在我日常的Web应用程序开发中，有许多情况需要从用户那里获取一些数字输入。
然后将这个数字输入传递给应用程序的服务层或DAO层。
由于它是一个数字（整数或浮点数），因此在某个阶段我们需要将其转换为整数，如下面的代码片段所示。
String cost = request.getParameter("cost");
if (cost !=null && !"".equals(cost) ){
Integer intCost = Integer.parseInt(cost);
List<Book> books = bookService . findBooksCheaperThan(intCost);
}
在上述情况下，我需要检查输入是否为空或者没有输入（空白），有时也可能存在非数字输入的可能性，例如blah，test等。
如何处理这种情况的最佳方式是什么？
只需捕获异常并进行正确的异常处理：
if (cost !=null && !"".equals(cost) ){
try {
Integer intCost = Integer.parseInt(cost);
List<Book> books = bookService . findBooksCheaperThan(intCost);
} catch (NumberFormatException e) {
System.out.println("This is not a number");
System.out.println(e.getMessage());
}
}
bookService的调用移出try块，因为这个特定的异常只适用于解析cost字符串。 - Alexander PogrebnyakparseInt(String) 之前像 StringUtils.isNumeric(String) 这样进行检查吗？ - Learner最近版本的Java中的异常并不昂贵，因此避免它们并不重要。使用人们建议的try/catch块；如果您在过程早期（即用户输入后）尽早捕获异常，则后续过程中就不会出现问题（因为它仍然是正确的类型）。
异常曾经比现在昂贵得多；在您知道异常实际上正在引起问题之前，请不要为性能进行优化（在这里它们不会引起问题）。
public class Main {
public static void main(String[] args) {
String number;
while(true){
try{
number = JOptionPane.showInputDialog(null);
if( Main.isNumber(number) )
break;
}catch(NumberFormatException e){
System.out.println(e.getMessage());
}
}
System.out.println("Your number is " + number);
}
public static boolean isNumber(Object o){
boolean isNumber = true;
for( byte b : o.toString().getBytes() ){
char c = (char)b;
if(!Character.isDigit(c))
isNumber = false;
}
return isNumber;
}
}
来自Apache Commons Lang的方法文档（从这里）：
Checks whether the String a valid Java number.
Valid numbers include hexadecimal marked with the 0x qualifier, scientific notation and numbers marked with a type qualifier (e.g. 123L).
Nulland empty String will returnfalse.Parameters:
`str` - the `String` to checkReturns:
`true` if the string is a correctly formatted number
isNumber 来自 java.org.apache.commons.lang3.math.NumberUtils：
public static boolean isNumber(final String str) {
if (StringUtils.isEmpty(str)) {
return false;
}
final char[] chars = str.toCharArray();
int sz = chars.length;
boolean hasExp = false;
boolean hasDecPoint = false;
boolean allowSigns = false;
boolean foundDigit = false;
// deal with any possible sign up front
final int start = (chars[0] == '-') ? 1 : 0;
if (sz > start + 1 && chars[start] == '0' && chars[start + 1] == 'x') {
int i = start + 2;
if (i == sz) {
return false; // str == "0x"
}
// checking hex (it can't be anything else)
for (; i < chars.length; i++) {
if ((chars[i] < '0' || chars[i] > '9')
&& (chars[i] < 'a' || chars[i] > 'f')
&& (chars[i] < 'A' || chars[i] > 'F')) {
return false;
}
}
return true;
}
sz--; // don't want to loop to the last char, check it afterwords
// for type qualifiers
int i = start;
// loop to the next to last char or to the last char if we need another digit to
// make a valid number (e.g. chars[0..5] = "1234E")
while (i < sz || (i < sz + 1 && allowSigns && !foundDigit)) {
if (chars[i] >= '0' && chars[i] <= '9') {
foundDigit = true;
allowSigns = false;
} else if (chars[i] == '.') {
if (hasDecPoint || hasExp) {
// two decimal points or dec in exponent
return false;
}
hasDecPoint = true;
} else if (chars[i] == 'e' || chars[i] == 'E') {
// we've already taken care of hex.
if (hasExp) {
// two E's
return false;
}
if (!foundDigit) {
return false;
}
hasExp = true;
allowSigns = true;
} else if (chars[i] == '+' || chars[i] == '-') {
if (!allowSigns) {
return false;
}
allowSigns = false;
foundDigit = false; // we need a digit after the E
} else {
return false;
}
i++;
}
if (i < chars.length) {
if (chars[i] >= '0' && chars[i] <= '9') {
// no type qualifier, OK
return true;
}
if (chars[i] == 'e' || chars[i] == 'E') {
// can't have an E at the last byte
return false;
}
if (chars[i] == '.') {
if (hasDecPoint || hasExp) {
// two decimal points or dec in exponent
return false;
}
// single trailing decimal point after non-exponent is ok
return foundDigit;
}
if (!allowSigns
&& (chars[i] == 'd'
|| chars[i] == 'D'
|| chars[i] == 'f'
|| chars[i] == 'F')) {
return foundDigit;
}
if (chars[i] == 'l'
|| chars[i] == 'L') {
// not allowing L with an exponent or decimal point
return foundDigit && !hasExp && !hasDecPoint;
}
// last character is illegal
return false;
}
// allowSigns is true iff the val ends in 'E'
// found digit it to make sure weird stuff like '.' and '1E-' doesn't pass
return !allowSigns && foundDigit;
}
[代码采用Apache许可证的第二版]
我建议做两件事情：
判断字符串是整数还是浮点数，并以更长的格式表示。
整数
String cost=Long.MAX_VALUE+"";
if (isNumeric (cost)) // returns false for non numeric
{
BigInteger bi = new BigInteger(cost);
}
public static boolean isNumeric(String str)
{
NumberFormat formatter = NumberFormat.getInstance();
ParsePosition pos = new ParsePosition(0);
formatter.parse(str, pos);
return str.length() == pos.getIndex();
}
String input = "123";
Scanner sc = new Scanner(input);
if (sc.hasNextInt())
System.out.println("an int: " + sc.nextInt());
else {
//handle the bad input
}
尝试将奖品转换为十进制格式...
import java.math.BigDecimal;
import java.math.RoundingMode;
public class Bigdecimal {
public static boolean isEmpty (String st) {
return st == null || st.length() < 1;
}
public static BigDecimal bigDecimalFormat(String Preis){
//MathContext mi = new MathContext(2);
BigDecimal bd = new BigDecimal(0.00);
bd = new BigDecimal(Preis);
return bd.setScale(2, RoundingMode.HALF_UP);
}
public static void main(String[] args) {
String cost = "12.12";
if (!isEmpty(cost) ){
try {
BigDecimal intCost = bigDecimalFormat(cost);
System.out.println(intCost);
List<Book> books = bookService.findBooksCheaperThan(intCost);
} catch (NumberFormatException e) {
System.out.println("This is not a number");
System.out.println(e.getMessage());
}
}
}
}
if (!"".equals(cost));) - Peter Lawrey