Question

我多年来一直在努力，检查谷歌的每一个可能的地方，有时候这样做会让我觉得脑子死了。

我这样做是为了它的乐趣，但我仍然坚持使用BLOB从MySQL数据库显示图像这一特定方面。

据我所知，我认为我正在正确上传图像，但每当我尝试显示图像时，它最终只会给我一个默认的图像图标和图像的名称，然后当我检查源，它为图像提出了胡言乱语。

有人可以告诉我我做错了吗？

show_profile.php

<html>
<body>
<a href="main_page.php">View Records</a>
<a href="logout.php">Logout</a>
<div id="body_show">
<link href="show_profile.css" rel="stylesheet" type="text/css">
    <div id="image" class="image" style="width: 152px; height:152px;">
    <?php
            include('database_connection.php');
            if(session_id() ==""){
                session_start();
            }
            $username = $_SESSION['myusername'];
            $query = "SELECT * FROM $dImage_table WHERE username='$username' and imageID = (SELECT MAX(imageID) FROM $dImage_table WHERE username='$username')";    
            $result=mysqli_query($DBConn, $query);
            if(!$result){
            $message =  "<p>
                    There was an error with the query.<br />\n" .
                    "The error was " .
                    htmlspecialchars(mysqli_error($DBConn), ENT_QUOTES) . 
                    ".<br />\nThe query was '" .
                    htmlspecialchars($query, ENT_QUOTES ) . 
                "'</P>\n";
            }
            else if (!mysqli_num_rows($result)){
                $message = "<p>Cannot find Image in the database.</p>\n";
            }
            else{
                while($report = mysqli_fetch_assoc($result)){

                    $imageData = $report['image'];
                    $imageName = $report['imageName'];
                    $imageType = $report['imageType'];
                    $imageID = $report['imageID'];
                    //echo "<img src="'data:image;base64,'.$imageData."' alt='Image'/>";
                    //echo "<img src=data:$imageType;base64,base64_encode($imageData) alt=$imageName width=152px height=152px/>";

                    echo '<img src="data:'.$imageType.';base64,'. base64_encode($imageData).'" alt="'.$imageName.'" width="152px" height="152px" />';
                    //echo "</br>";

                }

                //echo $imageData;
            }
    ?>
    </div>
    <div class="form_body">
    <form id="form_body"  action="pic_uploader.php" method="POST" enctype="multipart/form-data">
        <input type="file" name="image"><input type="submit" name="submit" value="Upload">
    </form>
    <div>
    <?php
        //echo $message;
    ?>
</div>  

</body>

pic_uploader.php

<?php
    include('database_connection.php');
        session_start();

        if(isset($_POST['submit']))
        {

            $username = $_SESSION['myusername'];
            $imageName = mysqli_real_escape_string($DBConn, $_FILES["image"]["name"]);
            $imageData = mysqli_real_escape_string($DBConn, file_get_contents($_FILES["image"]["tmp_name"]));
            $imageType = mysqli_real_escape_string($DBConn, $_FILES["image"]["type"]);
            $imageData= base64_encode($imageData);

            if(!substr($imageType,0,4) == "image")
            {
                $message = "<p>Only Images are allowed!</p>";
                include 'show_profile.php';
            }
            else{
                $query = 'INSERT INTO dImage_table (username, imageID, imageName, imageType, image) 
                VALUES("'.$username.'",
                       "'. "" .'",
                       "'.$imageName.'",
                       "'.$imageType.'",
                       "'.$imageData.'")';
                if(!mysqli_query($DBConn, $query))
                {
                    $message = "<p>
                    There was an error uploading the image.<br />\n" .
                    "The error was " .
                    htmlspecialchars(mysqli_error($DBConn), ENT_QUOTES) . 
                    ".<br />\nThe query was '" .
                    htmlspecialchars($query, ENT_QUOTES ) . 
                    "'</P>\n";
                }
                else
                {
                    $message = "<p>Image uploaded</p>";
                    include 'show_profile.php';
                }
            }
        }
?>

感谢您的输入

Answer 1

将图像上传到Web服务器并存储图像的URL，从未见过任何人将图像直接存储到mysql数据库中。这是一个关于如何上传文件的简单教程：http://www.w3schools.com/php/php_file_upload.asp

通过数据库显示图像php

1 个答案: