图片从数据库PHP的MySQL

Question

检索时显示空白我有四个文件：

1. main.php我的html提交表单，它提交的图像和文本与图像

2. storeinfo.php它将我的所有数据从html表单传送到数据库中，我的图像和文本已成功提交

3. image.php从数据库中提取图像，并具有头部函数以将aimagetype转换为任何格式图像是png，jpeg等。

4. show.php获取所有与图像一起发布的文本，并显示所有带有文本的图像但是图像不会显示，而是在图像无法显示时收到空白框。

我找不到我的错误，我猜它是与头功能image.php或当我试图显示与节目的HTML img标签的图像。 PHP。将图像（以blob存储）上载到数据库是成功的。 为什么不显示图像？

为了代码的每一页：

1. main.php HTML表单

   <form enctype="multipart/form-data" action="storeinfo.php" method="POST"> 
   
   <table border=0 align=center bgcolor=black width=100%> 
   <tr><td colspan=2><h2>&nbsp</h2></td></tr> 
   </table> 
   
   
   <table border=0 align=center bgcolor=grey> 
   <tr><td colspan=2><h2>Animal Information</h2></td></tr> 
   <tr> 
   <td>Name</td><td><input type=text name="aname"></td> 
   </tr> 
   <tr> 
   <td>Description</td><td><input type=text name="adetails"></td> 
   </tr> 
   <tr> 
   <td>Photo</td><td><input type=file name="aphoto"></td> 
   </tr> 
   <tr> 
   <td></td><td><input type=submit name="submit" value="Store Information"></td> 
   </tr> 
   </table> 
   </form> 
   

2. storeinfo.php

   <?php 
   $conn = mysql_connect("localhost","root",""); 
   if(!$conn) 
   { 
   echo mysql_error(); 
   } 
   $db = mysql_select_db("imagestore",$conn); 
   if(!$db) 
   { 
   echo mysql_error(); 
   } 
   $aname = $_POST['aname']; 
   $adetails = $_POST['adetails']; 
   $aphoto = addslashes (file_get_contents($_FILES['aphoto']['tmp_name'])); 
   $image = getimagesize($_FILES['aphoto']['tmp_name']);//to know about image type etc 
   
   $imgtype = $image['mime']; 
   
   $q ="INSERT INTO animaldata VALUES('','$aname','$adetails','$aphoto','$imgtype')"; 
   
   $r = mysql_query($q,$conn); 
   if($r) 
   { 
   echo "Information stored successfully"; 
   } 
   else 
   { 
   echo mysql_error(); 
   } 
   ?> 
   

3. image.php

   <?php 
   
   $conn = mysql_connect("localhost","root",""); 
   if(!$conn) 
   { 
   echo mysql_error(); 
   } 
   $db = mysql_select_db("imagestore",$conn); 
   if(!$db) 
   { 
   echo mysql_error(); 
   } 
   $id = $_GET['id']; 
   $q = "SELECT aphoto,aphototype FROM animaldata where id='$id'"; 
   $r = mysql_query("$q",$conn); 
   if($r) 
   { 
   
   $row = mysql_fetch_array($r); 
   $type = "Content-type: ".$row['aphototype']; 
   header($type); 
   echo $row['aphoto']; 
   } 
   else 
   { 
   echo mysql_error(); 
   } 
   
   ?> 
   

4. show.php

   <?php 
   //show information 
   
   
   $conn = mysql_connect("localhost","root",""); 
   if(!$conn) 
   { 
   echo mysql_error(); 
   } 
   $db = mysql_select_db("imagestore",$conn); 
   if(!$db) 
   { 
   echo mysql_error(); 
   } 
   
   $q = "SELECT * FROM animaldata"; 
   $r = mysql_query("$q",$conn); 
   if($r) 
   { 
   while($row=mysql_fetch_array($r)) 
   { 
   //header("Content-type: text/html"); 
   echo "</br>"; 
   echo $row['aname']; 
   echo "</br>"; 
   echo $row['adetails']; 
   echo "</br>"; 
   
   //$type = "Content-type: ".$row['aphototype']; 
   //header($type); 
   
   //$lastid = mysql_insert_id(); 
   // $lastid = $lastid; 
   //echo "Your image:<br /><img src=image.php?id=$lastid />"; 
   
   echo "<img src=image.php?id=".$row['id']." width=300 height=100/>"; 
   
   
   } 
   } 
   else 
   { 
   echo mysql_error(); 
   } 
   
   
   ?> 
   

Answer 1

所有我发现了如何做到这一点，你正试图在这里做的东西教程第一：你 http://www.mysqltutorial.org/php-mysql-blob/

第二应该使用mysql_escape_string（file_get_contents（$ _ FILES ['aphoto'] ['tmp_name']））而不是addshlashes。

根据这2条规则，你应该能够弄清楚你的代码有什么问题，你也可以尝试使用更小的图片。

Answer 2

您的代码有许多问题，但最值得注意的是您正在使用deprecated mysql functions，并且您的代码易受SQL injection attack影响。

我已将storeinfo.php和image.php改写为使用mysqli扩展名，并使用参数绑定来缓解SQL注入。我会留下重写show.php作为练习。

请注意，我对表的结构做了一些假设，因此您可能需要对SQL代码进行一些调整。

storeinfo.php

$aname = $_POST['aname']; 
$adetails = $_POST['adetails']; 
$aphoto = file_get_contents($_FILES['aphoto']['tmp_name']); 
$image = getimagesize($_FILES['aphoto']['tmp_name']);//to know about image type etc 
$imgtype = $image['mime']; 

$conn = new mysqli("localhost","root","", "imagestore"); 
if ($conn->connect_errno) { 
    echo "Failed to connect to MySQL: (" . $conn->connect_errno . ") " . $conn->connect_error; 
} 

if (!($stmt = $conn->prepare("INSERT INTO animaldata (aname, adetails, aphoto, aphototype) VALUES(?, ?, ?, ?)"))) { 
    echo "Prepare failed: (" . $conn->errno . ") " . $conn->error; 
} 
if (!$stmt->bind_param("ssbs", $aname, $adetails, $aphoto, $imgtype)) { 
    echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error; 
} 
$stmt->send_long_data(2, $aphoto); 

if (!$stmt->execute()) { 
    echo "Insert failed: (" . $conn->errno . ") " . $conn->error; 
} else { 
    echo "Information stored successfully"; 
} 

$conn = new mysqli("localhost","root","", "imagestore"); 
if ($conn->connect_errno) { 
    echo "Failed to connect to MySQL: (" . $conn->connect_errno . ") " . $conn->connect_error; 
} 

if (!($stmt = $conn->prepare("SELECT aphoto, aphototype FROM animaldata where id=?"))) { 
    echo "Prepare failed: (" . $conn->errno . ") " . $conn->error; 
} 
if (!$stmt->bind_param("i", $_GET['id'])) { 
    echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error; 
} 

if (!$stmt->execute()) { 
    echo "Select failed: (" . $conn->errno . ") " . $conn->error; 
} else { 
    $stmt->bind_result($aphoto, $aphototype); 
    $stmt->fetch(); 

    header("Content-type: ".$aphototype); 
    echo $aphoto; 
}