Question

我有两个下拉列表，其中的值从2个不同的表中选择。如何确保在点击提交按钮后，下拉列表orden会转到norm

行

<?php
    $servername = "localhost";
    $username = "root";
    $password = "Iamthebest1009";
    $dbname = "dktp";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }

    $normSql    = 'SELECT * FROM norm WHERE orden_id IS NULL';
    $normResult = $conn->query($normSql);

    $ordenSql    = 'SELECT * FROM orden';
    $ordenResult = $conn->query($ordenSql);

    function html($string) {
        return htmlspecialchars($string, ENT_QUOTES, 'UTF-8');
    }

    ?>
    <!DOCTYPE html>
    <html>
        <body>
            <form method="POST" action"">
                <select name="normID">
                    <option selected disabled>Choose norm</option>
                <?php while ($result = $normResult->fetch_assoc()): ?>
                    <option value="<?= html($result['id']); ?>"><?= html($result['norm_name']); ?></option>
                <?php endwhile; ?>
                </select>

                <select name="ordenID">
                    <option selected disabled>Choose orden</option>
                <?php while ($result = $ordenResult->fetch_assoc()): ?>
                    <option value="<?= html($result['id']); ?>"><?= html($result[ 'orden_name']); ?></option>
                <?php endwhile; ?>
                </select>

                <input type="submit" value="Insert">
            </form>
        </body>
    </html>

    <?php

    if(isset($_POST["submit"]))
    {

      $ordening=$_POST["ordenID"];

    $query = mysqli_query("INSERT INTO norm (orden_id)VALUES ('$ordening')");
                if($query)
                {
                    echo "Thank You! you are now registered.";
                }
    } 
    ?>

Answer 1

您的查询设置应如下所示，使用绑定参数进行安全性并执行查询：

只选择ID，而不是所有内容，您不需要所有内容：

<?php
    $servername = "localhost";
    $username = "root";
    $password = "Iamthebest1009";
    $dbname = "dktp";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }

    $normSql    = 'SELECT id FROM norm WHERE orden_id IS NULL';
    $normResult = $conn->query($normSql);

    $ordenSql    = 'SELECT id FROM orden';
    $ordenResult = $conn->query($ordenSql);

    function html($string) {
        return htmlspecialchars($string, ENT_QUOTES, 'UTF-8');
    }

    ?>
    <!DOCTYPE html>
    <html>
        <body>
            <form method="POST" action"">

    /* Not entirely sure why you need two selects? There's no reason you're not even using the value from this select? */

               /* <select name="normID">
                    <option selected disabled>Choose norm</option>
                <?php while ($result = $normResult->fetch_assoc()): ?>
                    <option value="<?= html($result['id']); ?>"><?= html($result['norm_name']); ?></option>
                <?php endwhile; ?>
                </select>*/

                <select name="ordenID">
                    <option selected disabled>Choose orden</option>
                <?php while ($result = $ordenResult->fetch_assoc()): ?>
                    <option value="<?= html($result['id']); ?>"><?= html($result[ 'orden_name']); ?></option>
                <?php endwhile; ?>
                </select>

                <input type="submit" value="Insert">
            </form>
        </body>
    </html>




Then, change your insert to be more secure:



    if(isset($_POST["submit"]))
    {

      $ordening= $_POST["ordenID"];    // get the order ID from your submitted from <select>

    $query = mysqli_query("INSERT INTO norm (orden_id) VALUES (?)");
       $stmt = $conn->prepare($query);
       if($stmt){ 
          $stmt->bind_param("s",$ordening");   // Bind the id
          $stmt->execute();                    // Execute query
          $affected = $stmt->affected_rows;    // Get the affected rows (inserted)
          $stmt->close();                     // Close the statement
       }
       if($affected > 0 {
         echo "Successful Entry";
       }

    }

在一个表中插入两个下拉列表值

1 个答案: