保存從一個表格到另一個表格的下拉菜單中顯示的數據

我是PHP-MySQL新手，當我嘗試從另一個表格中顯示dropdown字段值並將其保存到當前表格中時，會遇到問題。我在網上發現了一個完整的CRUD教程，但它只有簡單的創建，更新，顯示功能只使用輸入字段，沒有選擇字段，和其他複雜的東西。保存從一個表格到另一個表格的下拉菜單中顯示的數據

目前，我有2個表：

客戶，具有以下字段：

customer_id, customer_name, customer_address, customer_country。
訂單，與字段：

order_id, customer_id, product_name, product_price, product_qty。

具有以下orders.php我想用一個dropdown爲customer_id顯示customer_name從客戶表和選擇客戶名稱保存它在當前表的（訂單）customer_idcustomer_id場。

<?php 
    // creates the new record form 
    // since this form is used multiple times in this file, I have made it a function that is easily reusable 
    function renderForm($customerid, $productname, $productprice, $productqty, $error) 
    { 
?> 

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" 
    "http://www.w3.org/TR/html4/loose.dtd"> 
<html> 
<head> 
    <meta charset="utf-8"> 
    <title>ORDERS</title> 
</head> 

<body> 
    <form action="" method="post"> 
     <label for="customer_id">Customer ID:* </label> 
     <select name="customer_id"> 
     <?php 
      $sql = "SELECT * FROM customers ORDER BY customer_name ASC;"; 
      $res = mysql_query($sql); 
      while($row= mysql_fetch_assoc($res)) { 
       echo "<option value='" . $row['customer_id'] 
         . "'>" . $row['customer_name'] . "</option>"; 
      } 
     ?> 
     </select> 
    <label for="product_name">Product name:*</label> 
    <input type="text" name="product_name" value="<?php echo $productname; ?>" /><br/> 
    <label for="product_price">Price:* </label> 
    <input type="text" name="product_price" value="<?php echo $productprice; ?>" /><br/> 
    <label for="product_qty">Qty:* </label> 
    <input type="text" name="product_qty" value="<?php echo $productqty; ?>" /><br/> 
    <input type="submit" name="submit" class="btn-global" value="Save"> 
    </form> 
</body> 
</html> 
<?php 
    } 

    // connect to the database 
    include('connect.php'); 

    // check if the form has been submitted. If it has, start to process the form and save it to the database 
    if (isset($_POST['submit'])) 
    { 
     // get form data, making sure it is valid 
     $id = mysql_real_escape_string(htmlspecialchars($_POST['customer_id'])); 
     $product = mysql_real_escape_string(htmlspecialchars($_POST['product_name'])); 
     $price = mysql_real_escape_string(htmlspecialchars($_POST['product_price'])); 
     $qty = mysql_real_escape_string(htmlspecialchars($_POST['product_qty'])); 

     // check to make sure both fields are entered 
     if ($id == '' || $product == '' || $price == '' || $qty == '') 
     { 
      // generate error message 
      $error = 'ERROR: Please fill in all required fields!'; 

      // if either field is blank, display the form again 
      renderForm($customerid, $productname, $productprice, $productqty, $error); 

     } 
     else 
     { 
      // save the data to the database 
      mysql_query("INSERT orders SET customer_id='$id', product_name='$product', product_price='$price', product_qty='$qty'") 
      or die(mysql_error()); 

      // once saved, redirect back to the view page 
      header("Location: some.php"); 
     } 
    } 
    else // if the form hasn't been submitted, display the form 
    { 
     renderForm('','','','',''); 
    } 
?>

來源

2013-12-12 user3095221

你做了什麼錯誤？ –

你在混淆UPDATE和INSERT – Strawberry

你的SQL是錯誤的，INSERT應該是這樣的：

INSERT INTO orders (customer_id, product_name, product_price, product_qty) VALUES ('$id', '$product', '$price','$qty')

當你更新現有的記錄，你可以使用類似：

UPDATE orders SET product_name='$someValue', product_price='$someValue2' WHERE customer_id='$id'

來源

2013-12-12 12:33:49

保存從一個表格到另一個表格的下拉菜單中顯示的數據

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