I am trying to get all the image src's and rename the files using this code but it doesnt seem to work.
Any ideas ??
require_once('catalog/controller/forum/simple_html_dom.php');
$test = $data['description'];
$html = str_get_html($test);
foreach($html->find('img') as $element) {
$src = $element->src;
rename($src,$src.".jms");
}
All images files are local and this is an example image tag.
<img alt="Image" src="image/data/attaches/f7ff31f73f6d41f108ef31c01ea69228.png">
So i'm trying to rename the file from that to the same location but by adding .jms to the end.
Also i'm not sure how but i want to modify all the image tags in the string so i can put it back modified with the mysql update command.
The string $data['description'] contains other html not just image tags. Its varied.
rename() could only used to the local files in your server, no way to rename the remote files.
If the images are in your server, and you could know the real path from the src, then you need to convert the src to the real path of the image.
Related
I am trying to migrate some content from one resources into another and need to save some images (several hundred) located at a remote resource.
Suppose I have only the URL to an image:
https://www.example.com/some_image.jpg
And I would like to save it into the filesystem using PHP.
If I were uploading the image, I essentially would do the following:
<input type="file" name="my_image" />
move_uploaded_file($_FILES['my_image']['tmp_name'], '/my_img_directory');
But since I only have the URL, I would imagine something like:
$img = 'https://www.example.com/some_image.jpg';
$file = readfile($img);
move_uploaded_file($file, '/my_img_directory');
Which of course wouldnt work since move_uploaded_file() doesn't take an output buffer as a first argument.
Essentially, I would need to get $img into the $_FILES[] array under this approach. Or may some other approach?
You can use PHP's copy function to copy remote files to a location on your server:
copy("https://example.com/some_image.jpg", "/path/to/file.jpg");
http://php.net/manual/en/function.copy.php
$image = file_get_contents('http://www.url.com/image.jpg');
file_put_contents('/images/image.jpg', $image); //Where to save the image on your server
This is an odd question but I'm stuck on how I would achieve this and I am unable to find any methods of doing so.
I have a simple php script that takes variables (containing file names) from the URL, cleans then and then uses them to generate a single image from the inputted values. This works fine and outputs a new png to the webpage using:
imagepng($img);
I also have a facebook sharing script in PHP that takes a filepath as an input and then shares the image on the users feed where this statement is used to define the image variable:
$photo = './mypic.png'; // Path to the photo on the local filesystem
I don't know how I can link these two together though. I would like to use my generation script as the image to share.
Can anyone point me in the right direction of how to do this? I am not the master of PHP so go easy please.
-Tim
UPDATE
If it helps, here are the links to the two pages on my website containing the outputs. They are very ruff mind you:
The php script generating the image:
http://the8bitman.herobo.com/Download/download.php?face=a.png&color=b.png&hat=c.png
The html page with the img tag:
http://the8bitman.herobo.com/Share.html
Treat it as a simple image:
<img src="http://yourserve/yourscript.php?onlyImage=1" />
yourscript.php
if($_GET['onlyimage']) {
header('Content-type:image/png'); //or your image content type
//print only image
} else {
//print image and text too
}
How can I display an image and pass it as an input parameter in an executable in php without saving the image in a folder. The user gives the image path as input and I am using ajax to display the image when it is selected when I save it to a folder it works but how can I display it without saving it in a folder? My code now is
move_uploaded_file($_FILES["file"]["tmp_name"],"upload/".$_FILES["file"]["name"]);
//echo "Stored in "."upload/".$_FILES["file"]["name"];
echo "<img src='upload/".$_FILES["file"]["name"]."' class='preview'>";
I tried
<img src=$_FILES["file"]["tmp_name"]. class='preview'>
but it didnt work. As I will have thousands of input from thousands of user I dont want to save it. Is there any optimised and efficient method to do this?
I think, its not possible to show image without saving it. You could try to save the image in temp folder on the server side and clean this folder periodically to avoid much space consumption.
The src attribute of the <img> tag should be an URL accessible by the client.
You try to give a local path (ex: path/to/your/file.jpg) of a temporary file as URL, it will not working.
info: The uploaded image is save on the local disk on a temp directory, and could be deleted by PHP later.
If you want to show the image without moving it at a place reacheable by a URL, you can try to load its content as base64 content
$imagedata = file_get_contents($_FILES["file"]["tmp_name"]);
$base64 = base64_encode($imagedata);
and use in your HTML
<img src="data:image/png;base64, <?php echo $base64; ?>" />
I don't think you can show the image without saving it.
You need to save the file either to the filesystem or to memory if you want to later output
Your problem here is that $_FILES only exists in the script that the image was sent to. so when you initiate another http request for img source, php no longer has any clue what file your trying to read.
You need a way to tell which image to be read on http request.
One thing you can do is that you can save the file in a place accessible by the client and then just have php delete it after you output it. So once the image is outputted it will be deleted and no longer be existing in the file system.
Another approach would be to get the image from the memory by directly writing the contents to httpresponse.
You can do this way
$image = file_get_contents($_FILES["file"]["tmp_name"]);
$enocoded_data = base64_encode($image);
and when you show your image tag :
<img src="data:image/png;base64, <?php echo $enocoded_data ; ?>" />
Hope any of these helps you
Let's say I have a user enter the URL of an image.
After URL validation, etc. I want to get the image and store it in a PHP variable. Not the image path, but the actual image itself.
I am adding this image-holding variable in between other strings, so I cannot change use header() before echoing the image. Is it possible to use <img> HTML tags? I really don't want to copy the images to my own server...
How do I:
Store the image in a variable such that,
Echo the image from the variable without changing headers.
Edit:
I said above that I am putting this image inside another variable, e.g.:
$str = "blah blah" . $var_holding_img . "more text";
Is it possible to insert something in the string above that will be replaced with the images? Can parse the variable $str later to replace some random text like "abg30j-as" with the image...
I found an answer to my own question:
First, I created another PHP file, called img.php:
<?php
$url = $_GET['imgurl'];
/*
Conduct image verification, etc.
*/
$img_ext = get_ext($url); //Create function "get_ext()" that gets file extension
header('Content-type: image/' . $img_ext);
echo file_get_contents($url);
?>
Then, in the original PHP file, I used this PHP code:
<?php
$var_holding_img = '<img src="img.php?imgurl=http://example.com/image.png"/>';
$string = "This is an image:<br \>" . $var_holding_img . "<br \>displayed dynamically with PHP.";
echo $string;
?>
This way, the PHP file "img.php" can use the proper headers and the image can be inserted as HTML into any other PHP variable.
How do I:
Store the image in a variable such that,
Echo the image from the variable without changing headers.
You can do this in two ways.
In way one, you serialize the image in a string, and save the string in the session. Which is exactly the same as saving it server side, except that now the session GC should take care of clearing it for you. Then the IMG SRC you use will redirect to a script that takes the image and outputs it as image with proper MIME type.
In way two, if the image is small enough, you can encode it as BASE64 and output it into a specially crafted IMG tag:
http://www.sweeting.org/mark/blog/2005/07/12/base64-encoded-images-embedded-in-html
This saves you some time in connection, also. Of course the image must be reasonably small.
You can't save the actual image in a variable. Either you save the URL or copy the image (what you obvious don't want) to your server and save the path to the image
See answer 1, you can't echo the image itself, only link it
Edit: Okay obviously you can save images directly to a variable, but I don't recommend you to do this.
No, that isn't possible. If you want to serve something, it has to exist on the server.
This is a bit of a shot int he dark, but i dont even know how to google this to get started.
It's a very simple idea, but it would be extremely useful to designers and creatives for creating a scrapbook of sorts.
Baiscally i want to be able to drop an image I find on the internet onto an FTP folder which will just be a droplet on my desktop. The img gets uploaded to the folder. Then a Chron job on the server (I'm assuming that's what i'd need to set up?) checks to see if any new images are up, and then automatically creates code for the image and updates a single html file.
Really simple. But how would one go about that? PHP? I really dont know where to start as I'm only ok at a bit of jQuery and templating wordpress.
I mean maybe I could do this in wordpress too. That it just checks to see if any images are in the uploads folder and then creates a single post with no titles, comments and all that BS, but just inserts the image.
Any pointers would be immensely appreciated!
Best,
Marc
As I understand you correctly, you just want to have a page showing every image in a specific folder. That can easily be done with the glob() function:
foreach(glob('folder_with_images/*.jpg') as $image){
echo '<img src="'.$image.'" alt="" />';
}
That's all. Just set up the "query" for the globas you need it. You don't need to do any kind of cronjob or update. It simply show all available images "live" as you open the page. I use glob() and not opendir() because you can use filter like "*.jpg" which you can't do with opendir().
When you have multiple file types, you can use scandir() and a if-statement:
foreach (scandir('folder_with_images/') as $image) {
if(substr($image, -3) == 'jpg' || substr($image, -3) == 'gif' || substr($image, -3) == 'png'){
echo '<img src="'.$image.'" alt="" />';
}
}
Check out the function opendir in php :
http://us.php.net/opendir
You can just have a php page that does opendir on your ftp directory, then spits out an img tag for each .jpg file in the directory. This can read teh directory each time you open the page.