Integer.parseInt("1234");

确保字符串中没有非数字数据。