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)?
Related
I have img tags that have parameters as part of the url. These parameters are used for an ajax image crop. They look like this
<img src="/resize/45-800/files/myImage.jpeg" alt="Chania">
By adding /resize/45-800/ in front of the url
I am able to tell the ajax script to get that file from the files directory, and dump a cropped version into /resize/files/
My question is, is there a way to remove these parameters 45-800/, which can be different each time, for file reading purposes only, so that html will read for the image like this
<img src="/resize/files/myImage.jpeg" alt="Chania">
but the world will see the src still like this
<img src="/resize/45-800/files/myImage.jpeg" alt="Chania">
Is this possible using htaccess?
I know I could use a query string and just use the ajax script to read from the src like this
<img src="/resize/files/myImage.jpeg?params=45-800" alt="Chania">
I was just hoping for a prettier way. I don't really want to use the question mark '?'.
The /resize and /files will always be there every time.
Does anyone know of a way?
I just had an epiphany... I could just put the crop width and height parameters in data attributes instead of in the url... So this
<img src="/resize/files/myImage.jpeg" data-wh="45-800" alt="Chania">
Instead of this
<img src="/resize/45-800/files/myImage.jpeg" alt="Chania">
Then I can serve them from the resize/files/ directory for all images that have the resize parameter passed into the url; and I don't have to use an ugly query string.
"/resize/45-800/files/myImage.jpeg?ugly=45-800"
So I haven't tested this, but I think I found the legit answer to my original question. I know its possible as the people at the last job I worked at used to resize images based on src url params. Here is the link - http://sneak.co.nz/projects/img-resizing/
My plugin is already finished and working now 8/ This method looks a little nicer though, they walk you through the whole image resize process, and use image caching as well.
Hope this helps someone as much as it does me if it works! I'll leave a comment if I end up testing it out and it works well.
I have PHP code that generates HTML code which makes a grid of images. The images are taken from links that are generated for each new image, I also add 133x100 at the end of the image link to resize it on the page. My problem is that a seemingly random selection of images won't display, and I just get a broken image symbol. For example:
This is a link to an image that is generated on my page and is displayed.
This is a link to an image that will not be displayed.
I am only allowed to post 2 links, but removing the %20.%20/133x100 from the end of the last link will show what the picture should be.
Here is the part of the code for the image source:
function display_images(){
//This cycles through each image and displays it as HTML
while($row = $item->fetch()){
Echo "`<img src= '$link[Image_Link] . /133x100' />`"
}
}
It is then called here in a class which puts the images in a grid:
<ul class="rig columns-4">
<?php
display_images();
?>
</ul>
Seemingly about every 2/20 images won't work, and seeing all the links are in the same format, I don't understand why they won't work, and it just seems random.
EDIT: I have noticed that the links that work have 62fx62f at the end of them before the added %20.%20/133x100. If I add it to the raw link in the right place, it makes the image work. But using that generated link, the image still won't load on the page. So using a link with a working image will not work on the page. (This is the same with the raw link without %20.%20/133x100, that links to an image but also won't work on the website)
When visiting the links, the urls look like this:
http://www.example.com/image/randomcharacters%20.%20/133x100
The links work without the %20.%20 at the right dimensions, like so:
http://www.example.com/image/randomcharacters/133x100
This leads me to believe that it may work if you try using the following for the image source instead:
<img src= '$link[Image_Link]/133x100' />
The full code would look like this, for the while function:
while($row = $item->fetch()){
echo "<img src= '" . $link['Image_Link'] . "/133x100' />";
}
I am not aware of steamcommunity much but from the looks of it, i think you should try this.
Instead of putting
. /133x100
Use
/133fx100f
So your URLs would be
while($row = $item->fetch()){
echo "<img src= '" . $link['Image_Link'] . "/133fx100f' />";
}
Just did some trial and error and found out. No explanations for this though!!
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="' . $image . '" /> <br />';
}
?>
I am trying to call images in this function, but they all come up as question marks. Could anyone tell me what I did wrong?
if(!$currentvotes) $currentvotes = 0;
echo '<div class="vote vote'.$id.'"><span>'.$currentvotes.'</span>';
if($user_ID && !$alreadyVoted) echo '<br /><a post="'.$id.'" user="'.$user_ID.'"><img src="/images/thumbsup.png" WIDTH=25 HEIGHT=25></a>';
if($user_ID && $alreadyVoted) echo '<br /><span class="voted"><img src="/images/thumbsup.png" WIDTH=25 HEIGHT=25></span>';
echo '</div>';
if(!$user_ID) echo '<div class="signup"><p><a style="color:#4ec1f3;" href="'.get_bloginfo('url').'/wp-login.php?action=register"><img src="/images/thumbsup.png" WIDTH=25 HEIGHT=25></p></div>';
}
I am trying to call images in this function
As in the comments:
You can't 'call' an image. You can call functions, methods,
procedures, code but images are data and can't be 'called'. – Patashu
[...] they all come up as question marks
Right click on those question marks: choose "open image in new window". You should see a new page with a 404 error. Look at the link of that page: that's the path where your image is supposed to be (if you're getting a 404 error it simply means they are not there). Create the folder images and upload the file thumbsup.png. Now reload the page.
I suppose the path you are using has to be absolute eg (http://.../images/myimage.jpg) instead /images/myimage.jpg. This is common issue when you write custom code (don't use standard framework) and use .htaccess to rewrite urls.
You can set basepath as constant so you can easy change it if you switch domains.
define("BASE_PATH", "http://mydomain.com");
and then in code use it:
BASE_PATH."/images/myimage.jpg"
Or use base tag in your head in html
<base href="http://www.yourdomain" target="_blank">
And as Saturnix said in his answer you should have folder named "images" and filename you put in src.
If an image can't be found in the location you specify in your HTML, the browser will display a broken image link marker, maybe like one of these;
Chrome
IE10
Firefox 10
Each browser has its own marker. Maybe your browser uses a question mark to indicate a broken image link.
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