drra6593
2014-03-12 00:02
浏览 35
已采纳

PHP For Mysqli的循环问题

I'm trying to get this PHP script to paginate the data pulled from the mysql database.

It is going wrong somewhere in the second for loop. Instead of pulling the data through it is just returning blank or empty fields.

I need it to display the title, description and content fields from the database, along with the ID.

        require_once("../controls/config.php");
        $connection = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

        if ($connection->connect_errno) {
            printf("Connect failed: %s
", $connection->connect_error);
            exit();
}

        // number of results to show per page
        $per_page = 3;

        // figure out the total pages in the database
        $result = $connection->query("SELECT * FROM pages");
        $total_results = $result->num_rows;
        $total_pages = ceil($total_results / $per_page);

        // check if the 'page' variable is set in the URL (ex: view-paginated.php?page=1)
        if (isset($_GET['page']) && is_numeric($_GET['page']))
        {
                $show_page = $_GET['page'];

                // make sure the $show_page value is valid
                if ($show_page > 0 && $show_page <= $total_pages)
                {
                        $start = ($show_page -1) * $per_page;
                        $end = $start + $per_page; 
                }
                else
                {
                        // error - show first set of results
                        $start = 0;
                        $end = $per_page; 
                }               
        }
        else
        {
                // if page isn't set, show first set of results
                $start = 0;
                $end = $per_page; 
        }

        for ($i = 1; $i <= $total_pages; $i++)
        {
                echo "<a href='?page=$i'>$i</a><br>";
        }

        // loop through results of database query, displaying them in the table 
        for ($i = $start; $i < $end; $i++)
        {
                // make sure that PHP doesn't try to show results that don't exist
                if ($i == $total_results) { break; }

                echo $i["id"].' ';
                echo $i["title"].' ';
                echo $i["description"].' ';
                echo $i["content"].' ';
                echo '<a href="edit.php?id='.$i["id"].'">Edit</a> ';
                echo '<a href="delete.php?id='.$i["id"].'">Delete</a><br>';

        }

?>
<p><a href="new.php">Add a new record</a></p>

Can anyone point me in the right direction?

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1条回答 默认 最新

  • dtdr57046 2014-03-12 00:18
    已采纳

    It looks like you are trying to treat $i as an associative array. Though $i is only a integer. Additionally, you do not have an array the contains the results from your mysqli query. You should try:

    // this will loop through results and assign to $rows array
    while($row = $result->fetch_array())
    {
       $rows[] = $row;
    }
    
    // this will loop through $rows array and provide each column result
    foreach($rows as $row)
    {
        echo $row["id"];
        echo $row["title"];
        echo $row["description"];
        echo $row["content"];
    }
    

    For further information please refer to: http://us2.php.net/mysqli_fetch_array

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