如何将String转换为int?
"1234" → 1234
如何将String转换为int?
"1234" → 1234
当前回答
如果需要原语,请使用parseInt,否则使用Integer.valueOf()
其他回答
我有一个解决方案,但我不知道它有多有效。但它工作得很好,我认为你可以改进它。另一方面,我用JUnit做了几次测试,哪一步正确。我附上了功能和测试:
static public Integer str2Int(String str) {
Integer result = null;
if (null == str || 0 == str.length()) {
return null;
}
try {
result = Integer.parseInt(str);
}
catch (NumberFormatException e) {
String negativeMode = "";
if(str.indexOf('-') != -1)
negativeMode = "-";
str = str.replaceAll("-", "" );
if (str.indexOf('.') != -1) {
str = str.substring(0, str.indexOf('.'));
if (str.length() == 0) {
return (Integer)0;
}
}
String strNum = str.replaceAll("[^\\d]", "" );
if (0 == strNum.length()) {
return null;
}
result = Integer.parseInt(negativeMode + strNum);
}
return result;
}
使用JUnit进行测试:
@Test
public void testStr2Int() {
assertEquals("is numeric", (Integer)(-5), Helper.str2Int("-5"));
assertEquals("is numeric", (Integer)50, Helper.str2Int("50.00"));
assertEquals("is numeric", (Integer)20, Helper.str2Int("$ 20.90"));
assertEquals("is numeric", (Integer)5, Helper.str2Int(" 5.321"));
assertEquals("is numeric", (Integer)1000, Helper.str2Int("1,000.50"));
assertEquals("is numeric", (Integer)0, Helper.str2Int("0.50"));
assertEquals("is numeric", (Integer)0, Helper.str2Int(".50"));
assertEquals("is numeric", (Integer)0, Helper.str2Int("-.10"));
assertEquals("is numeric", (Integer)Integer.MAX_VALUE, Helper.str2Int(""+Integer.MAX_VALUE));
assertEquals("is numeric", (Integer)Integer.MIN_VALUE, Helper.str2Int(""+Integer.MIN_VALUE));
assertEquals("Not
is numeric", null, Helper.str2Int("czv.,xcvsa"));
/**
* Dynamic test
*/
for(Integer num = 0; num < 1000; num++) {
for(int spaces = 1; spaces < 6; spaces++) {
String numStr = String.format("%0"+spaces+"d", num);
Integer numNeg = num * -1;
assertEquals(numStr + ": is numeric", num, Helper.str2Int(numStr));
assertEquals(numNeg + ": is numeric", numNeg, Helper.str2Int("- " + numStr));
}
}
}
int foo = Integer.parseInt("1234");
确保字符串中没有非数字数据。
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
}
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;
}
}
整数代码
您还可以使用公共静态整数解码(Stringnm)抛出NumberFormatException。
它也适用于底座8和16:
// base 10
Integer.parseInt("12"); // 12 - int
Integer.valueOf("12"); // 12 - Integer
Integer.decode("12"); // 12 - Integer
// base 8
// 10 (0,1,...,7,10,11,12)
Integer.parseInt("12", 8); // 10 - int
Integer.valueOf("12", 8); // 10 - Integer
Integer.decode("012"); // 10 - Integer
// base 16
// 18 (0,1,...,F,10,11,12)
Integer.parseInt("12",16); // 18 - int
Integer.valueOf("12",16); // 18 - Integer
Integer.decode("#12"); // 18 - Integer
Integer.decode("0x12"); // 18 - Integer
Integer.decode("0X12"); // 18 - Integer
// base 2
Integer.parseInt("11",2); // 3 - int
Integer.valueOf("11",2); // 3 - Integer
如果要获取int而不是Integer,可以使用:
取消装箱:int val=Integer.decode(“12”);intValue():Integer.decode(“12”).intValue();