[英]This program throws an exception error when the user inputs a letter or word instead of the expected int (java)
[英]How to catch an error if user inputs a String instead of Int (JOptionPane)
int [] numbers = {1,2,3,4};
Random random = new Random();
int totalGood=0;
int totalFalse=0;
String titl = "title1";
String titl2 = "Question";
String titl3 = "Error!";
boolean ages = true;
String name = JOptionPane.showInputDialog(null,"Welcome Please enter your name:",titl,JOptionPane.PLAIN_MESSAGE,new ImageIcon(image0), null,"").toString();
while(ages == true){
int age = Integer.parseInt(JOptionPane.showInputDialog(null,"Welcome" + " " + name +"!" + " " + "How old are you?",titl2,3));
if(age <=28){
JOptionPane.showMessageDialog(null,String.format("Text" + " " + "Your age:" + " " +age,titl,2));
}else if(age >=29 && age <=40){
JOptionPane.showMessageDialog(null,String.format("Text" + " " + "Your age:"+ " "+age,titl,2));
}else if(age>=41){
JOptionPane.showMessageDialog(null,String.format("Text " + " " + "Your age:" + " "+age,titl,2));
}else{
// the part where I get stuck!
// What to write here to catch an error and return the user to again input int age if he types String instead?
continue;
}
}
我正在嘗試驗證用戶輸入以僅允許 Int 並且如果他鍵入字符串,他將收到錯誤,並且將再次彈出 window 年齡。
我嘗試了一切來捕捉錯誤,但我一直在失敗。 我無法使用age=Integer.parseInt(age)
或.hasNextInt
或其他類似的東西來完成此操作。 他們總是給我一個錯誤。
我看到了很多關於如何使用普通 Scanner 和system.println
的教程,但我無法弄清楚 JOptionPane 的分辨率。
我試過嘗試/捕捉,但我不明白,它永遠不會奏效。
你能幫我么? 我是 Java 的新手,我將不勝感激。
編輯:我修好了! 非常感謝安德魯 Vershinin!
while(ages == true){
Pattern AGE = Pattern.compile("[1-9][0-9]*");
String ageInput = JOptionPane.showInputDialog(null,"Welcome" + " " + name +"!" + " " + "How old are you?",titl2,3);
if(!AGE.matcher(ageInput).matches()){
JOptionPane.showMessageDialog(null,String.format("Please enter only numbers! :)"));
continue;
}
int age = Integer.parseInt(ageInput);
if(age <=28){
JOptionPane.showMessageDialog(null,String.format("Text" + " " + "Your age:" + " " +age,titl,2));
}else if(age >=29 && age <=40){
JOptionPane.showMessageDialog(null,String.format("Text" + " " + "Your age:"+ " "+age,titl,2));
}else if(age>=41){
JOptionPane.showMessageDialog(null,String.format("Text" + " " + "Your age:" + " "+age,titl,2));
}
使用try/catch
是一種有效的方法,但我建議使用正則表達式,通過java.util.regex.Pattern
實例表示: Pattern AGE = Pattern.compile("[1-9][0-9]*");
,可以翻譯為“從1到9的數字,后跟任意數量的任意數字”,因此它不能為零,也不能包含前導零。
像這樣使用它:
String ageInput = JOptionPane.showInputDialog(null,
"Welcome" + " " + name +"!" + " " + "How old are you?",titl2,3)
if (!AGE.matcher(ageInput).matches()) {
// handle invalid input
}
int age = Integer.parseInt(ageInput); // no exception there
Java 內置函數:
public static boolean isNumeric(String str){
for (int i = str.length();--i>=0;){
if (!Character.isDigit(str.charAt(i))){
return false;
}
}
return true;
}
正則表達式,最快:
public static boolean isInteger(String str) {
Pattern pattern = Pattern.compile("^[-\\+]?[\\d]*$");
return pattern.matcher(str).matches();
}
使用ASCII碼:
public static boolean isNumeric(String str){
for(int i=str.length();--i>=0;){
int chr=str.charAt(i);
if(chr<48 || chr>57)
return false;
}
return true;
}
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