[英]Check input string to int
I have this method: 我有这种方法:
public static int parseInt(String str) {
if (isValidNumber(str)) {
int sum = 0;
int position = 1;
for (int i = str.length() - 1; i >= 0; i--) {
int number = str.charAt(i) - '0';
sum += number * position;
position = position * 10;
}
return sum;
}
return -1;
}
which converts a string into a integer. 将字符串转换为整数。 And as you can see it is (at the moment) in a if-statement with a method which checks if the input is a valid input for my purpose: 正如您所看到的(当前),它处于if语句中,该方法使用一种检查输入是否对我而言有效的输入:
public static boolean isValidNumber(String str) {
for (int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
if(c >= '0' && c <= '9'){
return true;
}
}
return false;
}
I want the string to be number only (negative and positive) no other is allowed. 我希望字符串仅是数字(负数和正数),不允许其他数字。 At that time a string ie 1a1a
will be converted to a integer which it shouldn't whereas -1
will not be converted. 那时,一个字符串(即1a1a
将被转换为它不应该转换的整数,而-1
将不被转换。 I think you guys understand what I mean. 我想你们明白我的意思。 I don't know how to do that. 我不知道该怎么做。
Please help! 请帮忙!
Try this: 尝试这个:
CODE: 码:
public class validNumbers {
public static void main(String[] args) {
System.out.println(parseInt("345"));
System.out.println(parseInt("-345"));
System.out.println(parseInt("a-345"));
System.out.println(parseInt("1a5b"));
}
public static int parseInt(String str) {
String numberWithoutSign = removeSign(str);
if (isValidNumber(numberWithoutSign)) {
int sum = 0;
int position = 1;
for (int i = numberWithoutSign.length() - 1; i >= 0; i--) {
int number = numberWithoutSign.charAt(i) - '0';
sum += number * position;
position = position * 10;
}
if(isNegative(str)){
return -(sum);
}else{
return sum;
}
}
return -1;
}
/**
* Removes sign in number if exists
*/
public static String removeSign(String number){
if(number.charAt(0) == '+' || number.charAt(0) == '-'){
return number.substring(1);
}else{
return number;
}
}
/**
* Determines if a number is valid
*/
public static boolean isValidNumber(String number) {
for (int i = 0; i < number.length(); i++) {
char c = number.charAt(i);
if(c >= '0' && c <= '9'){
continue;
}else{
return false;
}
}
return true;
}
/**
* Determines if a number is negative or not
*/
public static boolean isNegative(String number){
if(number.charAt(0) == '-'){
return true;
}else{
return false;
}
}
}
OUTPUT: 输出:
345
-345
-1
-1
To check if a string is a real number you can use a method like this: 要检查字符串是否为实数,可以使用如下方法:
public static boolean isInteger(String str) {
try {
Integer.parseInt(str);
return true;
} catch (NumberFormatException nfe) {}
return false;
}
The problem is with your function isValidNumber
. 问题出在你的函数isValidNumber
。 It should return a false on first occurrence of a non numeric value, as follows: 第一次出现非数字值时,应返回false,如下所示:
public static boolean isValidNumber(String str) {
for (int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
if(!(c >= '0' && c <= '9')){
if (i > 0) {
return false;
}
//This will only be invoked when `i == 0` (or less, which is impossible in this for loop), so I don't need to explicitly specify it here, as I have checked for `i > 0` in the above code...
if (c != '-' && c != '+') {
return false;
}
}
}
return true;
}
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