将图片+图片信息从php表格上传到mysql数据库

时间:2015-03-23 20:59:27

标签: php mysql mysqli image-uploading

我有一个表单的代码,用于将图片上传到我的网站并将信息保存到mysql数据库:

<form method='post'>
        Album Name: <input type="text" name="title" /> 
        <input type="submit" name="submit" value="create" />
    </form>
<h4>Add Photo</h4>
<form enctype="multipart/form-data" method="post">
        <?php
        require_once 'config.php'; 
    $mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
    if(isset($_POST['upload'])){
        $caption = $_POST['caption'];
        $albumID = $_POST['album'];
        $file = $_FILES ['file']['name'];
        $file_type = $_FILES ['file']['type'];
        $file_size = $_FILES ['file']['size'];
        $file_tmp = $_FILES ['file']['tmp_name'];
        $random_name = rand();

        if(empty($file)){
            echo "Please enter a file <br>";
        } else{
             move_uploaded_file($file_tmp, 'uploads/'.$random_name.'.jpg');
    $ret = mysqli_prepare($mysqli, "INSERT INTO photos (caption, image_url, date_taken)
    VALUES(?, ?, NOW())");
    $filename = "uploads/" + $random_name + ".jpeg";
    mysqli_stmt_bind_param($ret, "ss", $caption, $filename);
    mysqli_stmt_execute($ret);
    echo "Photo successfully uploaded!<br>";
    }
    }
    ?>
    Caption: <br>
    <input type="text" name="caption">
    <br><br>
    Select Album: <br>
    <select name="album">
    <?php
    $mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
    $result = $mysqli->query("SELECT * FROM albums");
    while ($row = $result->fetch_assoc()) {
        $albumID = $row['albumID'];
        $title = $row['title'];
        echo "<option value='$albumID'>$title</option>";
    }
    ?>
    </select>
    <br><br>
    Select Photo: <br>
    <input type="file" name="file">
    <br><br>
    <input type="submit" name="upload" value="Upload">
</form>

这已成功将图片上传到我的上传内容中。文件夹以及我的mysql数据库,但是,我想放入图片网址&#34;上传/(随机名称生成).jpg&#34; 我没有用我当前的代码执行此操作,这些信息记录在&#39; image_url&#39;我照片表的列只是生成的随机数。没有&#34;上传/&#34;在开始和&#34; .jpg&#34;最后。

我应该提一下我照片表的架构是: caption,image_url,date_taken,imageID

任何帮助将非常感谢!! 提前谢谢你

1 个答案:

答案 0 :(得分:1)

您正在使用+(加号)符号连接,在此行中:

$filename = "uploads/" + $random_name + ".jpeg";

PHP使用点/句点来连接,而不是使用JS / C语言语法的加号:

$filename = "uploads/" . $random_name . ".jpeg";

错误检查会发出语法错误信号。