我希望代码在用户输入字符串而不是整数时捕获错误。您可以看到我尝试了一个try catch块，但仍然无法正常工作。除此之外，其他一切都很完美。我该怎么解决？
输出内容如下：
import datetime as dt
from sklearn import linear_model
df = pd.DataFrame(list, columns=['date', 'value'])
df['date_ordinal'] = pd.to_datetime(df['date']).map(dt.datetime.toordinal)
reg = linear_model.LinearRegression()
reg.fit(df['date_ordinal'].values.reshape(-1, 1), df['value'].values)
reg.coef_
array([0.80959405])
代码如下：
Welcome to the Squares and Cubes table
Enter an integer: five
Error! Invalid integer. Try again.
Enter an integer: -5
Error! Number must be greater than 0
Enter an integer: 101
Error! Number must be less than or equal to 100
Enter an integer: 9
Number Squared Cubed
====== ======= =====
1 1 1
2 4 8
3 9 27
4 16 64
5 25 125
6 36 216
7 49 343
8 64 512
9 81 729
Continue? (y/n): y
Enter an integer: 3
Number Squared Cubed
====== ======= =====
1 1 1
2 4 8
3 9 27
答案 0 :(得分：1)
为了避免引起混淆，我对您的代码做了一些更改并将其整体发布了：
public static void main(String[] args) {
// Welcome the user
System.out.println("Welcome to the Squares and Cubes table");
System.out.println();
Scanner sc = new Scanner(System.in);
String choice = "y";
do {
int integer = Integer.MAX_VALUE;
while (integer == Integer.MAX_VALUE) {
// Get input from the user
System.out.print("Enter an integer: ");
String input = sc.nextLine();
try {
integer = Integer.parseInt(input);
}
catch (NumberFormatException e) {
System.out.println("Error! Invalid integer. Try again.");
}
}
if(integer<0){
System.out.println("Error! Number must be greater than 0");
System.out.print("Enter an integer: ");
integer = sc.nextInt();
}
if(integer>100){
System.out.println("Error! Number must be less than or equal to 100");
System.out.print("Enter an integer: ");
integer = sc.nextInt();
}
// Create a header
String header = "Number " + "Squared " + "Cubed " + "\n"
+ "====== " + "======= " + "===== ";
System.out.println(header);
int square = 0;
int cube = 0;
String row = "";
for (int i = 1; i <= integer; i++)
{
square = i * i;
cube = i * i * i;
row = i + " " + square + " " + cube;
System.out.println(row);
}
// See if the user wants to continue
System.out.print("Continue? (y/n): ");
choice = sc.next();
System.out.println();
} while (!choice.equalsIgnoreCase("n"));
}
这个想法是在循环中创建另一个while并运行它，直到用户传递整数为止。
答案 1 :(得分：0)
Integer.parseInt方法是将String转换为int类型，如果无法将字符串转换为NumberFormatException类型则抛出int。
应该是这样的：
System.out.print("Enter an integer: ");
Scanner sc =new Scanner(System.in);
try {
int integer = Integer.parseInt(sc.nextLine());
} catch (NumberFormatException e) {
System.out.println("Error! Invalid integer. Try again.");
}
答案 2 :(得分：0)
您可以使用此方法仅测试输入的值是否为有效整数。根据此结果，您可以从其他验证开始
public boolean isInt(string input) {
try {
Integer.parseInt(text);
return true;
} catch (NumberFormatException e) {
return false;
}
}
答案 3 :(得分：-2)
使用此getInput(scanner);方法来获取用户的输入。这将处理异常并递归调用自身，直到用户输入数字为止。
public static int getInput(Scanner sc) {
int integer=0;
try {
System.out.print("Enter an integer: ");
integer = Integer.parseInt(sc.nextLine());
}
catch (Exception e) {
System.out.println("Error! Invalid integer. Try again.");
getInput( sc);
}
return integer;
}
对该函数的调用将类似于int integer = getInput(sc);
修改后，您的代码将如下所示，
public class cube2 {
public static int getInput(Scanner sc) {
int integer=0;
try {
System.out.print("Enter an integer: ");
integer = Integer.parseInt(sc.nextLine());
}
catch (Exception e) {
System.out.println("Error! Invalid integer. Try again.");
getInput( sc);
}
return integer;
}
public static void main(String[] args)
{
// Welcome the user
System.out.println("Welcome to the Squares and Cubes table");
System.out.println();
Scanner sc = new Scanner(System.in);
String choice = "y";
do
{
int integer = getInput(sc); // To get the Numeric input from Console
if(integer<0){
System.out.println("Error! Number must be greater than 0");
System.out.print("Enter an integer: ");
integer = sc.nextInt();
}
if(integer>100){
System.out.println("Error! Number must be less than or equal to 100");
System.out.print("Enter an integer: ");
integer = sc.nextInt();
}
// Create a header
String header = "Number " + "Squared " + "Cubed " + "\n"
+ "====== " + "======= " + "===== ";
System.out.println(header);
int square = 0;
int cube = 0;
String row = "";
for (int i = 1; i <= integer; i++)
{
square = i * i;
cube = i * i * i;
row = i + " " + square + " " + cube;
System.out.println(row);
}
// See if the user wants to continue
System.out.print("Continue? (y/n): ");
choice = sc.nextLine();
System.out.println();
}
while (!choice.equalsIgnoreCase("n"));
}
}
在您的代码中choice = sc.next();更改为choice = sc.nextLine();
输出：
Welcome to the Squares and Cubes table
Enter an integer: 9
Number Squared Cubed
====== ======= =====
1 1 1
2 4 8
3 9 27
4 16 64
5 25 125
6 36 216
7 49 343
8 64 512
9 81 729
Continue? (y/n): y
Enter an integer: hi
Error! Invalid integer. Try again.
Enter an integer: hello
Error! Invalid integer. Try again.
Enter an integer: 5
Number Squared Cubed
====== ======= =====
Continue? (y/n): y
Enter an integer: 12
Number Squared Cubed
====== ======= =====
1 1 1
2 4 8
3 9 27
4 16 64
5 25 125
6 36 216
7 49 343
8 64 512
9 81 729
10 100 1000
11 121 1331
12 144 1728
Continue? (y/n):