New to PHP, and found a solid code that's able to retrieve JSON data from Twitter API and echo it out. Here's the PHP code that retrieves the user-pic of #Guardian (newspaper)
<?php
$data = json_decode(file_get_contents('https://api.twitter.com/1/users/lookup.json?screen_name=guardian'), true);
echo $data[0]['profile_image_url'];
?>
The JSON data for the image is:
profile_image_url":"http://a0.twimg.com/profile_images/1857791844/G_twittermain_normal.jpg"
So obviously it echoes the url (text), i'd like to output the actual image and possibly have control over the sizing of it too if possible.
Thanks in advance!
<img src=" <?php echo $data[0]['profile_image_url'];?>" />
</body>
Related
First posting here. I know inline php is not preferred but I haven't converted all my scripts to echo json_encoded arrays to work in javascript on the client side...so for now, I have inline php.
I do not know the extension of the user uploaded media because it could be a jpg,mp4,etc and upon upload it goes into a media folder with the user id as an identifier.
When my user first loads the div (and html page), the php script cycles through an array and does a fetch_assoc from sql query to the database each time; It returns the (media_id #) and prints out an li with the respective media displayed next to some other values from the query.
I only know the (media_id) and the file path name without the extension. When the page first loads, everything works great and the file_exists function returns correctly.
THE PROBLEM
When I AJAX the div and do the query again, because the user added a row to the database, the new list prints out with all info, BUT the file_exists function doesn't recognize the exact same paths as before and I don't have an img or video on the page.
I copy/pasted the exact same code from the original div and put it in a file for ajax to re-query and print the new li's.
All variables are the same and when I hard code a test filepath, it prints fine. Maybe there's a caching issue?
THE CODE
<?php
$result=$conn->query($select);
$row=$result->fetch_assoc();
?>
<li>
<?php
if ($row['count']>0) {
echo "<div class='media-container'>";
$pathname = "uploads/".$row["id"]."media1";
$testjpg=$pathname.".jpg";
$testjpeg=$pathname.".jpeg";
$testpng=$pathname.".png";
$testmp4=$pathname.".mp4";
if (file_exists($testjpg)==TRUE || file_exists($testpng)==TRUE || file_exists($testjpeg)==TRUE) {
echo '<img src="'.$pathname.'">';
}if(file_exists($testmp4)==TRUE) {
echo "<video></video>";
}
echo "</div>";
}?>
</li>
I could use some advice on how to fix this and how to print appropriate media tags on unknown media types.
THE OUTPUT
<div class='media-container'>
</div>
DEBUGGING ATTEMPTS
echoing the exact file path of a known image in an <img> tag works fine. putting echo'test'; inside the file_exists case does nothing.
--
Solution (Kind of)
So I've used html's onerror before and I found a workaround, though I'd still like to know why I was getting an error. PSA this uses JQuery but javascript works too:
My Solution
<script>
function img2video(el, src) {
$( el ).replaceWith( '<video class="videoClass"><source src="'+src+'" type="video/mp4"></video>' );
}
</script>
<body>
<img style="width:100%" onerror="img2video(this,'<?php echo$pathname;?>')" src="<?php echo$pathname;?>">
</body>
Alright, so here's the final answer I made to best fit the problem using glob:
Javascript:
function img2video(el,src,place) {
if (place=='type') {
$( el ).replaceWith( '<video controls controlsList="nodownload" disablePictureInPicture style="width:100%;object-fit:contain;" preload="auto"><source src="'+src+'" type="video/mp4"></video>');
}
}
PHP:
<?php for ( $i=1; $i <= $limit; $i++) {
$path ="[DIRECTORY]/".$row["id"]."media".$i;
$path = (!empty(glob($path . '*.{jpg,png,jpeg,avi,mp4}', GLOB_BRACE)[0])) ? glob($path . '*.{jpg,png,jpeg,avi,mp4}', GLOB_BRACE)[0] : false;?>
<div>
<img onerror="img2video(this,'<?php echo$path;?>','type',<?php echo$row["id"];?>,<?php echo$i;?>)" src="<?php echo$path;?>">
</div>
<?php } ?>
I don't know how to mark as duplicate, if someone could help with that. My answer uses Glob_Brace from #Akif Hussain 's response on This Question.
I'm trying to scrape an image url from twitter e.g. 'https://pbs.twimg.com/media/BGZHCHwCEAACJ19.jpg:large' using php. I have found the following php code and file_get_contents is working but I don't think the regurlar expression is matching the url. Can you help debug this code? Thanks in advance.
Here is a snippet from twitter which contains the image:
<div class="media-gallery-image-wrapper">
<img class="large media-slideshow-image" alt="" src="https://pbs.twimg.com/media/BGZHCHwCEAACJ19.jpg:large" height="480" width="358">
</div>
Here is the php code:
<?php
$url = 'http://t.co/s54fJgrzrG';
$twitter_page = file_get_contents($url);
preg_match('/(http:\/\/p.twimg.com\/[^:]+):/i', $twitter_page, $matches);
$imgURL = array_pop($matches);
echo $imgURL;
?>
Something like this should provide a URL.
<?php
$url = 'http://t.co/s54fJgrzrG';
$twitter_page = file_get_contents($url);
preg_match_all('!http[s]?:\/\/pbs\.twimg\.com\/[^:]+\.(jpg|png|gif)!i', $twitter_page,$matches);
echo $img_url=$matches[0][0];
?>
Response is
https://pbs.twimg.com/media/BGZHCHwCEAACJ19.jpg
It appears that your regular expression is missing part of the beginning of the URI. It was missing the 'pbs' part, and was not able to determine if http or https.
preg_match('/((http|https):\/\/pbs.twimg.com\/[^:]+):/i', $twitter_page, $matches);
I have a php variable which contain the whole source code of image.What I want is to echo the tag with its attributes src,height & width and send the source code of image through mail (i.e ).
$imagename = abc.jpg;
$concatpath = SITE_URL."/uploads/affiliatesAdv/".$imagename;
$imagesrc = '<img src="'.$concatpath.'" height="200px" width="200px">';
Now when I echo $imagesrc it displays the image. How can I show the source code instead?
Try this,
echo "<pre>";
echo $imagesrc;
echo "</pre>";
You need to use html encoding.
Some thing like this
<img src="abc.jpg"/>
to
<img src="abc.jpg"/>
See the reference.
You can try this site to do it online..
I'm trying to display an image that is stored in my database. I know it is not best practice to store an image in the database, but just for this purpose its what I need. My problem is when I go to display the image, all I get is a load of scrambled code, which I guess is the image in code but not visually what I want. When I use the header tag to identify the image all I get is a thumbnail to mean it isn't reading it in right. Any help would be great.
Below is code I'm using. I've tried it in both ways using it to display in either the php or in html but can't display the image:
while($row = mysql_fetch_array($result))
{
$Long = $row['Long'];
$Lat = $row['Lat'];
$img = $row['file'];
echo "<b><center>Database Output</b><br><br>";
echo "<td>" .$row['file']."</td>";
echo "----";
echo "<td>" .$row['Lat']."</td></center>";
//echo "<td>" .$row['file']."</td>";
}
?>
<div id="map">
<img src="<? $img?>" alt="">
Thanks for any input
Either you embed the image into your html page as a data URI
<img src="data:image/jpeg;base64,<?php echo base64_encode($img) ?>" />
which is hideously NASTILY horrible for 'large' files - you make it impossible for the browser to cache the image, forcing the user download the base64-encoded data EVERY time they load the page.
or you have a sub-script to serve up the actual image, and have something more like:
<img src="getimage.jpg?id=XXX">
and
<?php
$data = get_image_from_db($_GET['id']);
header('Content-type: image/jpeg');
echo $data;
I wanna return a link inside to <img>. I dont know what is the problem.
categoria.php
<HTML>....
<img src="categoriaMAIN.php?type=celular">
</HTML>
categoriaMAIN.php
<?php
$varcate= $_GET['type'];
if ($varcate == "celular")
echo "http://ecx.images-amazon.com/images/I/41l1yyZuyXL._AA160_.jpg";
?>
categoriaMAIN.php
<?php
switch ($_GET['type'])
{
case 'celular':
header('Location: http://ecx.images-amazon.com/images/I/41l1yyZuyXL._AA160_.jpg');
break;
case '...':
header('Location: http://somesite.com/some/img/path/image.jpg');
break;
//...
}
Everyone else seemed to offer the readfile/grab and forward the content method. I thought I'd let the HTTP protocol do the work for us.
Well, the img tag is pointing to text, not an image.
Try:
header("Content-Type: image/jpg"); //tell the browser that this is an image.
$varcate= $_GET['type']; // you know this part
if ($varcate == "celular")
{
// readfile will grab the file and then output its contents without
// procressing it.
readfile("http://ecx.images-amazon.com/images/I/41l1yyZuyXL._AA160_.jpg");
}
Bit of a warning: if you don't output an image here, then the browser will probably complain about the image it is trying to load. You should add a default.
EDIT
Kristian made the point that this is a lot of work for the server and he is right. It would be much better if you could manage to make it so that the src of the img tag changed directly. The above, however, will get you where you are asking to go, though it may not be the best option.
The img tag has to point at the actual image - not a webpage containing the URL.
<img src="categoriaMAIN.php?type=celular">
has to get evalulated to
<img src="http://ecx.images-amazon.com/images/I/41l1yyZuyXL._AA160_.jpg">
It isn't being right now. How to accomplish this, greatly depends on the rest of your source code and what you actually are trying to accomplish.
This won't work! <img src="" searches for the image file, not its location! Try this:
categoria.php
<?php $varcate='celular'; ?>
<html>....
<img src="<?php include('categoriaMAIN.php'); ?>">
</html>
categoriaMAIN.php
<?php
if ($varcate == "celular")
echo "http://ecx.images-amazon.com/images/I/41l1yyZuyXL._AA160_.jpg";
?>
You give the img a PHP page as its src. So the browser is expecting that PHP page to return an image. Instead, you're having that page simply return a URL to the image. You'll have to change that so that you're actually echo ing the image data. I'm a little rusty will all this, but I think you'd do something like the following:
<?php
$varcate = $_GET['type'];
if ($varcate == 'celular') {
header('Content-type: image/jpg');
echo file_get_contents('http://ecx.images-amazon.com/images/I/41l1yyZuyXL._AA160_.jpg');
}
?>
If you really want to do it this way, your categoriaMAIN.php would need to look more like:
<?php
$varcate = $_GET['type'];
if ($varcate == "celular") {
header("Content-Type: image/jpeg");
echo file_get_contents("http://ecx.images-amazon.com/images/I/41l1yyZuyXL._AA160_.jpg");
}
?>
This will get the actual image data and return it to the browser as an image, which is what the browser needs.