根据URL生成PHP页面
Apologies if this has been asked before. Basically I did a search on Google and found nothing relevant.
I want to be able to create a template and use that template to generate dynamic content based on the URL.
So if I go to mydomain.com/mypage.php?img=source
It should generate my mypage.php template and change the img or any other content source to that of "source"
so if I go to mydomain.com/mypage.php?img=bird
my mypage.php will show and the image or other content whose variable needs to be changed will output bird
I hope that is clear
如果之前已经询问过此道歉。 基本上我在Google上进行了搜索,没有发现任何相关内容。 p>
我希望能够创建模板并使用该模板根据URL生成动态内容。 p> \ n
所以,如果我去mydomain.com/mypage.php?img=source
nn它应该生成我的mypage.php模板并更改img或任何其他内容源 对于“来源” p>
所以,如果我去mydomain.com/mypage.php?img=bird
nnn我的mypage.php会显示 并且需要更改其变量的图像或其他内容将输出鸟 p>
我希望这是明确的 p> div>
Try this:
if ($_GET['img'] == 'source') {
// do stuff
} elseif ($_GET['img'] == 'bird') {
// do other stuff
} else {
// do stuff if 'img' is not set or is empty
// or just none of the above
}
Ok, I keep adding stuff to this!
<?php
if(isset($_REQUEST['img'])) {
}
else {
$_REQUEST['img'] = "false";
$image = "false";
}
$image = $_REQUEST['img'];
$image = strtolower($image);
if ($image == 'false'){
$content = 'no image specified';
}
else{
$content = print '<img src="images/'.$image.'.jpg">';
};
?>
To change the image size, you could just add a width and height before src
.
Then, to print the image, all you have to do is add this where you want it to show:
<?php print $content; ?>
well. apparently in order to achieve what i wanted, i need to use MYSQL so that the information can be retrieved via a query. none of the other answers met my requirement.
this website helped a lot with a tutorial which i was able to tinker with and achieve my requirement -> http://www.vdesignourweb.com/cmsphpsqlb/cms_intro.html