Doubt with logic: How to find out if a time is in the range of hours (independent of the day)

Solution in PHP

In my view, since we're working with schedules, the easiest is to use the DateTime class. of PHP. First, let's define the tests shown in the statement:

$tests = [
  [
    "inicio" => "07:00:00", 
    "final"  => "09:00:00",
    "busca"  => "08:00:00",
    "saida"  => true
  ],[
    "inicio" => "19:00:00", 
    "final"  => "22:00:00",
    "busca"  => "23:00:00",
    "saida"  => false
  ],[
    "inicio" => "18:00:00", 
    "final"  => "03:00:00",
    "busca"  => "01:00:00",
    "saida"  => true
  ],
];

The three tests proposed, consisting of the initial, final, search and expected time. To run the tests, we use a basic loop :

foreach($tests as $test)
{
    // ...
}

First step is to define the time-related objects:

foreach($tests as $test)
{
  $ininio = new DateTime($test["inicio"]);
  $final  = new DateTime($test["final"]);
  $busca  = new DateTime($test["busca"]);
}

As stated in the statement it is clear that it should be independent of the day and that even the interval can start in one day and end in another, as in the case of the third test, we need to do a simple check: final is less than the initial, add at the end an interval of one day.

foreach($tests as $test)
{
  $ininio = new DateTime($test["inicio"]);
  $final  = new DateTime($test["final"]);
  $busca  = new DateTime($test["busca"]);

  if ($final <= $inicio) {
    $final->add(new DateInterval("P1D"));
  }
}

Read more about the class DateInterval in the documentation. This way, if the final time is less than the initial time, it is added 24h in it, and it will be the same time the next day.

The same logic applies to the search time: if it is less than the start time, it should be considered as the next day and therefore added 24 hours as well.

foreach($tests as $test)
{
  $ininio = new DateTime($test["inicio"]);
  $final  = new DateTime($test["final"]);
  $busca  = new DateTime($test["busca"]);

  if ($final <= $inicio) {
    $final->add(new DateInterval("P1D"));
  }

  if ($busca <= $inicio) {
    $busca->add(new DateInterval("P1D"));
  }
}

With this, just check the range:

foreach($tests as $test)
{
  $ininio = new DateTime($test["inicio"]);
  $final  = new DateTime($test["final"]);
  $busca  = new DateTime($test["busca"]);

  if ($final <= $ininio) {
    $final->add(new DateInterval("P1D"));
  }

  if ($busca <= $ininio) {
    $busca->add(new DateInterval("P1D"));
  }

  if ($busca >= $ininio && $busca <= $final) {
    echo "Sim";
  } else {
    echo "Não";
  }

  echo ", esperado " . ($test["saida"] ? "sim" : "não") . PHP_EOL;
}

I added the expected value for each test to the output message for comparison. When executing the code, we will have the output:

Sim, esperado sim
Não, esperado não
Sim, esperado sim

  

See the code working at Repl.it or Ideone .

Well, this case is a bit simpler if you set a time model, if you are using the 24hrs model (the clock starts at 00h and goes until 11h) then you only have to count the hours, let's imagine the following:

There is also the possibility of using the PHP date object itself, as in this topic: link

Yes, in this case, from 07:00 to 09:00,

Se horaUsuario >= horaInicio E horaUsuario <= horaFim

22:00 - 02:00

When I wrote the question, I believed that only with logic would it be possible to solve this pineapple, but I could not. Through Anderson's code I almost solved it, but it lacks some details. The real case is this: there can not be a record in the bank with period overlapping another one, that is, if it has '07: 00: 00 'as the start time and' 09: 00: 00 ', no other could be added period coincided with this period. Therefore '06: 00: 00 'at '10: 00: 00' could not be inserted during the 24 hours of the day, this regardless of the day.

Thanks in advance for your personal support.