I'm working on task of share a link on facebook. But I want to display my own static image (independent of url). With help of google and facebook, I come to know that I have to pass image path (which I want to display) in image parameter of sharer.php. But still, it displays image related to that site. I don't know that where is my mistake. If anyone know answer then please explain or suggest me link from where I can understand from beginning. Thank You.
Here is my code.
<?php
$title = urlencode("How to Create a Custom Facebook Share Button with a Custom Counter");
$url = urlencode("http://www.daddydesign.com/wordpress/how-to-create-a-custom-facebook-share-button-with-a-custom-counter/");
$summary = urlencode("Learn how to create a custom Facebook 'Share' button, complete with a custom counter, for your website!");
$image = urlencode("http://www.daddydesign.com/ClientsTemp/Tutorials/custom-iframe-share-button/images/thumbnail.jpg");
?>
<html>
<a id="button" onClick="window.open('http://www.facebook.com/sharer.php?s=100&p[title]=<?php echo $title; ?>&p[summary]=<?php echo $summary; ?>&p[url]=<?php echo $url; ?>&&p[images][0]=<?php echo $image; ?>', 'sharer', 'toolbar=0,status=0,width=550,height=400');" target="_parent" href="javascript: void(0)">
Click to Share
</a>
</html>
Your image http://www.daddydesign.com/ClientsTemp/Tutorials/custom-iframe-share-button/images/thumbnail.jpg is the problem - it's not actually an image but takes you to a regular webpage.
You need to set a specific image which fits facebook's size requirements, the minimum size is 200x200 pixels.
Related
I'm trying to connect to the database and view pictures from there. Despite the fact that all pictures are shown, it is not possible to download them. To download I use tag .
Here is my code.
while($row = mysqli_fetch_array($result))
{
echo '
<tr>
<td align="center">
<a href="data:image/jpeg;base64,'.base64_encode($row['Pilt'] ).'" download>
<img src="data:image/jpeg;base64,'.base64_encode($row['Pilt'] ).'" width=400 class="img-thumnail" />
</a>
</td>
</tr>
';
}
Here is a small part of the database where I get pictures. https://imgur.com/yZGsFzb . So I noticed that only one photo is being downloaded, which size is 150.4KiB.
I expect that by clicking on the image (any size) it will be downloaded. Right now when I click on image appears error like this https://imgur.com/yZrlIH8 whereas I would like to see something like this https://imgur.com/4vkUukp .
I will be very grateful for any help.
I recommend you stop printing large quantities of big images that way, they will be hardcoded in the HTML code and downloaded all at once to the browser's cache, 2 times (3 with the click included). This wastes RAM and bandwidth. The network error you get is probably caused by the abnormally big size of the href in your link (might even look a bit like malware to your browser).
Instead, you want to serve these images dynamically on a PHP script first (by setting the header to 'Content-Type: image/png' and printing the image contents as obtained from your database) and provide a direct URL to the script.
This example works just as described in your screenshots, I'm indicating where you should replace your SQL queries to obtain the image:
while(for all your rows)
{
echo '
<tr>
<td align="center">
<a href="image.php?id=' . $id . '" download>
<img src="image.php?link=' . $id . '" width=400 class="img-thumnail" />
</a>
</td>
</tr>';
}
And image.php would look something similar to:
<?php
$id = htmlspecialchars($_GET['id']);
header('Content-Type: image/png');
$content = store the image contents from the BD here ($row['Pilt'] in your code);
echo $content;
Notice that image.php serves images dynamically and I'm supposedly identifying each image with an id (perhaps your row id? You can think of more secure identifiers too if ids are not public already). This is a common security practice in PHP so that you're not providing absolute file paths or other information to attackers.
Good evening, My first message here, where I read so much and did a lot of research to find a solution but without a good result. As a title, a good way to block uploads, when accessing the page, is contained within the <iframe> tags.
An example of code is: <iframe> width = "560" height = "315" src = "https://www.youtube.com/embed/example" frameborder = "0" allowfullscreen> </iframe>
I have read several codes on how to enter the youtube iframe and allow the upload of external resources later, more precisely after clicking on a preview image, adapting html, csss, and javascript in sync with the youtube video id .
But I did not try this solution because the affected iframes are not just part of youtube but they come from other sources, always with the same arrangement as in the above example, and found nothing that could guide me towards the right direction.
Always hoping I'm well explained, do you have any hints on how to hide iframes load and overlap a page loading image at the click, disappear, showing the contents of the frame? Thank you. Forgive my very poor English.
You can use an example for YouTube and just replace the iframe source along with the other variables with your own data. Just skip the thumbnail image part. Replace the on-click event with your own event if you need to.
The technique remains the same. YouTube or not!
see this vanilla js example
Make sure your video's have unique ID's. Render the HTML like this.
<?php $i = 0; ?>
<?php foreach ($sources as $source): ?>
<?php $i++; ?>
<div class="video" id="video-<?php echo $i; ?>">
<div class="divise" id="divise-<?php echo $i; ?>" data-embed="<?php echo $source; ?>">
<div class="play-button"></div>
</div>
</div>
<?php endforeach; ?>
I'm trying to do some stuff with jquery/js (I don't know which one it is..), for example this script ; http://pastebin.com/V3HpU0NY - > http://blog.mainstreethost.com/light-youtube-embeds-faster-page-load#.VG9HKamkVSJ
But now comes the difficult part.
Url: http://erwin.my89.nl/stage/sieh/
This is the website when you go to the url.
This loads the "gallery (top)" and all content/ and kinda images(lazyload) for all the 'gallery' posts. If this has to be done another way please say so.
When you click a image you'll get this:
I hope you understand the idea.
So everything working ok
but now i go in to the inspector and i encounter this?!
I assume it is because i have the script defined in the loop of the content.. so when i click a gallery image, it loads another bunch of those text/css's.
So my question is: Is this the right way to do it? or should i be doing this different?
while ($loop->have_posts()) : $loop->the_post($post->ID);
if ($metas) {
foreach ($metas as $metakey) {
} elseif ($metakey['url']) {
preg_match(
'/[\\?\\&]v=([^\\?\\&]+)/',
$metakey['url'],
$matches
);
$id = $matches[1];
?>
<div class="youtube" id="<?php echo $id ?>" style="width: 560px; height: 315px;"></div>
<script type="text/javascript" src="<?php bloginfo('template_url'); ?>/js/youtube.js"></script>
<?php
^part of the code, if need more please say so.
Thanks in advance!
Can you put your CSS file/s in the head? That's where it normally goes.
Also, I don't see anywhere in your PHP where it could be loading these <style></style> tags. However it does look like it's going to import the same script at the end of each loop.
Could you provide more information on what your exact problem is?
Let's say I have a url of an image link location that doesn't have an extension (.jpg) but when I go to the url it shows me that the corresponding link is an image? How would echo this url in php?
For example the recaptcha image from recaptcha.net it encodes the string with a public key. When I make an html file and use it outputs the image on the page, but when I try to echo the same attributes in php it just outputs a blank image and doesn't display the google captcha image. I am trying to create a forum login page for vbulletin but in this case it should use the captcha implementation since I don't want to sign up for crawling with bots on the my page by google. Can anyone tell me how I can go about doing this?
So basically after the request is made to obtain the key from the image which is assigned to $key the echo request doesn't output the image:
$key = the data inside the GET parameter 'c' everytime the captcha is requested.
echo "<img src='www.google.com/recaptcha/api/image?c=".$key."'/>";
You can use a image regardless of the file extension. It's the MIME type that matters.
The reason why the following won't work is because of no specified protocol.
<img src="goo.gl/XWyGjO"/>
Working:
<img src="//goo.gl/XWyGjO"/>
<img src="http://goo.gl/XWyGjO"/>
<img src="https://goo.gl/XWyGjO"/>
// = inherit the current protocol.
I want to know is there an easy way in order to be able to place images in a loop as below in like a slider? The user sees first image in a slider and then by using arrows for example, be able to slide to next to next or previous image. If user is on last image and wants to go on next image, it displays the first image again. if user is on first image and click on previous, it goes on last image. So it stays continuous.
Below is code where if there are no images then it displays a blank but if there are images, then displays those images in a slider:
<?php
//start:procedure image
$img_result = '';
if(empty($arrImageFile[$key])){
$img_result = ' ';
}else{
?>
<?php foreach ($arrImageFile[$key] as $i) { ?>
<p><img alt="<?php echo $i; ?>" height="200" width="200" src="<?php echo 'ImageFiles/'.$i; ?>"></p>
<?php } ?>
<?php
}
//end:procedure image
A basic jquery slider is easy to implement with instructions of you fancy it: http://basic-slider.com/
Codrops is a nice site for front-end sollutions, I found this plugin that can maybe can help you, in this tutorial she teaches how to do it, and gives you a demo code too.
http://tympanus.net/codrops/2012/04/05/slideshow-with-jmpress-js/
Enjoy it!