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我在使用 PHP 向数据库添加内容时遇到问题。每次我刷新页面时，它都会再次添加完全相同的值。我怎样才能让它停止这样做？我不知道为什么会这样。

我希望它一次将值添加到数据库中并将它们存储在那里。创建表并添加值后，将在页面上使用它们。用户将能够添加更多值。

它还会添加用户在刷新时输入的任何内容。

有问题的部分是

$db_query = 'INSERT INTO `cars`' .
    ' (`id`, `make`, `model`, `year`, `color`, `engine`)' .
    ' VALUES' .
    ' (NULL, \'Citroen\', \'Saxo\', \'1997\', \'White\', 1.1),' .
    ' (NULL, \'Citroen\', \'C3\', \'2012\', \'Blue\', 1.4),' .
    ' (NULL, \'Volkswagen\', \'Golf\', \'2010\', \'Blue\', 1.8),' .
    ' (NULL, \'Ford\', \'Mondeo\', \'2009\', \'Black\', 1.6),' .
    ' (NULL, \'Renault\', \'Clio\', \'2010\', \'Silver\', 1.2);';
$result = mysqli_query($conn, $db_query);

完整代码：

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="utf-8">
    <title>Task 7</title>
</head>
<body>
<?php


# Connection to the database start
$host = 'localhost';
$user = 'root'; # Your username
$password = 'vertrigo'; # Your password
$databaseName = 'db2'; # Your database

$conn = mysqli_connect($host, $user, $password, $databaseName);

if (mysqli_connect_errno($conn)) {
    echo 'Failed to connect to MySQL: ' . mysqli_connect_error();
} else {
    echo 'have connection<br>';
}
# Connection to the database end (output messages created)


# Creating the table with rows start
$db_query = 'CREATE TABLE IF NOT EXISTS `cars` (' .
    '`id` int(11) NOT NULL AUTO_INCREMENT,' .
    '`make` varchar(50) NOT NULL,' .
    '`model` varchar(50) NOT NULL,' .
    '`year` varchar(4) NOT NULL,' .
    '`color` varchar(50) NOT NULL,' .
    '`engine` decimal(2,1) NOT NULL,' .
    'PRIMARY KEY (`id`));';
$result = mysqli_query($conn, $db_query);
# Creating table with rows end


# Putting information in each column start
$db_query = 'INSERT INTO `cars`' .
    ' (`id`, `make`, `model`, `year`, `color`, `engine`)' .
    ' VALUES' .
    ' (NULL, \'Citroen\', \'Saxo\', \'1997\', \'White\', 1.1),' .
    ' (NULL, \'Citroen\', \'C3\', \'2012\', \'Blue\', 1.4),' .
    ' (NULL, \'Volkswagen\', \'Golf\', \'2010\', \'Blue\', 1.8),' .
    ' (NULL, \'Ford\', \'Mondeo\', \'2009\', \'Black\', 1.6),' .
    ' (NULL, \'Renault\', \'Clio\', \'2010\', \'Silver\', 1.2);';
$result = mysqli_query($conn, $db_query);
# Putting information at each end


# Add cars START
@$cars = $_POST['cars'];
if (isset($cars)) {
    $db_query = "INSERT INTO cars (make, model, year, color, engine) VALUES ('" . $cars['0'] . "','" . $cars['1'] . "','" . $cars['2'] . "','" . $cars['3'] . "','" . $cars['4'] . "')";
    mysqli_query($conn, $db_query);
} else {
    echo 'Cars array does not exists and obtained from FORM.';
}
# Add cars END


# Select the cars table if the right output the headers start
$db_query = "SELECT * FROM cars WHERE id <> 0 ORDER BY year ASC";

if ($result = mysqli_query($conn, $db_query)) {
    echo  "    <table border='1'>";
    echo  "        <tr><td>Make</td><td>Year</td><td>model</td><td>color</td><td>engine</td></tr>";
    # Select cars table if the right output the headers end

    # while start, output the content of the column.
    while ($row = mysqli_fetch_assoc($result)) {
    ?>
        <tr>
            <td><?php echo $row['make']; ?></td>
            <td><?php echo $row['year']; ?></td>
            <td><?php echo $row['model']; ?></td>
            <td><?php echo $row['color']; ?></td>
            <td><?php echo $row['engine']; ?></td>
        </tr>
    <?php
    }
    # End while

    echo "</table>";
}
?>
    <form action="index.php" method="post" name="zatupok">
        <fieldset>
            <legend>MY CARS</legend>
            <label>Make</label>
            <input type="text" name="cars[]">
            <label>Model</label>
            <input type="text" name="cars[]">
            <label>Year</label>
            <input type="text" name="cars[]">
            <label>Color</label>
            <input type="text" name="cars[]">
            <label>Engine</label>
            <input type="text" name="cars[]">
            <input type="submit">
        </fieldset>
    </form>
</body>
</html>
4

1 回答 1

-1

为什么在每次页面访问时添加示例值

每次访问该页面时都会执行您的 PHP 代码。它生成页面（动态写入其中的一部分，只是复制外部的所有内容<?php … ?>），并且页面是脚本的输出。Apache，或者你使用的任何服务器软件，只是获取一个 HTTP 请求，执行你的脚本，获取它的输出并在一个 HTTP 响应中将它发送到客户端。

如果您的INSERT查询是无条件执行的（按原样执行），则每次访问该页面时也会执行该查询。这就是为什么一遍又一遍地添加这些值。

如何停止添加样本值

您可能需要检查cars表是否已创建并仅在以下情况下执行插入查询：

# Creating the table with rows start
# CHANGED: Notice that IF NOT EXISTS is not used anymore!
#          Fails if the table already exists, not touching its original contents.
$db_query = 'CREATE TABLE `cars` (' .
    '`id` int(11) NOT NULL AUTO_INCREMENT,' .
    '`make` varchar(50) NOT NULL,' .
    '`model` varchar(50) NOT NULL,' .
    '`year` varchar(4) NOT NULL,' .
    '`color` varchar(50) NOT NULL,' .
    '`engine` decimal(2,1) NOT NULL,' .
    'PRIMARY KEY (`id`));';
$result = mysqli_query($conn, $db_query);
# Creating the table with rows end


# Putting information in each column start
$db_query = 'INSERT INTO `cars`' .
    ' (`id`, `make`, `model`, `year`, `color`, `engine`)' .
    ' VALUES' .
    ' (NULL, \'Citroen\', \'Saxo\', \'1997\', \'White\', 1.1),' .
    ' (NULL, \'Citroen\', \'C3\', \'2012\', \'Blue\', 1.4),' .
    ' (NULL, \'Volkswagen\', \'Golf\', \'2010\', \'Blue\', 1.8),' .
    ' (NULL, \'Ford\', \'Mondeo\', \'2009\', \'Black\', 1.6),' .
    ' (NULL, \'Renault\', \'Clio\', \'2010\', \'Silver\', 1.2);';
# CHANGED: Executing the query only if previous query (`CREATE TABLE …`) succeeded.
if ($result) {
    $result = mysqli_query($conn, $db_query);
}
# Putting information at each end

POST问题后刷新

我们在调试过程中遇到的另一个问题是 POST 表单提交后刷新。浏览器会自动再次提交原始表单内容，这是标准行为。它导致将提交的值再次插入数据库，现在将它们存储在两个甚至更多副本中。

不过，这是可以避免的。请参阅停止浏览器要求在刷新时重新发送表单数据。接受的答案建议使用 POST-redirect-GET 模式，这就是我将在您的情况下使用的模式。谷歌搜索post data resent stackoverflow时，您可以获得更多关于该主题的信息。

于 2013-10-27T14:27:58.127 回答