Get value from cell1, cell2 and cell4

With this code you can print the name of the first four records of the result set:

    $i=0; 
    do{
        echo $fila["Nombre"]."<br>"; 
        $i++; 
      } while($i < 5 && $fila = mysqli_fetch_assoc($result));

Consists of reading them within a loop do... while which will end when the counter is less than 5 , that is, when it reaches the record 4 .

Note that I used mysqli_fetch_assoc because fetch_array produces two values (one associative and one numerical) for each column, which does not seem convenient from the point of view of optimization.

Attention to this

Keep in mind that while will iterate over your result set and if you stop it in record number 4, if for some reason you try to recover data again, the results pointer will be where you stopped it.

If you put twice the code of the do ... while of above you will see that in the second part it will print again four records, but starting with the last one, where the pointer remained.

If you want to use your data in several scenarios, you can then retrieve them all in one array and then use that array to read them in those different parts.

For example:

    while ($row = mysqli_fetch_assoc($result)){
         $arrDatos[]=$row;
    }
    /*Puedes cerrar el conjunto de resultados*/
    $result->close();

There is in $arrDatos all results, not dependent on the pointer of the result set, which you can close and use $arrDatos later.

For example, to read the first four you can do it like this:

    $i=0; 
    do{
        echo "-".$arrDatos[$i]["Nombre"].PHP_EOL; 
        $i++; 
      } while($i < 5 && $arrDatos);

Or with a foreach loop:

    $i=0;
    foreach ($arrDatos as $row){
        $i++;
        echo $row["Nombre"].PHP_EOL;
        if ($i==4){ break;}
    }

And if it offers to read all the data:

    foreach ($arrDatos as $row){
        echo $row["Nombre"].PHP_EOL;            
    }

EDITION

If you want to create variables using specific rows, you can access each row by the index of $arrDatos and then by the key, Nombre in this case.

For example:

$nameOne   = $arrDatos[0]["Nombre"]; //Posición 0 o fila 1
$nameTwo   = $arrDatos[1]["Nombre"]; //Posición 1 o fila 2 
$nameThree = $arrDatos[2]["Nombre"]; //Posición 2 o fila 3

Etc.

There three variables were created with the names of the first three people in the array. We are always assuming that that column is called Nombre in your database. If not, you must put the name you have to make it work.

What you need to show your columns is to travel through a loop. When you launch a query with mysqli_query($link,$sql); and you know that you are going to get several rows, just as you have it in your code, it will only show you your first row. Try doing with a loop like this:

while($fila = mysqli_fetch_array($result))
{
    echo $fila['Nombre'].'<br>';
}

The variable $fila['Nombre'] within that loop, will show you all your data from the Name field. If you want to show another field of your database it would be enough to put for example $fila['Apellidos'] .

I hope it has been understood correctly and can serve you.

Greetings.