Primärschlüssel als Dropdown-Liste als Fremdschlüssel in neuer Tabelle verwenden

Question

Ich versuche, den in einer Dropdown-Liste ausgewählten Primärschlüssel als Fremdschlüssel für eine neue Tabelle einzufügen. Ich benutze PHP und MySQL.

Ich bekomme immer notice:undefined index. Ich habe bereits 3 Tabellen erstellt (Typ, Marke und Modell) Modell Tabelle besteht aus Fremdschlüssel aus dem Typ und Markenmodell.

HTML/PHP, dass ich die Daten einfügen verwendet:

<div class="modal fade" id="addVehicleModel" role="dialog"> 
    <div class="modal-dialog"> 
    <div class="modal-content"> 
     <div class="panel panel-danger"> 
     <div class="panel-heading"> 
      <h4>Add New Vehicle Model</h4> 
     </div> 
     <div class="panel-body"> 
      <form role="form-horizontal" action="addVehicleModel.php" method="post"> 
      <div class="form-group"> 
       <label>Vehicle Model:</label> 
       <input class="form-control" type="text" id="vehicleModel" name="vehicleModel" placeholder="Enter New Vehicle Model" /> 
      </div> 
      <div class="form-group"> 
       <label for="vehicleType">Vehicle Type:</label> 
       <div> 
       <select class="form-control" onchange="change_vehicleType()" name="vehicleType" id="vehicleTypeID"> 
        <option value="">Select Vehicle Type</option> 
        <?php 
        $res=mysqli_query($link,"Select * from vehicleType"); 
        while ($row=mysqli_fetch_array($res)) { ?> 
        <option value=<?php echo $row[ 'id_vehicleType'];?>> 
         <?php echo $row['vehicle_Type'];?> 
        </option> 
        <?php 
        } ?> 
       </select> 
       </div> 
      </div> 
      <div class="form-group"> 
       <label for="vehicleType">Vehicle Brand:</label> 
       <div> 
       <select class="form-control" onchange="change_vehicleBrand()" name="vehicleBrand" id="vehicleBrandID"> 
        <option value="">Select Vehicle Brand</option> 
        <?php 
        $res=mysqli_query($link,"Select * from vehicleBrand"); 
        while ($row=mysqli_fetch_array($res)) { ?> 
        <option value="<?php echo $row['id_vehicleBrand'];?>"> 
         <?php echo $row['vehicle_Brand'];?> 
        </option> 
        <?php 
        } ?> 
       </select> 
       </div> 
      </div> 
      <br> 
      <button type="submit" class="btn btn-danger">Submit</button> 
      <button type="submit" class="btn btn-primary" data-dismiss="modal">Close</button> 
      </form> 
     </div> 
     <!--class panel body end--> 
     </div> 
     <!--class panel danger end--> 
    </div> 
    <!--class modal content end--> 
    </div> 
    <!--class modal dialog end--> 
</div> 
<!--class modal ADD VEHICLE MODEL end--> 

Php Aktion, die ich verwendet:

<?php 
$mysqli = new mysqli("localhost", "root", "root", "vms"); 
if ($mysqli === false) { 
    die("ERROR: Could not connect. " . $mysqli->connect_error); 
} 
$vehicleModel = $mysqli->real_escape_string($_REQUEST['vehicleModel']); 
$vehicleTypeID = $mysqli->real_escape_string($_REQUEST['id_vehicleType']); 
$vehicleBrandID = $mysqli->real_escape_string($_REQUEST['id_vehicleBrand']); 
$sql   = "INSERT INTO vehiclemodel (vehicle_Model, id_FKvehicleType,  id_FKvehicleBrand) VALUES ('$vehicleModel', '$id_vehicleType', '$id_vehicleBrand')"; 
if ($mysqli->query($sql) === true) { 
    echo "Records inserted successfully."; 
} else { 
    echo "ERROR: Could not able to execute $sql. " . $mysqli->error; 
} 
$mysqli->close(); 
?> 

Der Fehler, den ich

Notice: Undefined index: id_vehicleType in 
C:\xampp\htdocs\vms\addVehicleModel.php on line 13 

Notice: Undefined index: id_vehicleBrand in 
C:\xampp\htdocs\vms\addVehicleModel.php on line 14 

Notice: Undefined variable: id_vehicleType in 
C:\xampp\htdocs\vms\addVehicleModel.php on line 17 

Notice: Undefined variable: id_vehicleBrand in 
C:\xampp\htdocs\vms\addVehicleModel.php on line 17 ERROR: Could not 
able to execute INSERT INTO vehiclemodel (vehicle_Model, 
id_FKvehicleType, id_FKvehicleBrand) VALUES ('Saga', '', ''). Cannot 
add or update a child row: a foreign key constraint fails 
(`vms`.`vehiclemodel`, CONSTRAINT `vehiclemodel_ibfk_1` FOREIGN KEY 
(`id_FKvehicleType`) REFERENCES `vehicletype` (`id_vehicleType`))Yes, 
vehicleModel is setN0, id_vehicleBrand is not setN0, id_vehicleType is 
not set 

Answer 1

bekam Bitte versuchen Sie die Code unten, dass ich richtig, Sie bekommen, wo Sie Fehler machen. Ich hoffe, so

form role="form-horizontal" action="addVehicleModel.php" method="post"> 

Fahrzeugmodell:

<label for="vehicleType">Vehicle Type:</label> 
<div> 
     <select class="form-control" onchange="change_vehicleType()" name="vehicleType" id="vehicleTypeID"> 
<option value="">Select Vehicle Type</option> 

     <option value="1">hello</option> 

     </select> 

<label for="vehicleType">Vehicle Brand:</label> 
<div> 
     <select class="form-control" onchange="change_vehicleBrand()" name="vehicleBrand" id="vehicleBrandID"> 
      <option value="">Select Vehicle Brand</option> 
         <option value= "2" >BMW</option> 
     </select> 

Senden Schließen

<?php 
$mysqli = new mysqli("localhost", "root", "root", "vms"); 

if($mysqli === false){ 
die("ERROR: Could not connect. " . $mysqli->connect_error); 
} 




$vehicleModel = $mysqli->real_escape_string($_REQUEST['vehicleModel']); 
$vehicleTypeID = $mysqli->real_escape_string($_REQUEST['vehicleType']); 
$vehicleBrandID = $mysqli->real_escape_string($_REQUEST['vehicleBrand']); 


    $sql = "INSERT INTO vehiclemodel (vehicle_Model, id_FKvehicleType,id_FKvehicleBrand) VALUES ('$vehicleModel', '$vehicleTypeID', '$vehicleBrandID')"; 
if($mysqli->query($sql) === true){ 
echo "Records inserted successfully."; 
    } else{ 
echo "ERROR: Could not able to execute $sql. " . $mysqli->error; 
} 



    $mysqli->close(); 
?>