如何将String转换为int?
"1234" → 1234
如何将String转换为int?
"1234" → 1234
当前回答
手动执行:
public static int strToInt(String str){
int i = 0;
int num = 0;
boolean isNeg = false;
// Check for negative sign; if it's there, set the isNeg flag
if (str.charAt(0) == '-') {
isNeg = true;
i = 1;
}
// Process each character of the string;
while( i < str.length()) {
num *= 10;
num += str.charAt(i++) - '0'; // Minus the ASCII code of '0' to get the value of the charAt(i++).
}
if (isNeg)
num = -num;
return num;
}
其他回答
可以通过七种方式实现:
import com.google.common.primitives.Ints;
import org.apache.commons.lang.math.NumberUtils;
String number = "999";
Ints.tryParse:int result=Ints.tryParse(数字);NumberUtils.createInteger:整数结果=NumberUtils.createInteger(数字);应用到内部的数字:int result=NumberUtils.toInt(数字);整数值:整数结果=Integer.valueOf(数字);整数.分析整数:int result=Integer.parseInt(数字);整数代码:int result=Integer.decode(数字);整数.分析未签名:int result=Integer.parseUnsignedInt(数字);
将字符串转换为int比仅转换数字更复杂。您已经考虑了以下问题:
字符串是否只包含数字0-9?字符串之前或之后的-/+怎么了?这是可能的吗(指会计数字)?MAX_-/MIN_INFINITY怎么了?如果字符串为99999999999999999999,会发生什么?机器可以将此字符串视为int吗?
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
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。
Google Guava有tryParse(String),如果无法解析字符串,则返回null,例如:
Integer fooInt = Ints.tryParse(fooString);
if (fooInt != null) {
...
}