如何在PHP中循环几个数字？

I have a question to looping several numbers together.

I've tried to do with two function below:

for ($x=28; $x <= 31; $x++){
    echo $x.'<br>';
}

for($i=1; $i<6; $i++){
    echo $i.'<br>';
}

But I want the script with a single step like:

for(){
  // code
  for(){
     // code
  }
}

And the final result, I try to implement it on the input option:

<option>28</option>
<option>29</option>
<option>30</option>
<option>31</option>
<option>1</option>
<option>2</option>
<option>3</option>
<option>4</option>
<option>5</option>

Hi if need to print only

`<option>28</option>`
    <option>29</option>
    <option>30</option>
    <option>31</option>
    <option>1</option>
    <option>2</option>
    <option>3</option>
    <option>4</option>
    <option>5</option>
    <option>6</option>

then can use below code 
$start = 28;
$count = 31;

for ($x = $start; $x <=$count; $x++){
       echo '<option>'.$x.'</option>';

    if($x==31){
        $count=6;
         $x=0;

    }
}

Increase your first loop by 1 and run the second loop within it

for ($x=28; $x <= 32; $x++)
{

    if($x <= 31)
    {
        echo $x.'<br>';
    }
    else
    {
       for($i=1; $i<6; $i++)
        {
          echo $i.'<br>';
        }
    }

}

If you "... want to make a date in option 1-31, but I cut it from the middle for example 28-31 then 1-27 ... ", you may try with the next example, using just one for loop:

<?php
$start = 28;
$count = 31;

for ($x = 0; $x < $count; $x++){
    $n = ($start + $x <= $count) ? ($start + $x) : ($start + $x - $count); 
    echo  $n.'<br>';
    // or more complex output
    //echo "<option>".$n."</option>";
}
?>