显示数据库中的图像不起作用

Question

我尝试了多种方法，但我的代码仍然没有显示我网站上数据库中的图像。当我点击上传时，我只得到文件名和文件详细信息的输出，但没有显示照片。

这是我必须显示图像的代码。

 <main>
     <section align='center'>
     <h1 id="rcorner2"align='center'style="font-size:30px; position:fixed;">Photo Library</h1>
     <br><br>
         <div class="wrapper">
             <!--h2 align='left'>Photos</h2-->

          <div class="photo-container"> 

<?php
include_once 'dbh.php';
$sql = "SELECT * FROM photos ORDER BY orderPhotos DESC";
$stmt = mysqli_stmt_init($conn);

if (!mysqli_stmt_prepare($stmt, $sql)) {
    echo "Error updating photo library!";
}else{
    mysqli_stmt_execute($stmt);
    $result = mysqli_stmt_get_result($stmt);
            
    while ($row = mysqli_fetch_assoc($result)) {
        echo '<a href="#">
                  <div style="background-image: url(../libraries/photos/'.$row["imageFullName"].');"></div>
                  <h3>'.$row["filetitle"].'</h3>
                  <p>'.$row["filedescription"].'</p>
               </a>';
    }
}
?>
          </div>
         </div>
     </section>
 </main>

连接数据库

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "portal uploads";

$conn = mysqli_connect($servername, $username, $password, $dbname);

?>

这是来自 html 表单的数据库连接。

<?php


if(isset($_POST['upload'])) {
    $newFileName = $_POST['filename'];
    if(empty($newFileName)){
        $newFileName = "photo";
    }else{  
        //Replacing spaces in filename with underscores
        $newFileName = strtolower(str_replace(" ", "-", $newFileName));
    }
    $filetitle = $_POST['filetitle'];
    $filedescription = $_POST['filedescription'];
    
    $file = $_FILES['file']; 

    $fileName = $file["name"];
    $fileType = $file["type"];
    $fileTempName = $file["tmp_name"];
    $fileError = $file["error"];
    $fileSize = $file["size"];

    $fileExt = explode(".", $fileName);
    $fileActualExt = strtolower(end($fileExt));

    $allowed = array("jpg", "jpeg", "png");

    //Error handling for allowed file types
    if(in_array($fileActualExt, $allowed)) {
        if ($fileError === 0) {
            if($fileSize < 10000000) {
                //Make file unique through naming
                $imageFullName = $newFileName . "." . uniqid("", true) . "." . $fileActualExt;
                $fileDestination = "../libraries/photos/" . $imageFullName;

                include_once "dbh.php";

                //Checking for error handling when fields have been left empty
                if(empty($filetitle) || empty($filedescription)) {
                    header("location:photos_edit.php?upload=empty");
                    exit();
                } else {
                    $sql = "SELECT * FROM photos;";
                    $stmt = mysqli_stmt_init($conn); 
                    if (!mysqli_stmt_prepare($stmt, $sql)) {
                        echo "SQL statement failed!";
                    }else{
                        mysqli_stmt_execute($stmt);
                        $result = mysqli_stmt_get_result($stmt);
                        $rowCount = mysqli_num_rows($result);
                        $setPhotoOrder = $rowCount + 1;

                        $sql = "INSERT INTO photos (filetitle, filedescription, imageFullName, orderPhotos) VALUES (?, ?, ?, ?);";
                        if (!mysqli_stmt_prepare($stmt, $sql)) {
                            echo "SQL statement failed!";
                        }else{
                            mysqli_stmt_bind_param($stmt, "ssss", $filetitle, $filedescription, $imageFullName, $setPhotoOrder);
                            mysqli_stmt_execute($stmt);

                            move_uploaded_file($fileTempName, $fileDestination);
                            header("location: photos_edit.php?upload=success");
                        }
                    }
                }
            }else{
                echo "Photo is too big!";
                exit();
            }
        }else{
            echo "An error ocurred while uploading your image!";
            exit();
        }
    }else{
        echo "File type not supported";
        exit();
    }
}
?>

Answer 1

例如，如果您使用此代码，则可以从 DB (MySQL) 加载图像:)

<?php
       $connection =mysql_connect("localhost", "root" , "");
       $sqlimage = "SELECT * FROM userdetail where `id` = '".$id1."'";
      $imageresult1 = mysql_query($sqlimage,$connection);

      while($rows = mysql_fetch_assoc($imageresult1))
    {       
       echo'<img height="300" width="300" src="data:image;base64,'.$rows['image'].'">';
    }
    ?>