Question

我正在尝试将图像信息插入数据库，网址，名称，类型和大小。我创建了我的数据库，就像你的名字一样，

Idimage 名称尺寸类型网址

我所做的代码会在我尝试获取数据库中的信息之后将图像上传到文件夹，但它不起作用。

代码：

<?php
    if(isset($_POST['upload']))
    {
        $dbhost = 'localhost';
        $dbuser = 'root';
        $dbpass = 'root';
        $conn = mysql_connect($dbhost, $dbuser, $dbpass);
        if(! $conn )
        {
          die('Could not connect: ' . mysql_error());
        }

        $name = $_FILES['file']['name'];
        $type = $_FILES['file']['type'];
        $size = $_FILES['file']['size'];

        $sql =  "INSERT INTO". 
                " `image`(`name`, `url`, `size`, `type`) VALUES " . 
                " (`$name`,". 
                "`images/contentImages/"."`$name`"."`$type`""".
                ",$size".
                ",`$type`)";

        mysql_select_db('vgv');
        $retval = mysql_query( $sql, $conn );
        if(! $retval )
        {
          die('Could not update data: ' . mysql_error());
        }
        echo "Updated data successfully\n";
        mysql_close($conn);
    }
?>




<?php include ('includes/header.php') ?>
    <?php 
        use foundationphp\UploadFile;

        $max = 1024 * 1024;
        $result = array();
        if (isset($_POST['upload'])) {
            require_once 'src/foundationphp/UploadFile.php';
            $destination = __DIR__ . '/images/contentImages/';
            try {
                $upload = new UploadFile($destination);
                $upload->setMaxSize($max);
                $upload->allowAllTypes();
                $upload->upload();
                $result = $upload->getMessages();
            } catch (Exception $e) {
                $result[] = $e->getMessage();
            }   
        }
    ?>
    <?php include ('includes/update_image.php'); ?>
    <div id=box>
        <div id="box-header">

        </div>
        <div id="box-container">
            <div id="box-content">

            </div>
            <div id="box-footer"></div>
        </div>
    </div>
    <div id="image-info">
        <div id="image">

        </div>
        <div>
            <table>
                <tr>
                    <td class="benaming">Naam</td>
                    <td class="naam"></td>
                </tr>
                <tr>
                    <td class="benaming">Type</td>
                    <td class="naam"></td>
                </tr>
                <tr>
                    <td class="benaming">Afmeting</td>
                    <td class="naam"></td>
                </tr>
                <tr>
                    <td class="benaming">Grote(MB)</td>
                    <td class="naam"></td>
                </tr>
                <tr>
                    <td class="benaming">URL</td>
                    <td class="naam"></td>
                </tr>
            </table>
        </div>
        <div id="knoppen">
            <form action="" method="post" enctype="multipart/form-data">
                <input type="file" name="filename[]" id="filename" multiple>
                <input type="submit" name="upload" id="submit" value="Upload File">
            </form>
            <?php 
                if ($result) { 
                    foreach ($result as $message) {
                        echo '<script>alert($message);<script>';
                    }
                } 
            ?>
        </div>
    </div>

<?php include ('includes/footer.php') ?>

问题在于它不是$ _POST ['sumbit']而是$ _POST ['upload']

Answer 1

$url =  "`images/contentImages/". $name . "." . $type . "";
$sql =  "INSERT INTO". 
            " `image`(`name`, `url`, `size`, `type`) VALUES " . 
            " (`$name`". 
            ", `$url`".
            ", `$size`".
            ",`$type`)";

放入数据库的图像信息（名称，类型，大小，URL）

1 个答案: