ไม่ทราบว่าจะเปลี่ยนเป็นใช้ ข้อมูลจาก db ยังไงครับ คือ โค้ดตัวนี้มันใช้ภาพจาก dir โดยตรงเลย จึงอยากสอบถามท่านเซียนๆ php ว่าจะใช้โค้ดยังไงครับ
Code (PHP)
<?php
/* Configuration Start */
$thumb_directory = 'img/thumbs';
$orig_directory = 'img/original';
$stage_width=600; // How big is the area the images are scattered on
$stage_height=400;
/* Configuration end */
$allowed_types=array('jpg','jpeg','gif','png');
$file_parts=array();
$ext='';
$title='';
$i=0;
/* Opening the thumbnail directory and looping through all the thumbs: */
$dir_handle = @opendir($thumb_directory) or die("There is an error with your image directory!");
$i=1;
while ($file = readdir($dir_handle))
{
/* Skipping the system files: */
if($file=='.' || $file == '..') continue;
$file_parts = explode('.',$file);
$ext = strtolower(array_pop($file_parts));
/* Using the file name (withouth the extension) as a image title: */
$title = implode('.',$file_parts);
$title = htmlspecialchars($title);
/* If the file extension is allowed: */
if(in_array($ext,$allowed_types))
{
/* Generating random values for the position and rotation: */
$left=rand(0,$stage_width);
$top=rand(0,400);
$rot = rand(-40,40);
if($top>$stage_height-130 && $left > $stage_width-230)
{
/* Prevent the images from hiding the drop box */
$top-=120+130;
$left-=230;
}
/* Outputting each image: */
echo '
<div id="pic-'.($i++).'" class="pic" style="top:'.$top.'px;left:'.$left.'px;background:url('.$thumb_directory.'/'.$file.') no-repeat 50% 50%; -moz-transform:rotate('.$rot.'deg); -webkit-transform:rotate('.$rot.'deg);">
<a class="fancybox" rel="fncbx" href="'.$orig_directory.'/'.$file.'" target="_blank">'.$title.'</a>
</div>';
}
}
/* Closing the directory */
closedir($dir_handle);
?>
Tag : - - - -
Date :
2010-07-16 18:35:56
By :
gusses
View :
743
Reply :
2
No. 1
Guest
เห็นโพสไว้นานแล้วไงก็ลองดูแล้วกัน Code (PHP)
/* Configuration Start */
$thumb_directory = 'img/thumbs';
$orig_directory = 'img/original';
$stage_width=600; // How big is the area the images are scattered on
$stage_height=400;
/* Configuration end */
$allowed_types=array('jpg','jpeg','gif','png');
$file_parts=array();
$ext='';
$title='';
$i=0;
include("connect_db.php");
$sql="select * from table";
if($query=@mysql_query($sql)){
while($rs= @mysql_fetch_array($query)){
$left=rand(0,$stage_width);
$top=rand(0,400);
$rot = rand(-40,40);
if($top>$stage_height-130 && $left > $stage_width-230)
{
/* Prevent the images from hiding the drop box */
$top-=120+130;
$left-=230;
}
/* Outputting each image: */
echo '
<div id="pic-'.($i++).'" class="pic" style="top:'.$top.'px;left:'.$left.'px;background:url('.$thumb_directory.'/'.$sr["ฟิวชื่อรูป"].') no-repeat 50% 50%; -moz-transform:rotate('.$rot.'deg); -webkit-transform:rotate('.$rot.'deg);">
<a class="fancybox" rel="fncbx" href="'.$orig_directory.'/'.$sr["ฟิวชื่อรูป"].'" target="_blank">'.$title.'</a>
</div>';
}
}else echo "Can't query sql";
ประมาณนี้อ่ะมั้งแหะๆ
Date :
2010-07-19 12:36:31
By :
Chineji
No. 2
Guest
น่าจะตกตัวแปล $title = ชื่อรูปจาก DB จับมาใส่เองแล้วกัน
Load balance : Server 04
