我通常使用以下习语来检查String是否可以转换为整数。
public boolean isInteger( String input ) {
try {
Integer.parseInt( input );
return true;
}
catch( Exception e ) {
return false;
}
}
是我的错觉，还是你觉得这有点粗俗?还有什么更好的办法吗?
看看我的回答(带有基准，基于CodingWithSpike之前的回答)，看看为什么我改变了立场，接受了Jonas Klemming对这个问题的回答。我认为大多数人都会使用这个原始代码，因为它实现起来更快，也更易于维护，但是当提供非整数数据时，它就慢了几个数量级。
当前回答
I recently (today) needed to figure out a quick way to do this and of course I was going to use the exception approach for ease when the monkey on the shoulder (conscience) woke up so it took me down this old familiar rabbit hole; no exceptions are not that much more expensive in fact sometimes exceptions are faster (old AIX multiprocessor systems) but regardless it’s to elegant so I did something that the younger me never did and to my amazement nobody here did either (apologize if someone did and I missed it I honestly did not find) : so what did I think we all missed; taking a look at how the JRE implemented it, yes they threw an exception but we can always skip that part.
10年前年轻的我可能会觉得这有失他的身份，但话又说回来，他是一个大嘴巴的炫耀者，性情不好，有一种神的情结，所以就是这样。
我把这些放在这里，是为了方便将来来这里的人。以下是我的发现:
public static int parseInt(String s, int radix) throws NumberFormatException
{
/*
* WARNING: This method may be invoked early during VM initialization
* before IntegerCache is initialized. Care must be taken to not use
* the valueOf method.
*/
if (s == null) {
throw new NumberFormatException("null");
}
if (radix < Character.MIN_RADIX) {
throw new NumberFormatException("radix " + radix +
" less than Character.MIN_RADIX");
}
if (radix > Character.MAX_RADIX) {
throw new NumberFormatException("radix " + radix +
" greater than Character.MAX_RADIX");
}
int result = 0;
boolean negative = false;
int i = 0, len = s.length();
int limit = -Integer.MAX_VALUE;
int multmin;
int digit;
if (len > 0) {
char firstChar = s.charAt(0);
if (firstChar < '0') { // Possible leading "+" or "-"
if (firstChar == '-') {
negative = true;
limit = Integer.MIN_VALUE;
} else if (firstChar != '+')
throw NumberFormatException.forInputString(s);
if (len == 1) // Cannot have lone "+" or "-"
throw NumberFormatException.forInputString(s);
i++;
}
multmin = limit / radix;
while (i < len) {
// Accumulating negatively avoids surprises near MAX_VALUE
digit = Character.digit(s.charAt(i++),radix);
if (digit < 0) {
throw NumberFormatException.forInputString(s);
}
if (result < multmin) {
throw NumberFormatException.forInputString(s);
}
result *= radix;
if (result < limit + digit) {
throw NumberFormatException.forInputString(s);
}
result -= digit;
}
} else {
throw NumberFormatException.forInputString(s);
}
return negative ? result : -result;
}
其他回答
另一个选择:
private boolean isNumber(String s) {
boolean isNumber = true;
for (char c : s.toCharArray()) {
isNumber = isNumber && Character.isDigit(c);
}
return isNumber;
}
我不喜欢方法与regex，因为regex不能检查范围(整数。MIN_VALUE Integer.MAX_VALUE)。
如果你希望在大多数情况下使用int值，而不是int值是不常见的，那么我建议使用Integer。valueOf或Integer。parseInt与NumberFormatException捕获。这种方法的优点-你的代码有很好的可读性:
public static boolean isInt(String s) {
try {
Integer.parseInt(s);
return true;
} catch (NumberFormatException nfe) {
return false;
}
}
如果你需要检查String是否为整数，并且关心性能，那么最好的方法是使用java jdk实现的integer。parseInt，但很少修改(用return false替换throw):
该功能性能良好，保证正确的结果:
public static boolean isInt(String s) {
int radix = 10;
if (s == null) {
return false;
}
if (radix < Character.MIN_RADIX) {
return false;
}
if (radix > Character.MAX_RADIX) {
return false;
}
int result = 0;
boolean negative = false;
int i = 0, len = s.length();
int limit = -Integer.MAX_VALUE;
int multmin;
int digit;
if (len > 0) {
char firstChar = s.charAt(0);
if (firstChar < '0') { // Possible leading "+" or "-"
if (firstChar == '-') {
negative = true;
limit = Integer.MIN_VALUE;
} else if (firstChar != '+')
return false;
if (len == 1) // Cannot have lone "+" or "-"
return false;
i++;
}
multmin = limit / radix;
while (i < len) {
// Accumulating negatively avoids surprises near MAX_VALUE
digit = Character.digit(s.charAt(i++), radix);
if (digit < 0) {
return false;
}
if (result < multmin) {
return false;
}
result *= radix;
if (result < limit + digit) {
return false;
}
result -= digit;
}
} else {
return false;
}
return true;
}
我在这里看到了很多答案，但他们中的大多数都能够确定字符串是否为数字，但他们无法检查数字是否在整数范围内…
因此我的目的是这样的:
public static boolean isInteger(String str) {
if (str == null || str.isEmpty()) {
return false;
}
try {
long value = Long.valueOf(str);
return value >= -2147483648 && value <= 2147483647;
} catch (Exception ex) {
return false;
}
}
这是对Jonas代码的修改，用于检查字符串是否在转换为整数的范围内。
public static boolean isInteger(String str) {
if (str == null) {
return false;
}
int length = str.length();
int i = 0;
// set the length and value for highest positive int or lowest negative int
int maxlength = 10;
String maxnum = String.valueOf(Integer.MAX_VALUE);
if (str.charAt(0) == '-') {
maxlength = 11;
i = 1;
maxnum = String.valueOf(Integer.MIN_VALUE);
}
// verify digit length does not exceed int range
if (length > maxlength) {
return false;
}
// verify that all characters are numbers
if (maxlength == 11 && length == 1) {
return false;
}
for (int num = i; num < length; num++) {
char c = str.charAt(num);
if (c < '0' || c > '9') {
return false;
}
}
// verify that number value is within int range
if (length == maxlength) {
for (; i < length; i++) {
if (str.charAt(i) < maxnum.charAt(i)) {
return true;
}
else if (str.charAt(i) > maxnum.charAt(i)) {
return false;
}
}
}
return true;
}
Integer.valueOf(string);
大部分时间对我都管用!
