这是在PHP中显示图像的最合理方式吗？ [关闭]

Question

I have a MySQL database

I have a column id called "picture"

The picture column stores file names

The files corresponding to the file names are stored on my htdocs root directory

I would like to display the image data stored in these file names.

The code below (finally) works buy my question is .... Is this the most efficient way to display the image. When I say efficient I mean is it the quickest way to code this..

require_once("database_connect.php");

$getinfo = mysql_query( "SELECT picture FROM historylist ORDER BY date_played LIMIT 0, 1");  // i think this gets the information
$answer = mysql_fetch_array($getinfo); // i think this stores the information into the variable $answer

{
    echo "<img src=\"/{$answer['picture']}\" alt=\"\" />";
}

Answer 1

If there is only one picture connected to one DB row, I wouldn't bother with storing it at all, but rather name it after row's id. But if there are many, it won't work.

Nevertheless, the way of getting data is indeed inefficient, as well as echoing HTML out

What have it have to be:

<?php
require_once("bootstrap.php");
$pic = $db->getOne("SELECT picture FROM historylist ORDER BY date_played LIMIT 0, 1");
?>

<img src="<?=$pic?>" alt="" />

by closing PHP tag and writing HTML as is you can get rid of all these annoying \\\s which will make your writing indeed efficient.
To get rid of all repetitive DB handling code is also a good idea.