Table content keeps adding when refreshing PHP page

StackOverflow https://stackoverflow.com/questions/19618172

Pergunta

I have a problem with PHP adding content to the database. Every time I refresh the page it adds the exact same values over again. How can I make it stop doing that? I have no idea why it does that.

I want it to add the values to the database once and store them there. When the table is created and values added, they are used on the page. User will have the ability to add more values.

It also adds whatever the user has entered when refreshing.

The problematic part is

$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);

Full code:

<!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>
Foi útil?

Solução

Why the sample values are added on each page visit

Your PHP code is executed every time you visit the page. It generates the page (writes parts of it dynamically, just copies everything outside <?php … ?>), and the page is the output of your script. Apache, or whatever server software you use, just gets an HTTP request, executes your script, takes its output and sends it to the client in an HTTP response.

If your INSERT query is executed unconditionally (as it is), it is also executed every time you visit the page. That's why the values are added over and over again.

How to stop adding the sample values

You may want to check if the cars table was created and execute the insert query only if it was:

# 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

Refresh after a POST problem

Another problem we met during debugging the issue is refresh after POST form submission. Browser automatically submits the original form contents again, which is standard behavior. It results in inserting the submitted values to the database again, now storing them in two or even more copies.

This can be prevented, though. See e.g. Stop browsers asking to resend form data on refresh. The accepted answer recommends the POST-redirect-GET pattern, which is what I would use in your case. You can get much more on that topic when googling post data resent stackoverflow.

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