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.
Related
I am trying to use the get_option() function in Wordpress to set the SRC of an image in a theme file. The file is index.php, and represents the homepage.
The current image code is:
<img class="outside-collage-image" <?php echo 'src="'.get_option('article-image-1').'"'; ?>>
However, the image doesn't display, and when I inspect it, it simply says src(unknown) where it should have the proper SRC
I have tried a myriad of fixes, including leaving the SRC out of the php area and simply calling the function, but it doesn't seem to work.
The oddest part is, if I put on the page <p><?php echo 'src="'.get_option('article-image-1').'"'; ?></p> to check the output, it outputs into the <p> the proper code: src="http://image-site.com/my-image.jpg" (obviously the link to the image is fake, but the point is, it's a valid link.)
Any idea what could be causing this?
You can try this to get the image article-image-1.jpg in your root.
<img class="outside-collage-image" src="<?php echo get_site_url().'/article-image-1.jpg'; ?>">
It turns out that I screwed up, and the above function works just fine. I was simply applying it to the wrong image, and didn't realize that until looking at it a day later, when it wasn't 1:30 AM and I wasn't so tired. Thanks to all that answered.
headdesk
I'm completely new to PHP and I'm writing a page which displays some images from my hard drive.
I've tried different syntaxes and none of them worked out for me ...
I've tested
echo '<img src="/var/www/netbeans/PhpProject2/Plate.jpg" alt ="Test" />';
and It didn't show anything for me .... I've tested this code Also and it didn't worked for me too, I've attached My monitor screen page below. as you can see it seems that my images are not found. I've pasted every address of my images in my browser separately and I was able to see them but when I add them in src argument of img function I'm not able to see them!
What do you think might be the source of problem .... I need to work on a project which needs to be fast enough in working with images.. what do you think I should use ? is "
Try to remove the /var/www. Path should be the http path not the file physical path.
echo '<img src="/netbeans/PhpProject2/Plate.jpg" alt ="Test" />';
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
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)?