如何将String转换为int?
"1234" → 1234
如何将String转换为int?
"1234" → 1234
当前回答
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;
}
}
其他回答
另一种解决方案是使用Apache Commons的NumberUtils:
int num = NumberUtils.toInt("1234");
Apache实用程序很好,因为如果字符串是无效的数字格式,则始终返回0。因此,节省了try-catch块。
Apache NumberUtils API 3.4版
除了前面的答案,我还想添加几个功能。以下是使用时的结果:
public static void main(String[] args) {
System.out.println(parseIntOrDefault("123", 0)); // 123
System.out.println(parseIntOrDefault("aaa", 0)); // 0
System.out.println(parseIntOrDefault("aaa456", 3, 0)); // 456
System.out.println(parseIntOrDefault("aaa789bbb", 3, 6, 0)); // 789
}
实施:
public static int parseIntOrDefault(String value, int defaultValue) {
int result = defaultValue;
try {
result = Integer.parseInt(value);
}
catch (Exception e) {
}
return result;
}
public static int parseIntOrDefault(String value, int beginIndex, int defaultValue) {
int result = defaultValue;
try {
String stringValue = value.substring(beginIndex);
result = Integer.parseInt(stringValue);
}
catch (Exception e) {
}
return result;
}
public static int parseIntOrDefault(String value, int beginIndex, int endIndex, int defaultValue) {
int result = defaultValue;
try {
String stringValue = value.substring(beginIndex, endIndex);
result = Integer.parseInt(stringValue);
}
catch (Exception e) {
}
return result;
}
如果需要原语,请使用parseInt,否则使用Integer.valueOf()
自定义算法:
public static int toInt(String value) {
int output = 0;
boolean isFirstCharacter = true;
boolean isNegativeNumber = false;
byte bytes[] = value.getBytes();
for (int i = 0; i < bytes.length; i++) {
char c = (char) bytes[i];
if (!Character.isDigit(c)) {
isNegativeNumber = (c == '-');
if (!(isFirstCharacter && (isNegativeNumber || c == '+'))) {
throw new NumberFormatException("For input string \"" + value + "\"");
}
} else {
int number = Character.getNumericValue(c);
output = output * 10 + number;
}
isFirstCharacter = false;
}
if (isNegativeNumber)
output *= -1;
return output;
}
另一种解决方案:
(使用string charAt方法,而不是将字符串转换为字节数组)
public static int toInt(String value) {
int output = 0;
boolean isFirstCharacter = true;
boolean isNegativeNumber = false;
for (int i = 0; i < value.length(); i++) {
char c = value.charAt(i);
if (!Character.isDigit(c)) {
isNegativeNumber = (c == '-');
if (!(isFirstCharacter && (isNegativeNumber || c == '+'))) {
throw new NumberFormatException("For input string \"" + value + "\"");
}
} else {
int number = Character.getNumericValue(c);
output = output * 10 + number;
}
isFirstCharacter = false;
}
if (isNegativeNumber)
output *= -1;
return output;
}
示例:
int number1 = toInt("20");
int number2 = toInt("-20");
int number3 = toInt("+20");
System.out.println("Numbers = " + number1 + ", " + number2 + ", " + number3);
try {
toInt("20 Hadi");
} catch (NumberFormatException e) {
System.out.println("Error: " + e.getMessage());
}
您可以有自己的实现,例如:
public class NumericStringToInt {
public static void main(String[] args) {
String str = "123459";
int num = stringToNumber(str);
System.out.println("Number of " + str + " is: " + num);
}
private static int stringToNumber(String str) {
int num = 0;
int i = 0;
while (i < str.length()) {
char ch = str.charAt(i);
if (ch < 48 || ch > 57)
throw new NumberFormatException("" + ch);
num = num * 10 + Character.getNumericValue(ch);
i++;
}
return num;
}
}