I'm using cakephp 2 and I'm trying to allow my users to upload an image which is then used as the background for their page.
I have the images uploading fine and saving to a (webroot)files/User/$userid/$bodybgimage. I've also got the name of the image in the database.
Trouble is I want because I don't know in advance what user id is going to be requesting their background image I can't put their background in the stylesheet, so I'm trying to get it added in the default layout. I've tried to add with jquery: (the image filename is in $bodybgimage)
<?php if(!$bodybgimage == '') {?>
$('body').css('background-image', 'url('<?php echo '../files/User/'.$userId.'/'.$bodybgimage; ?>')');
<?php }?>
But this seems to just remove everystyle I had for the body!
I also tried doing adding it straight to the body tag like so:
<body<?php if(!$bodybgimage == '') echo ' style="background-image:url(../files/User/'.$userId.'/'.$bodybgimage.')"';?>>
But again to no avail. Can anyone help me out. I know the image and all's there because if I put the background-image style in the stylesheet the image pops up ok. There must be a sensible way to do this that I'm missing. Plz help!
Please can you try to omit the "../" before the files as i assume that your images founded in a folder named files beside your script not ?
Related
I have this problem, i have this script in php that creates a image on the fly, the problem is that the outputted image on the browser is allright, but i need to change it's name.
Ex: Index.php
<?php $url = "http://www.somesite.com/cls/image_scrc?_aaa=qJ7VgSxWBLG3FfNQVB%2BKI58kfHQulPHZLuLYaZAG6Tk%3D&_ab=3ctGTf547wdsAGjY%2F5FASE%2BpBnbQEAcrhbJzCHQ7mGs%3D&_acb=89e62acf3b4d254abf0c3ab30d6ebb33" ?>
<img src="<?php echo $url ?>" />
The image_scrc.php is the file that creates the image, and as you can see i have several data that is passed by the get method.
In the image_scrc.php i have tryed
header('Content-type: image/jpg');
header('Content-Disposition:inline; filename="'.$random_name_jpeg.'"');
but the html link is is always appearing like this
http://www.somesite.com/cls/image_scrc?_aaa=qJ7VgSxWBLG3FfNQVB%2BKI58kfHQulPHZLuLYaZAG6Tk%3D&_ab=3ctGTf547wdsAGjY%2F5FASE%2BpBnbQEAcrhbJzCHQ7mGs%3D&_acb=89e62acf3b4d254abf0c3ab30d6ebb33
if i select the image on browser and then select copy image link it copies just like this also.
however, when I save the image it assumes the random_name.jpg, but only on save!
i've tried everything, even htaccess rules but nothing seems to work !!
it's this possible to acomplish? transform this
http://www.somesite.com/cls/image_scrc?_aaa=qJ7VgSxWBLG3FfNQVB%2BKI58kfHQulPHZLuLYaZAG6Tk%3D&_ab=3ctGTf547wdsAGjY%2F5FASE%2BpBnbQEAcrhbJzCHQ7mGs%3D&_acb=89e62acf3b4d254abf0c3ab30d6ebb33
to this
http://www.somesite.com/cls/random_name.jpg
i cant have the image on the server side! and must be displayed on the fly
Thanks in advance.
I've been looking for a way to make a custom direct access images on wp-content uploads folder since a year ago, but I still didn't found how to serve / display image in wp-content/uploads folder using custom script (maybe php)
Please take a look at my screensshot here : http://prntscr.com/30sdb8
you can see this wp-content/uploads are still have some additional code, and even this page source are hidden with an image (the same image)
is there anyone know how to do that on a worpdress website?
Thanks for answering my question
Updated
because of some unclearly information in my previous question here I try to explain as clear as I can.
In a default wordpress website if a user direct access an image from wp-content/uploads/ directory (for an example : www.domain.tld/wp-content/uploads/2014/04/image-name.jpg )
there will be only an image and a blank background, but in my screenshot example you can see that there are some additional code in header and footer.
my question is how to make modification like that in wordpress? so I can display a header and of footer on my wp-content/uploads/ Url pages
ps : the website I mean is lincah.com , you can go to google image, site:lincah.com then click on 1 image on the search result you'll be brought to the page I mean.
I hope thats clearly enough.
Thank you
What I could make out is you need a PHP script that could fetch all the images from uploads folder and display on the page.
<?php
$images_list = glob("wp-content/uploads/" . "*");
foreach($images_list as $image) {
echo '<img src="' . $image . '" /> <br />';
}
?>
I know there has to be a better way to do this.
Currently I have a php script which will generate a random image from a certain directory when called.
I have div's calling the background.php file in the stylesheet under the div's background setting
background:url(randomimagescript.php);
There are a lot of little div's on this page right now, all calling separate random image php scripts... is there a way I could use a variable when calling the file, so I can just use one script? I still need to have good styling control over the image, so i'm not sure if there is a better option than calling the script as a background image for a div.
If anyone has any ideas, let me know!
try this (it might not be optimal):
background:url("randomimagescript.php?folder=myfolder");
and in randomimagescript.php:
<?php
$folder = #_$REQUEST['folder'];
$url = "galleries/$folder/thumbs/image.jpg"; // ie, compose image
http_redirect($url); // go and find the image.
?>
It sounds a little crazy, but you could actually make your stylesheet be generated by PHP, and just fill in the blank, so to speak.
background:url(<? echo pickRandomImage(); ?>)
set the background for all divs once on the page using jquery
var image = <?echo randomimagescript.php?>
$("div").css('background', 'url('+ image+ ')');
I am trying to do the following; dynamically pick a server with the image on it, and then show said image in img src="". Yeah I know, I am horrible at explaining stuff like this but this should clear it up:
dl-main.php (on server0.domain.com)
$url = 'http://server2.domain.com/offerimage.php?f='.$_GET["f"];
header( 'Location: '.$url ) ;
offerimage.php (on server2.domain.com)
//Lots of link-protection stuff here
$f = "/".$_GET["f"];
$url = 'http://server2.domain.com'.$uri_prefix.$m.'/'.$t_hex.$f;
echo' <img src="'.$url.'"></img> ';
dl.php (on many other servers)
img src="http://server0.domain.com/dl-main.php?f=lalala.gif"
So it pretty much goes like this: Random person adds img src directing to dl-main.php?f=filename on server0. server0 then decides which server will provide the image. In the above example I am using only one server; server2
Now I simply want dl.php to show the photo hosted on server2.domain.com .
As it stands when I directly visit dl-main.php it succesfully redirects me to dl.php, which then succesfully shows me the image I requested. But when I use dl-main.php in a img src it doesn't show the image. I didn't expect it to work but it was worth a shot, but now I don't know what to do anymore :o
I hope this failed attempt is a good example of what I'm trying to accomplish here.
Thanks!
Here's the problem. You call image from server0 using:
<img src="http://server0.whatever/dl-main.php?f=thatimage.something" />
Where the dl-main.php code redirects to server2. Here, you do:
echo' <img src="'.$url.'"></img> ';
So basically the original img tag would get another img tag instead of the image data. That's why the browser can't render the image. You should echo the content of the image instead of an img tag.
Try using your browser's developer tools and check the request to server2 to verify my guess.
It can't work, your second script (offerimage) is producing text/plain, you should produce image/...in order to use img
I have some thumbnail images with its larger version.I placed the thumbnail images in a page.Now for link I just gave a link
<img src="thumbnail1.jpg>
but for this I have to make different pages for showing larger one.I want to give a link to show them in a single page.means whenever I will click the thumbnail it will open the larger one in a page with the same url but with its name like
imagegallery.php?news=images/largerimage1/13.jpg
imagegallery.php?news=images/largerimage1/14.jpg
so how to do that?
Pretty basic stuff, I suggest you get to read some PHP tutorials on the internet to get some knowledge on one thing and another.
The ?news= part in your URL is a parameter that can be read by PHP. This type is known as $_GET. To get this part you would need $_GET['news'] so if we'd use your first link and place this inside a script: echo $_GET['news']; the page would say images/largerimages1/13.jpg.
In order to get the image loaded on your website we need some simple steps, I'm changing the news parameter into image, that suits better for your script since it ain't news items:
<?php
// Define the path (used to see if an image exists)
$path = 'your/absolute/path/to/public_html/'; # or wwwroot or www folder
// First check if the parameter is not empty
if($_GET['image'] != "") {
// Then check if the file is valid
if(file_exists($path . $_GET['image'])) {
// If an image exists then display image
echo '<img src="'. $_GET['image'] . '" />;
}
}
?>
Below this script you can put all your thumbnails the way you want. Ofcourse, also for these thumbnails there are some automated options. But I strongly suggest you get a good look at the script above and some beginner PHP tutorials so you completely understand the example given. This still isn't the best method, but it's kicking you in the right direction.
if your imagegallery.php is in root of your domain, you can just add slash as a first char to links like this:
<img src="thumbnail1.jpg>
else you will have to write some php function which it returns BaseUrl of your web. Then it should looks like this:
<img src="thumbnail1.jpg>
maybe you can something like this,
Techincally, there is no thumbnail image, just a stretch version of the regular image
I don't understand which part you don't know how to do:
- the link part?
it should look like
<img src="thumbnail1.jpg>
- or the PHP part (the file called imagegallery.php)?