Xtykutl

I been working on a site to add, edit, delete, and see the data from a few different tables in mysql using php, so far I been able to add data to them with dropdown menus, but to update is causing me a bit of a problem. It worked before I added the dropdown, but decided to add it to limit the user to use data that exist and not input a Carnet that doesn't exist or does not follow the format, then I added the dropdown.

<form method="post" class="text-center">

 <select name="alumno" class="form-control mx-auto" style="width:auto;" required>

    <option value="" disabled selected>Seleccione un Alumno</option>

    <?php

    $cnn=mysqli_connect('localhost', 'root', 'CC2CRO','control_de_notas')

        or die('No se pudo conectar');  



    $sqlcg = "SELECT Carnet, Nombres, Apellidos FROM datos";

    $result = mysqli_query($cnn, $sqlcg);

            while ($row = mysqli_fetch_array($result))

        {

            echo "<option value='".$row['Carnet']."'>".$row['Nombres']." ".$row['Apellidos']."</option>";       

        }       

        ?>

    </select>

<button type="submit" name="buscar">Actualizar</button><br>

</form>

This dropdown works for another form I have to add information from the table datos into another table that has the information of 3 different tables, for example the same information from the table datos, the Carnet is the value I take in another form to input into a bigger table that has codes from smaller tables to join them together. But what I can't seem to fix is that when I want to edit the table it doesn't show any data even if the Carnet exist in the table

<?php

if (isset($_POST["buscar"]))

{   

    $cnn=mysqli_connect('localhost', 'root', 'CC2CRO','control_de_notas')

        or die('No se pudo conectar');





    $cod=strtoupper($_POST["alumno"]);

    $sql2 = "SELECT Carnet,Apellidos,Nombres,Direccion FROM datos WHERE Carnet='".$cod."'";

    $contar=mysqli_num_rows(mysqli_query($cnn,$sql2));

    if ($contar!=0)

    {

    $result2 = mysqli_query($cnn,$sql2);

    echo "<form name='mostrar' method='post' class='text-center'>";

    echo "<div align='center' class='table-responsive-sm'><table border='3' class='table table-sm table-hover' style='width:auto'>

             <tr>

                <td class='table-secondary text-center' style='font-size:110%;'><b>Codigo</b></td>

                <td class='table-secondary text-center' style='font-size:110%;'><b>Nombre</b></td>  

                <td class='table-secondary text-center' style='font-size:110%;'><b>Direccion</b></td>

              </tr>";

        $row2 = mysqli_fetch_row($result2);         

              echo "<tr>

              <td class='table-info><input type='text' name='mcodigo' value='".$row2[0]."' size='12' placeholder='Carnet' readonly/></td>

              <td class='table-info><input type='text' name='mnombre' value='".$row2[2]." ".$row2[1]."' size='50' placeholder='Digitar Nombre' Maxlength='50' readonly></td>

              <td class='table-info><input type='text' name='dir' value='".$row2[4]."' size='75' placeholder='Digitar Nombre' Maxlength='75' required/></td>

              </tr></table>

              <button type='submit' name='actualizar'>Actualizar</button>

              </form></div>"; 

    }

    else

    {

        echo ("<SCRIPT LANGUAGE='JavaScript'>window.alert('Problema con el Sistema / Registro NO encontrado')</SCRIPT>");   

    }   

}



if (isset($_POST["actualizar"]))

{

    $cod=strtoupper($_POST["mcodigo"]);

    $nom=trim(ucwords($_POST["dir"]));

    $sql3="update datos set Direccion='$nom' where Codigo ='$cod'";

    $mod = mysqli_query($cnn,$sql3);

    if ($mod)

    {

        echo ("<SCRIPT LANGUAGE='JavaScript'>window.alert('Cambio Exitoso !!!!')</SCRIPT>");

    }

    else

    {

        echo ("<SCRIPT LANGUAGE='JavaScript'>window.alert('NO se pudo realizar la actualizacion')</SCRIPT>");

    }

}

?>

If I select a entry that exist in the dropdown it doesn't show when I click

<button type='submit' name='actualizar'>Actualizar</button>

it did worked when I didn't have the dropdown and instead had an input

<input type="text" name="alumno" size="15" placeholder="Carnet a Modificar"  Maxlength="8" required autofocus />

The same issue happens to my other tables when I try to update, my most concerning one is the table that I have 3 dropdown menus of the different tables, as I mentioned it worked when I had the input but I want to make it so you can select information that already exists, as for this table and the others I'm not editing the primary key or i the case of the big table I'm not editing the foreign keys.

For the big table to show data I use 2 dropdown to select 2 different cells of the table to get 1 result in common and run the query same as above

$sql2 = "SELECT Carnet,Materia,Docente,Laboratorio,Parcial,Promedio FROM datos WHERE Carnet='".$_POST['alumno']."' AND Materia='".$_POST['materia']."'";

This code is similar to the one above the only difference between this is that the first 2 code blocks are to update 1 table and this last code block is to update a table but I have to select data from 2 cells and I use 2 dropdown menus for it.