how is a value returned from within a caught exception loop?

how is a value returned from within a caught exception loop?

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how is a value returned from within a caught exception loop?
Tag : javascript , By : James Lupiani
Date : November 28 2020, 12:01 PM

I wish did fix the issue. Using filter returns an array. So, tempArray, being an array, doesn't have firstname and lastname properties.
For this, I suggest you switching filter() to find() and remove [] in your desconstructing:
let traveler = [{
    timestamp: 'qualia',
    firstname: 'unborn',
    lastname: 'child',
    location: 'null'
    timestamp: 1000,
    firstname: 'Olivia',
    lastname: 'Kirshner',
    location: 'Titan'
    timestamp: 1001,
    firstname: 'James',
    lastname: 'Cole',
    location: 'Emerson Hotel'

// function noerror(splinter = traveler[0]) {
//     return [{firstname, lastname}] = splinter
// }

function restructure(time) {
    // if no array is returned on the lookup, and the value flags as undefined in restructure()
    let missingLink = traveler[0]
    let tempArray = traveler
        .find((item) => item.timestamp === time)
    try {
        if (tempArray.firstname !== 'undefined' || tempArray.firstname !== 'null') { // check does not work
          return { firstname, lastname } = tempArray

        // else {
        //     return [{firstname, lastname}] = missingLink
        // }
    } catch (e) {
        // e instanceof TypeError // boolean error type check
        switch (e.name) {
            case 'TypeError':
                console.error(`Could not complete your request: ${e.message}`);
                return {firstname, lastname} = missingLink

console.log(`firstname: ${firstname}, lastname: ${lastname}`)

console.log(`firstname: ${firstname}, lastname: ${lastname}`)

lastname = 'Redforester' // temporary assignment to the variable, by value, not to the Object
console.log(`firstname: ${firstname}, lastname: ${lastname}`)


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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.

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
        } 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.");

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);

How to NOT INCREMENT in for loop after an exception was caught?

Tag : java , By : Ram
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);
        i++; // added here
    } catch (IllegalArgumentException e) {
        System.out.println("Invalid Input, Try Again!"); // don't throw just print
if(variable == false){...}

In calling method why Exception can be caught without throwing and why subclass of exception cannot be caught without th

Tag : java , By : Feroz
Date : March 29 2020, 07:55 AM
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