I got problem with my code, can't figure out how to make my script work. I wanted to make on my HTML site, select dropdown list to choose parents for Kid, that imports data from my database, and this works, i can see things listed.

This is code to connect DB and determine $resultimie

<?php
$con=mysqli_connect('localhost','root','','aplikacja_kolonijna');
    if(!$con)
    {
        echo 'Data base not found';
    }

    $resultimie=$con->query("select concat(imie_opiekun,' ',nazwisko_opiekun) as full_name_opiekun from opiekun");


?>

And this is for show variables as options in select, it works. BUT.. check below

Opiekun Prawny:<select id="text_box_od" name="fullname_opiekun" >
  <?php
      while($rows = $resultimie->fetch_assoc())
    {
         $full_name_opiekun = $rows['full_name_opiekun'];
         echo "<option value='.$full_name_opiekun'>$full_name_opiekun</option>";            
    }
  ?>
</select>

But also made form to upload information about "KID" and also includes foreign key about Parent ID, so i made action in php, I won't post it because it's really long, all i can say is that similar form ( insert) worked on every other site i've made, but not this one with this select option.

This is my Action i use for FORM.

<?php

    $con=mysqli_connect('localhost','root','','aplikacja_kolonijna');
    if(!$con)
    {
        echo 'Data base not found';
    }

    $imie_dziecko = $_POST['imie_dziecko'];
    $nazwisko_dziecko = $_POST['nazwisko_dziecko'];
    $pesel_dziecko = $_POST['pesel_dziecko'];
    $data_urodzenia_dziecko = $_POST['data_urodzenia_dziecko'];
    $nr_kontaktowy_dziecko = $_POST['nr_kontaktowy_dziecko'];
    $miejscowosc_zamieszkania_dziecko = $_POST['miejscowosc_zamieszkania_dziecko'];
    $ulica_zamieszkania_dziecko = $_POST['ulica_zamieszkania_dziecko'];
    $kod_pocztowy_dziecko = $_POST['kod_pocztowy_dziecko'];
    $diety_dziecko = $_POST['diety_dziecko'];
    $choroby_dziecko = $_POST['choroby_dziecko'];
    $lekarstwo_dziecko = $_POST['lekarstwo_dziecko'];
    $fullname_opiekun = $_POST['fullname_opiekun'];

    $choose_opiekun = "(SELECT id_opiekun FROM opiekun where concat(imie_opiekun,' ',nazwisko_opiekun) like '%$fullname_opiekun%')";

    $sql = "Insert into 'dziecko' (id_opiekun,imie_dziecko,nazwisko_dziecko,pesel_dziecko,data_urodzenia_dziecko,nr_kontaktowy_dziecko,miejscowosc_zamieszkania_dziecko,ulica_zamieszkania_dziecko,kod_pocztowy_dziecko,diety_dziecko,choroby_dziecko_dziecko,lekarstwa_dziecko) values ('$choose_opiekun','$imie_dziecko','$nazwisko_dziecko','$pesel_dziecko','$data_urodzenia_dziecko','$nr_kontaktowy_dziecko','$miejscowosc_zamieszkania_dziecko','$ulica_zamieszkania_dziecko','$kod_pocztowy_dziecko','$diety_dziecko','$choroby_dziecko','$lekarstwo_dziecko')";
    mysqli_query($con,$sql);

    if(!$sql)       
        echo 'nie dodano wpisu'


?>

<script type="text/javascript">
    alert('Dodano wpis!');
    </script>

<?php
header("refresh:0.1; url=add_dziecko_admin.php");
mysqli_close($con);
?>

There is $chose_opiekun where i declared select to determine parent's (opiekun) ID via Concat of name and surname, in phpmy admin, this select works, and shows ID, so i don't know why this doesnt work in $sql query. There is no Error (even if i change "refresh" to 10). It says like it all works, also there is pop up that data was entered into database... but it doesn't , there is no new entry.

So my question is.. What can i possibly do, to make this form works, it's really important for me.

Thanks for every answer. P.S. I also tried without $choose_opiekun, and paste it directly into insert query, as id_opiekun value.