插入 MySql PHP [已关闭]

问题描述 投票：0回答：2

我正在尝试使用 PHP 将数据添加到 MySql 数据库中。

代码：

<?php
    session_start();
    include('config.php');


    if(isset($_POST['Submit'])) {
        $city = $_POST['city'];
    
        $from_station = $_POST['from_station'];
        $from_date = $_POST['from_date'];

        // Prepare and execute the SQL query
        $sql = "INSERT INTO journeys (city, from_station, from_date) VALUES ('$city', '$from_station', '$from_date')";
        $result = $conn->query($sql);

        if (mysqli_query($conn, $sql)) {
            echo "New record created successfully";
        } else {
                echo "Error: " . $sql . "<br>" . mysqli_error($conn);
        }

    }
?>



<!DOCTYPE html>
<html lang="pt_PT">
    <head>
        <meta charset="UTF-8">
        <meta name="viewport" content="width=device-width, initial-scale=1.0">
        <title>IR PLANNER</title>
        <link rel="stylesheet" href="Styles\general_styles.css">
        <link rel="stylesheet" href="Styles\citymanager_styles.css">
    </head>


    <body>        
        <form class="modal" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
            <label for="city">City:</label>
            <input type="text" name="city" id="city">
            <label for="from_station">From Station:</label>
            <input type="text" name="from_station" id="from_station">
            <label for="from_date">From Date:</label>
            <input type="date" name="from_date" id="from_date">
            <input type="submit" value="Submit" id="Submit">
        </form>
       
    </body>

</html>

按下提交按钮时，不会显示错误。然而，当检查数据库表时，什么也没有出现。

我已阅读文章，并且 sql 连接（config.php）工作正常（在其他页面和文件中工作）并且 sql 查询也很好。

有人遇到这个问题吗？

php mysql mysqli sql-insert
2个回答
0
投票

您需要在提交按钮中传递 name 属性，直到您无法从中获取任何值。在您的条件下，它总是返回 false。所以不会更新任何内容，也不会生成错误。只需更新这里是更新的代码。

<?php
        session_start();
        include('config.php');
    
    
        if(isset($_POST['Submit'])) {
            $city = $_POST['city'];
        
            $from_station = $_POST['from_station'];
            $from_date = $_POST['from_date'];
    
            // Prepare and execute the SQL query
            $sql = "INSERT INTO journeys (city, from_station, from_date) VALUES ('$city', '$from_station', '$from_date')";
            $result = $conn->query($sql);
    
            if (mysqli_query($conn, $sql)) {
                echo "New record created successfully";
            } else {
                    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
            }
    
        }
    ?>
    
    
    
    <!DOCTYPE html>
    <html lang="pt_PT">
        <head>
            <meta charset="UTF-8">
            <meta name="viewport" content="width=device-width, initial-scale=1.0">
            <title>IR PLANNER</title>
            <link rel="stylesheet" href="Styles\general_styles.css">
            <link rel="stylesheet" href="Styles\citymanager_styles.css">
        </head>
    
    
        <body>        
            <form class="modal" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
                <label for="city">City:</label>
                <input type="text" name="city" id="city">
                <label for="from_station">From Station:</label>
                <input type="text" name="from_station" id="from_station">
                <label for="from_date">From Date:</label>
                <input type="date" name="from_date" id="from_date">
                <input type="submit" name="Submit" value="Submit" id="Submit">
            </form>
           
        </body>
    
    </html>
0
投票

您可以采取一些措施来分析问题/优化代码。

1.要调试代码：回显一个值以检查是否进入 if 语句

例如：

    if(isset($_POST['Submit'])) {
        var_dump(__LINE__);

现在运行代码以查看到达了哪些行。

2. if 语句中的代码不会被执行，因为：

此代码：

<input type="submit" value="Submit" id="Submit">

需要 

name
属性，才能作为数据发送到服务器：

<input type="submit" name="Submit" value="Submit" id="Submit">

3.使用mysqliprepare准备sql语句

代替当前的插入代码，制作它

$stmt = $conn->prepare("INSERT INTO journeys (city, from_station, from_date) VALUES (?, ?, ?)");
$stmt->bind_param("sss", $city, $from_station, $from_date);
$stmt->execute();

这样就不会发生SQL注入了。

4.在您的代码中，您还执行了两次 SQL。

在这些线上：

$result = $conn->query($sql);
if (mysqli_query($conn, $sql)) {

替换为第 3 点的代码。

生成的代码

<?php
session_start();
include('config.php');

if(isset($_POST['Submit'])) {
    $city = $_POST['city'];
    $from_station = $_POST['from_station'];
    $from_date = $_POST['from_date'];

    // Prepare and execute the SQL query
    $stmt = $conn->prepare("INSERT INTO journeys (city, from_station, from_date) VALUES (?, ?, ?)");
    $stmt->bind_param("sss", $city, $from_station, $from_date);
    $stmt->execute();
}
?>

<!DOCTYPE html>
<html lang="pt_PT">
    <head>
        <meta charset="UTF-8">
        <meta name="viewport" content="width=device-width, initial-scale=1.0">
        <title>IR PLANNER</title>
        <link rel="stylesheet" href="Styles\general_styles.css">
        <link rel="stylesheet" href="Styles\citymanager_styles.css">
    </head>

    <body>        
        <form class="modal" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
            <label for="city">City:</label>
            <input type="text" name="city" id="city">
            <label for="from_station">From Station:</label>
            <input type="text" name="from_station" id="from_station">
            <label for="from_date">From Date:</label>
            <input type="date" name="from_date" id="from_date">
            <input type="submit" name="Submit" value="Submit" id="Submit">
        </form>
    </body>
</html>
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