如何像Twitter一样在后台显示数据库中存储的重复图像？

Question

I want to display an image on background (by using CSS repeat) the same way as on Twitter but the problem is that I am retrieving it from a MySQL database and CSS cannot handle the src tag on background.

How can I retrieve an image from the MySQL database (I know how to display image by retrieving from database but here I want to display it differently), ie display the image as backround on body and the image should repeat as if CSS repeat statement were used. Like on twitter.com.

My code is is in PHP and MySQL. I need to get the URL for as this image is retrieved from a database.

Answer 1

You cannot repeat <img src="#" /> so for that you need to use CSS in your PHP document, you can do it like

<style>
body {
   background-image: url('<?php echo $whatever; ?>') !important;
   background-repeat: repeat;
}
</style>

Either you can do is

Make a stylesheet with a .php extension.

Than <link rel='stylesheet' type='text/css' href='css/stylesheet.php' />

At the top of the page mention this

<?php
    header("Content-type: text/css; charset: UTF-8");
?>

Now you can set variables accordingly

<?php
    header("Content-type: text/css; charset: UTF-8");
    $background = "$imgsrc"; /* Retrieve your image here */

    /*Now simply use the $background variable for setting the body background */
?>

Answer 2

You can do this in two steps.

Create a PHP script that accepts a parameter to identify through a unique ID, which Row has the image to display. This script will extract the image from database and send the codes with appropriate mime-type, so that browser understands. This way, apply a class to the container (or body tag) and display the background like:

.backgroundTile { background-image: url('/path/to/php-image-render-script.php?image_id=1212') !important; background-repeat: repeat; }

Example PHP Script (Source- http://cookbooks.adobe.com/post_Display_an_image_stored_in_a_database_ PHP -16637.html ) :

<?php require_once('Connections/testConn.php'); ?>
<?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  if (PHP_VERSION < 6) {
    $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
  }

  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? doubleval($theValue) : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}

$colname_getImage = "-1";
if (isset($_GET['image_id'])) {
  $colname_getImage = $_GET['image_id'];
}
mysql_select_db($database_testConn, $testConn);
$query_getImage = sprintf("SELECT mimetype, image FROM images WHERE image_id = %s", GetSQLValueString($colname_getImage, "int"));
$getImage = mysql_query($query_getImage, $testConn) or die(mysql_error());
$row_getImage = mysql_fetch_assoc($getImage);
$totalRows_getImage = mysql_num_rows($getImage);
header('Content-type: ' . $row_getImage['mimetype']);
echo $row_getImage['image'];
mysql_free_result($getImage);
?>