如何将String转换为int?
"1234" → 1234
如何将String转换为int?
"1234" → 1234
当前回答
String myString = "1234";
int foo = Integer.parseInt(myString);
如果您查看Java文档,您会注意到“陷阱”是此函数可以引发NumberFormatException,您可以处理该异常:
int foo;
try {
foo = Integer.parseInt(myString);
}
catch (NumberFormatException e) {
foo = 0;
}
(此处理方法默认将格式错误的数字设置为0,但如果您愿意,可以执行其他操作。)
或者,您可以使用Guava库中的Ints方法,该方法与Java 8的Optional相结合,为将字符串转换为int提供了一种强大而简洁的方法:
import com.google.common.primitives.Ints;
int foo = Optional.ofNullable(myString)
.map(Ints::tryParse)
.orElse(0)
其他回答
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);
}
}
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());
}
如上所述,Apache Commons的NumberUtils可以做到这一点。如果无法将字符串转换为int,则返回0。
您还可以定义自己的默认值:
NumberUtils.toInt(String str, int defaultValue)
例子:
NumberUtils.toInt("3244", 1) = 3244
NumberUtils.toInt("", 1) = 1
NumberUtils.toInt(null, 5) = 5
NumberUtils.toInt("Hi", 6) = 6
NumberUtils.toInt(" 32 ", 1) = 1 // Space in numbers are not allowed
NumberUtils.toInt(StringUtils.trimToEmpty(" 32 ", 1)) = 32;
import java.util.*;
public class strToint {
public static void main(String[] args) {
String str = "123";
byte barr[] = str.getBytes();
System.out.println(Arrays.toString(barr));
int result = 0;
for(int i = 0; i < barr.length; i++) {
//System.out.print(barr[i]+" ");
int ii = barr[i];
char a = (char) ii;
int no = Character.getNumericValue(a);
result = result * 10 + no;
System.out.println(result);
}
System.out.println("result:"+result);
}
}
嗯,需要考虑的一个非常重要的点是Integer解析器抛出了Javadoc中所述的NumberFormatException。
int foo;
String StringThatCouldBeANumberOrNot = "26263Hello"; //will throw exception
String StringThatCouldBeANumberOrNot2 = "26263"; //will not throw exception
try {
foo = Integer.parseInt(StringThatCouldBeANumberOrNot);
} catch (NumberFormatException e) {
//Will Throw exception!
//do something! anything to handle the exception.
}
try {
foo = Integer.parseInt(StringThatCouldBeANumberOrNot2);
} catch (NumberFormatException e) {
//No problem this time, but still it is good practice to care about exceptions.
//Never trust user input :)
//Do something! Anything to handle the exception.
}
在尝试从拆分参数中获取整数值或动态解析某些内容时,处理此异常非常重要。