Struggling to display blob image with php

问题

I am building a simple website, I want to allow the users to upload and change their avatars. At present I have been able to upload images to a mysql database, stored as blobs with the code as follows:

//connected to DB, userID fetched

$image = $FILES['fileToUpload']['tmp_name'];
$fp = fopen($image, 'r');
$content = fread($fp, filesize($image));
$content = addslashes($content);
fclose($fp);

$sql = "UPDATE tbUsers SET profileImage = '".$content."' WHERE userID = ".userID;
$result = mysql_query($sql) or die (mysql_error());

When I download the files from phpmyadmin after upload they are saved as .bin files, but can be viewed normally. I'm not sure if this is correct or not. My code to display the images is as follows:

HTML:

<?php echo '<img src ="showPic.php?q='.$_SESSION['profile'].'"/>'; ?>

PHP:

if (!empty($_GET['profile']) && is_numeric($_GET['profile']))
{
  $con = mysql_connect("localhost", "root", "");
  $mysql_select_db("projectDB");
  $sql = "SELECT profileImage FROM tbUsers WHERE userID = ". $_GET['profile'];
  $result = mysql_query($sql) or die (mysql_error());

  header('Content-type: image/jpeg');
  $row = mysql_fetch_object($result);
  echo $row['image_data'];
}

I am unsure if I am attempting to display the image in the correct way, any help (corrections/alternative solutions) would be greatly appreciated :)

回答1:

You can do this :

if (!empty($_GET['profile']) && is_numeric($_GET['profile']))
{
  $con = mysql_connect("localhost", "root", "");
  $mysql_select_db("projectDB");
  $sql = "SELECT profileImage FROM tbUsers WHERE userID = ". $_GET['profile'];
  $result = mysql_query($sql) or die (mysql_error());

  $content = mysql_result($result,0,"file_content");
  $name = mysql_result($result,0,"file_name");
  $type = mysql_result($result,0,"file_type");
  $size = mysql_result($result,0,"file_size");
  header("Content-type: $type");

  echo $content

}

Note : You should have these column in you table where you save your BLOB data

file_name = for save filename

$_FILES['file']['name']

file_type = for save file type

$_FILES['file']['type']

file_size = for save file size

$_FILES['file']['size']

回答2:

You select this

$sql = "SELECT profileImage FROM tbUsers WHERE userID = ". $_GET['profile'];

and refer to not selected column

echo $row['image_data'];

来源：https://stackoverflow.com/questions/12663608/struggling-to-display-blob-image-with-php

标签

php

html

mysql

image

blob