将Blob图片(存储在数据库中)下载到我的计算机上 [英] download blob image (stored in the database) to my computer

问题描述

我的Blob图片存储在数据库中，我想将其下载到设备中.

i have blob images that are stored in the database I want to download it to my device.

• 该图像将被下载(作为***.jpg)，但已损坏. 这是我的download.php代码

• the image will be download (as ***.jpg) but it is corrupted . this is my download.php code

<?php
$servername = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "text_blob1";
$con= new mysqli($servername, $dbusername, $dbpassword, $dbname); 

$id=$_GET["id"];
$sql = "select * from table1 where id=$id "; // 1
$res = $con->query($sql);
while($row = $res->fetch_assoc())
 { 
 $name = $row['name'];
  echo "<br>";
 $size =  $row['size'];
  echo "<br>";
 $type = $row['extension'];
  echo "<br>";
 $image = $row['image'];
 }

  header("Content-type: ".$type);
  header('Content-Disposition: attachment; filename="'.$name.'"');
  header("Content-Transfer-Encoding: binary"); 
  header('Expires: 0');
  header('Pragma: no-cache');
  header("Content-Length: ".$size);

  echo $image;
  exit($image);   
   ?>

感谢< 3

推荐答案

您输出到页面的所有内容均视为文件内容.我想您不希望"<br>"出现在文件的内容中.

Everything you output to the page is considered a file contents. I suppose you don't want "<br>" to be in the contents of your file.

第二点-在设置标题之前不要进行任何输出.

Second point - do not do any output before setting headers.

第三点-exit('string')输出'string'，因此您可以使用echo和exit输出文件两次:

Third point - exit('string') outputs 'string', so you output content of your file twice: with echo and with exit.

因此，您的代码应如下所示:

So, your code should look like:

$id=$_GET["id"];
$sql = "select * from table1 where id=$id "; // 1
$res = $con->query($sql);
while($row = $res->fetch_assoc())
{ 
    $name = $row['name'];
    $size =  $row['size'];
    $type = $row['extension'];
    $image = $row['image'];
}

header("Content-type: ".$type);
header('Content-Disposition: attachment; filename="'.$name.'"');
header("Content-Transfer-Encoding: binary"); 
header('Expires: 0');
header('Pragma: no-cache');
header("Content-Length: ".$size);

echo $image;
exit();

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