圖片從數據庫PHP的MySQL

Question

檢索時顯示空白我有四個文件：

1. main.php我的html提交表單，它提交的圖像和文本與圖像

2. storeinfo.php它將我的所有數據從html表單傳送到數據庫中，我的圖像和文本已成功提交

3. image.php從數據庫中提取圖像，並具有頭部函數以將aimagetype轉換爲任何格式圖像是png，jpeg等。

4. show.php獲取所有與圖像一起發佈的文本，並顯示所有帶有文本的圖像但是圖像不會顯示，而是在圖像無法顯示時收到空白框。

我找不到我的錯誤，我猜它是與頭功能image.php或當我試圖顯示與節目的HTML img標籤的圖像。 PHP。將圖像（以blob存儲）上載到數據庫是成功的。 爲什麼不顯示圖像？

爲了代碼的每一頁：

1. main.php HTML表單

   <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

所有我發現瞭如何做到這一點，你正試圖在這裏做的東西教程第一：你 http://www.mysqltutorial.org/php-mysql-blob/

第二應該使用mysql_escape_string（file_get_contents（$ _ FILES ['aphoto'] ['tmp_name']））而不是addshlashes。

根據這2條規則，你應該能夠弄清楚你的代碼有什麼問題，你也可以嘗試使用更小的圖片。

Answer 2

您的代碼有許多問題，但最值得注意的是您正在使用deprecated mysql functions，並且您的代碼易受SQL injection attack影響。

我已將storeinfo.php和image.php改寫爲使用mysqli擴展名，並使用參數綁定來緩解SQL注入。我會留下重寫show.php作爲練習。

請注意，我對錶的結構做了一些假設，因此您可能需要對SQL代碼進行一些調整。

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; 
}