图片作为MySQL数据库中的链接？ [英] Images as links in MySQL database?

问题描述

我想将图像链接存储到数据库中，然后将它们回显到页面上。...这可能吗，我该怎么做？ （回显实际图像，而不是图像链接）

I want to store image links into a database and then echo them onto a page.... Is this possible and how do I do it? (echo the actuall images and not the image link)

您可以看到我的代码回显了：$ link，$ description和$ title。我想添加另一个回显 $ imagelink的元素。

As you can see my code echo's out: The $link, $description and $title. I want to add another element which echos $imagelink.

echo "
<div class=\"pagination\" style=\"display:inline\"><ul style=\"background-color:#\"><li><div   class=\"span3_search\"><h2><a href='$link'><b>$title</b></a></h2><br><img id=\"result_img\"   src=\"img/lvmo.png\" /><br />
$description<br />
<a href='$link'>$link<br /><br /></a><p></div></li></ul></div>
";

推荐答案

简短答案

在 src 属性中仅将 echo 移出图像的URL，就会显示该图像

Short answer

Simply echo out the URL of the image in the src attribute and the image will be displayed.

<img id="result_img" src="<?php echo $imagelink; ?>" />

有效解决方案：

表设计：

Worked Solution:

Table Design:

images

| id | title      | link                  | imagelink    | description |
------------------------------------------------------------------------
| 1  | Some Title | http://www.google.com | myimage1.png | Some text   |
------------------------------------------------------------------------
| 2  | My Title   | http://www.yahoo.com  | myimage2.png | Some text   |         

SQL查询

SELECT title, link, imagelink, description FROM images WHERE id = ?

PHP

<?php

$stmt = $mysqli->prepare("SELECT title, link, imagelink, description FROM images WHERE ID = ?");
$stmt->bind_param("i", $ID);
$stmt->execute();
$stmt->bind_result($title,$link,$imagelink,$description);
$stmt->fetch();

?>
<div class="pagination" style="display:inline">
  <ul style="background-color:#">
    <li><div   class="span3_search">
    <h2><a href='<?php echo $link; ?>'><b><?php echo $title; ?></b></a></h2>
    <br />
    <img id="result_img" src="<?php echo $link; ?>" />
    <br />
    <?php echo $description; ?>
    <br />
    <a href='<?php echo $link; ?>'><?php echo $link; ?><br /><br /></a>
    <p></div>
    </li>
  </ul>
 </div>

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