For loop not continuing after first iteration - php

Question

So I've written the for loop as shown below, and for some reason, after the first time through the loop, the loop stops. In this case, I am trying to send the value 4 through, and it fails the if statement that checks for a string "true" from the checkRes function. That is all that function does is return strings. It is my understanding of loops that it will keep going through the loop until the statement at the top is met, or it is exited.
What am I doing wrong below?
// assume $avaliable = 4;
for ($i=$avaliable;10>$i;$i++) {
$check = checkRes($i, $people_no, $booking_date);
if ($check === "true") {
switch($i) {
case 0: $newResTime = "6 PM"; break;
case 1: $newResTime = "6:30 PM"; break;
case 2: $newResTime = "7 PM"; break;
case 3: $newResTime = "7:30 PM"; break;
case 4: $newResTime = "8 PM"; break;
case 5: $newResTime = "8:30 PM"; break;
case 6: $newResTime = "9 PM"; break;
case 7: $newResTime = "9:30 PM"; break;
case 8: $newResTime = "10 PM"; break;
case 9: $newResTime = "10:30 PM"; break;
case 10: $newResTime = "11 PM"; break;
}
// Replace next line with your return from the chatbot
echo "We're sorry, that time isn't avaliable, but a reservation at $newResTime has been made!";
exit;
}
}

Answer 1

You are currently using exit as a "default" solution. But it always terminates execution of the program (since that is what exit does).
Instead do this
switch($i) {
case 0: $newResTime = "6 PM"; break;
case 1: $newResTime = "6:30 PM"; break;
case 2: $newResTime = "7 PM"; break;
case 3: $newResTime = "7:30 PM"; break;
case 4: $newResTime = "8 PM"; break;
case 5: $newResTime = "8:30 PM"; break;
case 6: $newResTime = "9 PM"; break;
case 7: $newResTime = "9:30 PM"; break;
case 8: $newResTime = "10 PM"; break;
case 9: $newResTime = "10:30 PM"; break;
case 10: $newResTime = "11 PM"; break;
default: echo "..."; break;