如何在第1列和第2列的交叉点上填充第3列结果，这些列都来自用户在下拉列表中选择的同一个表？

Question

I have a column vehicle_name and I would like 2 dropdown lists of my 2 other columns namely, vehicle_type and vehicle_color. 我有一个列vehicle_name，我想要2个其他列的下拉列表，即vehicle_type和vehicle_color。 When these 2 dropdown values are selected and submitted, I would like their intersection to print out the values from vehicle_name. 当选择并提交这两个下拉值时，我希望它们的交集打印出来自vehicle_name的值。 So far my code only generates a dropdown list for vehicle_type, I would need another dropdown for vehicle_colour. 到目前为止，我的代码只为vehicle_type生成一个下拉列表，我需要另一个vehicle_colour下拉列表。 Which on submissions populates the intersected values for the vehicle_name. 在提交时填充了vehicle_name的相交值。 How can I achieve this? 我怎样才能做到这一点？

<!DOCTYPE html>
<html>
<body>
<?php
echo "<br>";
echo "<br>";
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "mydb";


$db = new mysqli($servername, $username, $password, $dbname);


if (!$db) {
    exit('Connect Error (' . mysqli_connect_errno() . ') '
        . mysqli_connect_error());
}
?>

<br>
<div class="label">Select vehicle type:</div>

<select name="payment_method">
    <option value = "">---Select---</option>
<?php
$queryusers = "SELECT DISTINCT vehicle_type FROM orders";
$db = mysqli_query($db, $queryusers);
while ($d=mysqli_fetch_assoc($db)) {
    echo "<option value='{".$d['vehicle_type']."}'>".$d['vehicle_type']."</option>";
}

?>
</select>


<br>
<div class="label_for_time">Select color:</div>

<select name="vehicle_color">
    <option value = "">---Select---</option>

<?php
$query_for_color = "SELECT DISTINCT vehicle_color FROM orders";
$db = mysqli_query($db, $query_for_date);
while ($a=mysqli_fetch_assoc($db)) {
    echo "<option value='{".$a['vehicle_color']."}'>".$a['vehicle_color']."</option>";
}

?>
</select>
<br>
<br>


<button class="go-btn" type="submit">Go</button>
</body>
</html>

Answer 1

As I don't see any AJAX / client-side code in your above example I assume that this is a pure backend-side filtering you are performing. 由于我在上面的示例中没有看到任何AJAX /客户端代码，我认为这是您正在执行的纯后端端过滤。 Your code is currently missing parts of the required elements we would need but let's try to figure this out together: 您的代码目前缺少我们需要的部分元素，但让我们一起尝试解决这个问题：

1. Form around your inputs 1.围绕您的输入形成

Add a <form method="POST" target="path-to-your-script.php"> where "path-to-your-script.php" has to be changed to your PHP file name or rewritten URL path. 添加<form method="POST" target="path-to-your-script.php">其中“path-to-your-script.php”必须更改为PHP文件名或重写的URL路径。 This has to be around the <select> boxes. 这必须在<select>框周围。

You may also use PHP_SELF to set this automatically, this should work in most cases. 您也可以使用PHP_SELF自动设置它，这在大多数情况下都适用。 I used html_entities($var) to avoid any code injections via manipulated URL. 我使用html_entities($var)来避免通过操纵的URL进行任何代码注入。

<form name="test" action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post">

2. Check for POST'ed variable 'vehicle_type' 2.检查POST变量'vehicle_type'

In your form, check if a search for available colors has been performed: 在表单中，检查是否已搜索可用颜色：

<?php
  $query_for_color = "SELECT DISTINCT vehicle_color FROM orders";

  // check if the form variable 'vehicle_type' is available; if so, filter entries.
  if (isset($_POST['vehicle_type'])) {
    $vType= filter_var($_POST['vehicle_type'], FILTER_SANITIZE_STRING);
    $query_for_colors .= ' WHERE vehicle_type = \''.$vType.'\'';
  } 
  $db = mysqli_query($db, $query_for_date);
  while ($a=mysqli_fetch_assoc($db)) {
      echo "<option value='{".$a['vehicle_color']."}'>".$a['vehicle_color']."</option>";
  }
?>

Edit: 编辑：

As pointed out by one user in the comment, filter_var($var, FILTER_SANITIZE_STRING) won't be enough to avoid potential SQL injections. 正如评论中的一个用户所指出的， filter_var($var, FILTER_SANITIZE_STRING)不足以避免潜在的SQL注入。 This was just a recommendation and was not part of the question at all. 这只是一个建议，根本不是问题的一部分。 If you have to work with user data, do more than using filter_var(), instead use either prepared statements or properly escape the user data. 如果必须使用用户数据，请使用filter_var（），而不是使用预准备语句或正确转义用户数据。 There are many tutorials like this one out there that will guide you to safe queries. 有很多这样的教程可以指导您进行安全查询。

如何在第1列和第2列的交叉点上填充第3列结果，这些列都来自用户在下拉列表中选择的同一个表？

问题描述

1 个解决方案

解决方案1
0 已采纳 2019-05-23 14:21:01

如何在第1列和第2列的交叉点上填充第3列结果，这些列都来自用户在下拉列表中选择的同一个表？

问题描述

1 个解决方案

解决方案1 0 已采纳 2019-05-23 14:21:01

解决方案1
0 已采纳 2019-05-23 14:21:01