Question

我有这段代码可以帮助我查看文章。这篇文章包含一张照片。照片已保存在数据库中，但正在检索中。它仍然是二进制文件。可能是什么问题？这是以下代码： view_aboutme.php

                <?php
            include ("../connect/connection.php");
            require_once("../connect/connection.php");
            ?>
            <html xmlns="http://www.w3.org/1999/xhtml">
            <head>
            <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
            <title>Admin</title>
            <!----Favicon---->
            <link rel="icon" href="../img/favicon.ico" />
            <body>
            <?php
            $sql= mysqli_query($link,"SELECT * FROM about") or die(mysqli_error($link));
            while($row=mysqli_fetch_assoc($sql)){
            $id=$row['About_ID'];
            $title=$row['About_title'];
            $imagetmp=$row['About_photo'];
            $story=$row['About_post'];

            echo '<tr>
            <td>'.$title.'</td>
            <td>'.$imagetmp.'</td>
            <td>'.$story.'</td>
            <td> <img src = aboutfetch_image.php?About_ID='.$id.' width=100 height=100 ></td>
            <td><a href="edit_aboutme.php?id='.$id.'">Edit</a></td>
            <td><a href="delete_aboutme.php?id='.$id.'">Delete</a></td>
            </tr>';  
            }
            ?>

应该转换图像的文件称为 aboutfetch_image.php 。代码如下：

                <?php
            include ("../connect/connection.php");
            require_once("../connect/connection.php");
            header("content-type:image/jpeg");

            $id=$_GET['About_ID'];

            $sql="select About_photo from about where About_ID=".$id."";

            $var=mysql_query($sql);
            //echo($sql);

            if($row=mysql_fetch_array($var))
            {
             $imagetmp=$row['About_photo'];
            }
            echo $imagetmp;
            ?>

查看保存在Longblob中的图像

0 个答案: