在一个网页问题中两次插入更新

Question

I have two insert updates on my webpage, i have done this in the past without any issue, however, today i am having major issues, the first insert update works perfectly, but the second doesn't (on pressing the submit button nothing appears in the db table) even through the exactly the same script is being used.

I am not sure if its the two insert updates in one webpage, or if it is the fact that i need to close the db after the first insert update, which i haven't.

The db table i am trying to insert into has a unique key, i am also connected to my db or a complete different problem all together. Extremely frustrated. Please help

The second insert update

   $Restaurant_id = mysqli_real_escape_string($dbc, $_GET['restaurant_id']);
                 if (isset($_POST['Save_changes'])) {
        $code = $_POST['Pcode'];


        $insert_ = "INSERT INTO Delivery_Pcode
  (Pcode,Restaurant_ID) VALUES(?,?)
   ON DUPLICATE KEY 
    UPDATE 
    Pcode  = ?
    ,Restaurant_ID  = ?";


        $r_query = mysqli_prepare($dbc, $insert_);
        //new
        // $stmt = mysqli_prepare($dbc, $insert_c);
        //debugging
        //$run_query = $dbc->prepare($insert_Delex);

        $r_query->bind_param('sisi', $code,$Restaurant_id,$code,$Restaurant_id);
        // THIS now executes the above transaction, returns TRUE if successful - issdissd duplicate update
        if (!$r_query->execute()) {
            $insertEr = "There was an error inserting data: " . $r_query->error;
        }

        print "affeted rows:" . $r_query->affected_rows; //how many records affected? 
        $r_query->free_result(); // Frees memory on completion 
        $r_query->close(); //closes this action 
    }

HTML

  <form id="delivery_pcodes" action ="Franchise_postcodes.php?restaurant_id=<?php echo $_GET['restaurant_id'] ?>&Franchise=<?php echo $_GET['Franchise']; ?>" method="POST">
            <input type="text" name="pcode" id="pcode"  value="<?php echo $pcodes; ?>"  placeholder="e.g LS1,LS2,LS3" tabindex="2">
            <input type="number" name="pcode_price" id="pcode_price"  value=""  placeholder="e.g 3.50" tabindex="2">

            <button type="submit" id="Save_changes" name="Save_changes" value="Save Changes">
                <span class="glyphicon glyphicon-ok"></span> Save changes
            </button>
        </form>

Answer 1

在此代码中，您将两次执行$code,$Restaurant_id 。

 $r_query->bind_param('sisi', $code,$Restaurant_id,$code,$Restaurant_id);