So when users sign up, they get an email, and I'm trying to add an image to that email. That image is dynamically generated using php, so its not a static url. The code works, but when I add it to the email code, the image shows up blank. I know the code works because I have echo'd the body, and the image showed up fine in my browser but empty in the email (i have a border around the image, the border shows up, inside is empty).
$img = "<img src='../inc/barcode.php?id=".$regid."' style='padding:5px; border:2px dashed '/>";
elsewhere
$body .= $img;
So does anyone know why the image won't display, and how I could make it so it does?
Also, another question to kind of go off this, how do I display a generic url for the image so that instead of /img.php?id="" it just reads /img.php or something.
You are using relative links, not absolute.
$img = '<img src="http://www.mydomain.com/inc/barcode.php?id='.$regid.'" style="padding:5px; border:2px dashed" />';
Related
How can I resize an image on mouse hover using PHP with $_GET? I know that this can easily be done with css or js but I want to know how to do it with php. I don't want to upload a file either. So let's say I have html code with an <img> tag and an image, how can I use PHP or $_GET to resize that image on hover?
EDIT:
What i mean is to put the image between <a> tags and then when clicked it refreshes the page but adds the the name of the image to the url so then I could use $get to access and echo a style to resize it.
Resize question on Stack Overflow
PHP is server side and you would have to refresh your page to display a resized image on hover(although I'm not sure if you can implement action on hover without js) and if you don't want to refresh you need to use AJAX(and that requires Javascript)
Going off of your edit, are you looking for something like this?
<?php
if(isset($_GET['image'])){
echo "<img src={$_GET['image']} style='height: 100px; width: 100px;' />";
}
?>
<img src="image_name.jpg" />
I have PHP code that generates HTML code which makes a grid of images. The images are taken from links that are generated for each new image, I also add 133x100 at the end of the image link to resize it on the page. My problem is that a seemingly random selection of images won't display, and I just get a broken image symbol. For example:
This is a link to an image that is generated on my page and is displayed.
This is a link to an image that will not be displayed.
I am only allowed to post 2 links, but removing the %20.%20/133x100 from the end of the last link will show what the picture should be.
Here is the part of the code for the image source:
function display_images(){
//This cycles through each image and displays it as HTML
while($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= '" . $link['Image_Link'] . "/133fx100f' />";
}
Just did some trial and error and found out. No explanations for this though!!
I'm using php to display image(s) and accompanying text (This is done using a WHILE loop) See below:
While (true) {
echo "
<h2>$post_title</h2>
<img src ='new_images/$post_image' width='200' height='200'/> // Image
<div>$post_content</div> "; // Text
}
I have used CSS to float the image left. Therefore,the accompanying text appears to the right of the image.
However, the second image and accompanying text is NOT sitting below the first. It's trying to force itself into any space to the right of the first image
Can I use CSS to ensure the images are stacked??
I hope that above makes sense?
add clear:both to your images and it should work just fine
I have an ul with a li that cannot find a .PNG image associated with the list item 'li' in the code here:
$pngFilename= 'C:/xampp/htdocs/myProj/' . 'just_a.png';
echo 'pngFilename is "', $pngFilename, '" -- that was the .png image filename.';
// within a 'ul' is this li item that displays an image (the ul code is simplified
// to only show the item that (needs to but doesn't!) display a .PNG image.)
echo '<ul>';
echo '<li>';
echo '<a href="http://localhost/myProj/just_an.htm">';
echo '<img src="', $pngFilename, '"';
echo 'alt="http://localhost/myProj/the_other.png"';
// NOTE -- I left out the close /> to the img statement when I copied my code here but
// it was in fact in my source code.
echo '/>';
echo '</a>';
echo '</li>';
echo '</ul>';
Before I wrote the above php code, I tested just the raw html and it was fine -- my .PNG file called
C:/xampp/htdocs/myProj/just_a.png was displayed correctly.
But when I switched to php server-side generation of the html, the C:/xampp/htdocs/myProj/just_a.png image does not appear, only the 'cant find it' default small image appears that looks like a piece of paper torn horizontally.
Any ideas? The .png file exists and so does the directory and the html correctly displayed the image, but when I put the html into php 'echo' calls I must be screwing something up, just not sure what.
To make sure I have the correct path and filename you'll see I echo it out at the top of the code.
The php variable $pngFilename is displayed when I do 'View Source' in the browser as:
pngFilename is "C:/xampp/htdocs/myProj/just_a.png" -- that was the .png image filename.
The only other thing to mention is that the 'alt' link, the 'alt="http://localhost/myProj/the_other.png"' -- this link (not the image) shows as blue underlined link text.
Why did this work in my html but breaks when I use the 'echo' in php? After all, the 'echo' simply sends the html to the client side -- and that .png file is 100% definitely there and displays fine when I run the above html outside of php's "echo" command.
This is because the path to the image changes depending on how your viewing the page. in a local context or from a server context; and since you are using an absolute path instead of a Relative path the system can't adjust for the change in the location of the image. Unlike when using a PHP function that calls on the php file system functions that do use the internal file system. what your doing is having it send a text file to the browser which isn't rendered as HTML code until after PHP has finished. because of that it has no access to the php file system to resolve the path to the image on the server. the way to fix it would be to use the path to the image Relative to the PHP script or use the web accessible path to the image
$pngFilename= 'C:/xampp/htdocs/myProj/' . 'just_a.png';
Should for example be
$pngFilename= 'http://localhost/myProj/' . 'just_a.png';
or if the image and the php file are in the same directory you could just do
$pngFilename= 'just_a.png';
You should try to close your img tag:
echo '<img src="', $pngFilename, '"';
echo 'alt="http://localhost/myProj/the_other.png">';
You are generating a image tag like:
<img src="C:/xampp/htdocs/myProj/just_a.png" alt="http://localhost/myProj/the_other.png" />
You cannot use this path, you need to change it to http://localhost/myProj/just_a.png or something like file:///C:/xampp/htdocs/myProj/just_a.png.
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