PHP/Mysqli：为什么此代码会使插入行加倍？

我的代码需要帮助；不知何故，我的代码创建了两个房间（它一次将两行插入表中），我不知道为什么。

（我需要每个插入都需要一个 id 才能知道我们在哪个房子中创建一个新房间。我的数据库包含表“house”和表“room”。表“room”有一个字段“house_id”，它是一个外键表“house”中的字段“id”。）

那是我的 php 页面：

<?php 
            // turn autocommit off
            mysqli_autocommit($con, FALSE);

            // fetch the houses so that we have access to their names and id
            $query = "SELECT name, id
                    FROM house";
            $result = mysqli_query($con, $query);
            // check query returned a result
            if ($result === false) {
                echo mysqli_error($con);
            } else {
                $options = "";
               // create an option
                while ($row = mysqli_fetch_assoc($result)) {
                    // $options .= "".$row['name']."";
                     $options .= "<option value='".$row['id']."'>";
                     $options .= $row['name'];
                     $options .= "</option>";
                }
            }

            include('templates/add_room.html');

            if ($_SERVER["REQUEST_METHOD"] == "POST") {
                $price = mysqli_real_escape_string($con, $_POST["price"]);
                $house = mysqli_real_escape_string($con, $_POST["house_id"]);

                $query = "INSERT INTO room (price, house_id)
                VALUES ('$price', '$house')";

                // run the query to insert the data
                $result = mysqli_query($con, $query);

                // check if the query went ok
                if ( $con->query($query) ) {
                    echo "<script type= 'text/javascript'>alert('New room created successfully with the id of {$con->insert_id}');</script>";
                    mysqli_commit($con);

                } else {
                    echo "There was a problem:<br />$query<br />{$con->error}";
                    mysqli_rollback($con);
                }
            }

            //free result set
                mysqli_free_result($result);

?>

这是我的 html 模板，其形式为：

<h2>Add new room</h2>
<form action='' method='POST'>
            <fieldset>
                <label for='price'>Price:</label>
                <input type='number' name='price'>
            </fieldset>
            <fieldset>
            <label for='house_id'>House:</label>
                <select name='house_id' required>
                    <option value='' disabled selected>Select house</options>
                    <?php echo $options; ?>
                </select>
            </fieldset>
            <button type='submit'>Add</button>
</form>

由于您使用查询函数两次，它插入了 2 行：

$result = mysqli_query($con, $query);

                // check if the query went ok
                if ( $con->query($query) ) {

因此，您需要将该条件语句更改为：

if ($result)  {

顺便说一句，使用准备好的语句，它比real_escape_string():

• https://en.wikipedia.org/wiki/Prepared_statement