I don't know if anybody's going to know about pretty photo, but it's a image gallery box that opens images in a window like facebook.
http://www.no-margin-for-errors.com/projects/prettyphoto-jquery-lightbox-clone/
Basically a user will click on a persons profile and it will open a page in this pretty photo box. It is capable of loading images and pages but when i do this i get an error saying "image can not be loaded make sure the path is correct."
I'm trying it with google for now to see if i can get the page to open but it won't even work with google. Does anyone know where i'm going wrong and can anyone show me where to fix this?
<?
echo
"<img width=90px height= 90px src=\"data/photos/{$chat['id']}/_default.jpg\"/>"; ?>
've tried putting the image path in the hyperlink?
<?php
echo "<img width=90px height= 90px src=\"data/photos/{$chat['id']}/_default.jpg\"/>";
?>
Related
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 have problem to call image with link.
i implementing url management in Yii. it working fine. but. when call image from database. image and link ie if i hover on the image. pages gets render. so dont want to render or refresh the page. i added the code following please where i did mistake. please suggest me how to call the image only
<?php $image = '<img class="img-responsive" src="'.Yii::app()->baseUrl.'/img/'.$small_recipe.'" />';
echo CHtml::link($image, array('name'=>$reciep_name,)); ?>
Try this way
$imghtml=CHtml::image('gallery/'.$data->gallery'.jpg', $data->picturenumber.'.jpg');
echo CHtml::link($imghtml, array('view', 'id'=>$data->id));
I'm using cakephp 2 and I'm trying to allow my users to upload an image which is then used as the background for their page.
I have the images uploading fine and saving to a (webroot)files/User/$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 ?
I am trying to create a function where the user can click an image and a bigger one will load in a small popup window. I already have the bigger image in the system so it merely needs to load the image but in a window the right size!
Any ideas how I can achieve this?
Thanks.
You might want to look into using one of many js lightbox solutions
http://leandrovieira.com/projects/jquery/lightbox/ for example
Look into window.open. That will let you open a new window of a specified height and width, you just need to do something like:
window.open("<?php echo $url; ?>", "_blank",
"height=<?php echo $height;?> width=<?php echo $width; ?>")
You can get the image size in PHP with getimagesize
I created a responsive javascript only lightbox (no jquery needed) where you can pass links to the bigger image. So your thumbnail HTML should look like this, where your thumbnail-picture goes into the src attribute and the link to the bigger picture goes into the data-jslghtbx attribute:
<img class="jslghtbx-thmb" src="img/lightbox/thumbnail-picture.jpg" alt="" data-jslghtbx="img/big-picture.jpg">
You can also use the gallery function via the data-jslghtbx-group attribute to show multiple pictures, but be sure to hide all image elements (except the thumbnail which triggers the lightbox) via display: none;. Visit github for full documentation. Hope this helps!
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