what is wrong to this code?
header('Content-type: application/x-shockwave-flash');
$video_id = $_REQUEST['id'];
$content = readfile("http://www.youtube.com/watch?v=$video_id");
echo $content;
Why this code is not working? How should look the code?
i've edited like this:
header('Content-type: application/x-shockwave-flash');
$video_id = $_REQUEST['id'];
$content = readfile("http://www.youtube.com/v/$video_id");
echo $content;
and if i get http://localhost/media.php?id=pkyRRD9f0ts in browser it works but if i add it in jwplayer it doesn't work :(
That's because you're pulling the entire webpage as you would view it in a browser, not just the video file (youtube specifically doesn't allow that).
Try echoing an embed html tag instead.
echo <embed src="http://www.youtube.com/watch?v=$video_id">;
Related
after no one answered at this question Php Rss feed use img in CDATA -> content:encoded i try to do something else solving this problem...
how can i load an image from a given url directly into my homepage?
<?php
$url = "...";
$image = file_get_contents("$url");
echo $image;
?>
*i don't want to save the image anywhere... just load the image from the url and show it in my own homepage.
Try this code,work fine on my machine.
<?php
$image = 'http://www.google.com/doodle4google/images/d4g_logo_global.jpg';
$imageData = base64_encode(file_get_contents($image));
echo '<img src="data:image/jpeg;base64,'.$imageData.'">';
?>
You are almost there. When you download the contents of the file you need to encode it to base64 if you do not plan to store it on server.
<?php
$url = '...';
$image = base64_encode(file_get_contents($url));
?>
Then you can display it:
<img src="data:image/x-icon;base64,<?= $image ?>">
how to read a php file and echo it in a html file?
i try to use readfile() , file() file_get_content() to read a php file.
but when i echo file_string its parsed and then show.
how i can prevent to pars stirng var that included php codes.
here my code:
<?php
$path = '..../ex.php';
$source = fopen($path , "r");
echo fread($source,filesize($path ));
fclose($source);
?>
how to echo $source without compiled or parsed.
With this function
<?php
echo htmlspecialchars($text);
?>
php.net/htmlspecialchars
fread should works fine. Remember that when you use echo it prints <?php opening tag and in rendered page it can be not visible.
To test it, just try with var_dump:
$content = fread($source,filesize($path));
var_dump($content);
I'll go out on a limb and guess that the code is not showing up completely in your HTML page, because the browser is trying to interpret <?php as HTML tags. The solution is to HTML encode any text which may contain characters with a special meaning in HTML:
echo htmlspecialchars(file_get_contents('..../ex.php'));
See The Great Escapism (Or: What You Need To Know To Work With Text Within Text).
use this
$path = 'ex.php';
$source = fopen($path , "r");
echo "<textarea style='border:0px; overflow: hidden; width:100%; height:100% '>";
echo fread($source,filesize($path ));
echo "</textarea>";
fclose($source);
i want get the html source of a page but give a error!!
i use this code for give:
<?php
header('Content-Type: text/html; charset=utf-8');
$html = file_get_html("http://www.google.com/");
echo $html;
when I want to get the source from here I don't correct response and I get something like these characters
����i�[S$%ٲ�9������
Use the function file_get_contents instead:
http://www.php.net/manual/en/function.file-get-contents.php
you should do like this,note: Download simple_html_dom class here
<?php
include_once('simple_html_dom.php');
$html = file_get_html('http://www.google.co.in');
echo $html;
?>
Hello there i have a php file with the included:
The image shows properly when i access the PHP file, however when I try to show it in the HTML template, it shows as the little img with a crack in it, so basically saying "image not found"
<img src="http://konvictgaming.com/status.php?channel=blindsniper47">
is what i'm using to display it in the HTML template, however it just doesn't seem to want to show, I've tried searching with next to no results for my specific issue, although I'm certain I've probably searched the wrong title
adding code from the OP below
$clientId = ''; // Register your application and get a client ID at http://www.twitch.tv/settings?section=applications
$online = 'online.png'; // Set online image here
$offline = 'offline.png'; // Set offline image here
$json_array = json_decode(file_get_contents('https://api.twitch.tv/kraken/streams/'.strtolower($channelName).'?client_id='.$clientId), true);
if ($json_array['stream'] != NULL) {
$channelTitle = $json_array['stream']['channel']['display_name'];
$streamTitle = $json_array['stream']['channel']['status'];
$currentGame = $json_array['stream']['channel']['game'];
echo "<img src='$online' />";
} else {
echo "<img src='$offline' />";
}
The url is not an image, it is a webpage with the following content
<img src='offline.png' alt='Offline' />
Webpages cannot be displayed as images. You will need to edit the page to only transmit the actual image, with the correct http-headers.
You can probably find some help on this by googling for "php dynamic image".
Specify in the HTTP header that it's a PNG (or whatever) image!
(By default they are interpreted as text/html)
in your status.php file, where you output the markup of <img src=... change it to read as follows
$image = file_get_contents("offline.png");
header("Content-Type: image/png");
echo $image;
Which will send an actual image for the request instead of sending markup. markup is not valid src for an img tag.
UPDATE your code modified below.
$clientId = ''; // Register your application and get a client ID at http://www.twitch.tv/settings?section=applications
$online = 'online.png'; // Set online image here
$offline = 'offline.png'; // Set offline image here
$json_array = json_decode(file_get_contents('https://api.twitch.tv/kraken/streams/'.strtolower($channelName).'?client_id='.$clientId), true);
header("Content-Type: image/png");
$image = null;
if ($json_array['stream'] != NULL) {
$channelTitle = $json_array['stream']['channel']['display_name'];
$streamTitle = $json_array['stream']['channel']['status'];
$currentGame = $json_array['stream']['channel']['game'];
$image = file_get_contents($online);
} else {
$image = file_get_contents($offline);
}
echo $image;
I suppose you change the picture dynmaclly on this page.
Easiest way with least changes will just be using an iframe:
<iframe src="http://konvictgaming.com/status.php?channel=blindsniper47"> </iframe>
I am trying to link a tumblr feed to a website. I found this code (As you can see, something must be broken with it as it doesnt even format correctly in this post):
<?php
$request_url = “http://thewalkingtree.tumblr.com/api/read?type=post&start=0&num=1”;
$xml = simplexml_load_file($request_url);
$title = $xml->posts->post->{‘regular-title’};
$post = $xml->posts->post->{‘regular-body’};
$link = $xml->posts->post[‘url’];
$small_post = substr($post,0,320);
echo ‘<h1>’.$title.’</h1>’;
echo ‘<p>’.$small_post.’</p>’;
echo “…”;
echo “</br><a target=frame2 href=’”.$link.”’>Read More</a>”;
?>
And i inserted the tumblr link that I will be using. When I try to preview my HTML, i get a bunch of messed up code that reads as follows:
posts->post->{'regular-title'}; $post = $xml->posts->post->{'regular-body'}; $link = $xml->posts->post['url']; $small_post = substr($post,0,320); echo '
'.$title.'
'; echo '
'.$small_post.'
'; echo "…"; echo "Read More"; ?>
Any help would be appreciated. Thank you!
That is PHP, not HTML. You need to process it with a PHP parser before delivering it to a web browser.
… it should also be rewritten so it can cache the remote data, and escape special characters before injecting the data into an HTML document.