Question

以下是用于显示数据库中长图像格式的图像的代码。

upload.php的：

if(isset($_FILES['files'])){
    $servername = "localhost";
    $username = "root";
    $password = "123456";
    $dbname = "photost";
    $conn=mysqli_connect($servername, $username, $password, $dbname);
    $user_id=$_POST['user_id'];
    if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
    }
  $errors="";
    foreach($_FILES['files']['tmp_name'] as $key => $tmp_name ){
        $file_name = $key.$_FILES['files']['name'][$key];
        $file_size =$_FILES['files']['size'][$key];
        $file_tmp =$_FILES['files']['tmp_name'][$key];
        $file_type=$_FILES['files']['type'][$key];
        $file_content = file_get_contents($file_tmp);
        $maxsize = 10000000; 

    if($_FILES['files']['tmp_name'][$key]==UPLOAD_ERR_OK) {
            if(is_uploaded_file($_FILES['files']['tmp_name'][$key])) {
                if( $_FILES['files']['tmp_name'][$key] < $maxsize) {
                    $finfo = finfo_open(FILEINFO_MIME_TYPE);
                    if(strpos(finfo_file($finfo, $_FILES['files']['tmp_name'][$key],"image"))==0) {
                        $imgData = addslashes(file_get_contents($_FILES['files']['tmp_name'][$key]));
                        $query="INSERT INTO `images`(`id`, `u_id`, `image`, `status`, `file_name`, `file_size`) VALUES ( NULL,'$user_id','{$imgData}',0,'$file_name','$file_type');";
                        if (mysqli_query($conn, $query)) {
                            echo "New record created successfully";
                            } else {
                                echo "Error: <br>" . mysqli_error($conn);
                }
                else
                                echo "<p>Uploaded file is not an image.</p>";
                                }
                                }
                            }
                            }
                            }



<body>
    <div style="width:100%;height:40%">
    </div>
<div align=center>
<form action="" method="POST" enctype="multipart/form-data">
    <table style="width:40%">
    <tr>
        <th align="left">User Id:</th>
        <th align="left"><input type='text' name='user_id'></th>
    </tr>
    <tr>
        <th></th>
        <th align="left"><input type="file" name="files[]" multiple/></th>
    </tr>
    <tr>
        <th></th>
        <th ><input type="submit"/></th>
    </tr>

    </table>

    </form>
</div>
</body>

test.php的：

<body>
<img src="getImage.php?id=1" />
</body>

getImage.php：

<?php
$id = $_GET['id'];
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "photost";
$conn=mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
$query="SELECT * FROM `images` WHERE `id`=$id";
$result =mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);

$image =$row['image'];

/** check if the image is db */
if($image!=null)
{
    $db_img = imagecreatefromstring($image);
    Header("Content-type: image/jpeg");
    imagejpeg($db_img);

}
mysql_close($conn);
?>

我也试过这个：

  $row = mysql_fetch_assoc($result);

header("content-type: image/jpeg");

  echo $row['image'];
    mysql_close($conn);

我得到像这样的输出

https://i.stack.imgur.com/RDkdM.png

我可以通过从phpmyadmin下载并以jpeg格式保存图像来查看图像。

Answer 1

mysqli_fetch_assoc函数返回每行的数组。现在试试

$image = $row[0]['image'];

代替：

$image = $row['image'];

无法显示图像从Mysql尝试了这么多解决方案似乎没什么用

1 个答案: