Resume loop after caught exception
Tag : java , By : fstender
Date : March 29 2020, 07:55 AM
should help you out You left menuNumber equal to 4, which is the termination condition of your loop. Of course your loop will end.
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Trouble continuing loop when exception caught
Tag : java , By : jaredsmiller
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , The problem is that the Scanner class does not clear the input if an exception occurs, so it keeps feeding itself with a letter, if one was typed. Also, the payrate variable does not need to be used for loop control. String value;
while (true) { // Repeat until a valid payrate is entered
try {
System.out.print("Enter payrate: "); // Ask for payrate
value = input.nextLine(); // Get payrate as string
payrate = Float.parseFloat(value); // Convert the string to float
if (payrate > 0) { // Valid payrate number (> 0), done
break;
} else { // Invalid payrate number (<= 0), repeat
System.out.println("Payrate must be > 0.");
}
} catch (NumberFormatException e) { // Payrate not a number, repeat
System.out.println("Payrate must be a number.");
}
}
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Exception caught in try-catch, returned in ajax error state not success
Tag : php , By : n1ckless_id
Date : March 29 2020, 07:55 AM
should help you out I see some solutions: [GOOD] Disable connect warnings and throw custom exception, to example https://stackoverflow.com/a/14049169/1559720 [BAD] Bufferize output using ob_start @$mysql = new mysqli($host, $user, $password, $database);
if ($mysql->connect_errno) {
throw new \Exception($mysql->connect_error, $mysql->connect_errno);
}
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How to NOT INCREMENT in for loop after an exception was caught?
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Instead of incrementing your i in the third part of the for loop with the unary ++ operator, just do it at the end of the try block: for (int i = 0; i < arrLength;) { // removed here
try {
System.out.println("Enter an issuer ID# (6 digits) for element #" + i);
String issuerId = scanner.next();
System.out.println("Enter an account # (9 digits) for element #" + i);
String accountNum = scanner.next();
CreditCardNumber obj = new CreditCardNumber(issuerId, accountNum);
obj.changeId(issuerId);
i++; // added here
} catch (IllegalArgumentException e) {
System.out.println("Invalid Input, Try Again!"); // don't throw just print
}
}
if(variable == false){...}
if(!variable){...}
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In calling method why Exception can be caught without throwing and why subclass of exception cannot be caught without th
Date : March 29 2020, 07:55 AM
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