I like the photo that I keep stored in my view using the following code but the problem is I can not play because I need the header('Content-Type: image/jpeg'); implementation would then display., But I have orders sure before header('Content-Type: image/jpeg'); What should I do to display images with no header.
If you're only outputting the image, then it doesn't matter if you have other PHP instructions before it since the only output should be the image.
If you are trying to output an image directly into an HTML document, consider using a data URI:
<img src="data:image/jpeg;base64,<?php echo base64_encode($imagedata); ?>" />
Related
I have a PHP function that does on-the-fly image resizing for thumbnail creation.
I am having trouble as it's just displaying raw image stream instead of the actual image.
My code is using a function called thumbnail:
$thumbnail = thumbnail($item['filename'], 209, 137);
imagejpeg($thumbnail);
I've tried putting in:
header("Content-type: image/jpeg");
However, this just expects the full page to be an image. I have absolutely no idea where to go from here, been working at it for a while. I'd rather not save the image to disk although it's looking like this might be necessary.
You either
Do it the normal way
This mean you point at one url, and serve the contents of one image:
<img src="myimage.php">
and myimage.php is a script that looks like:
header('Content-type: image/jpeg');
imagejpeg($thumbnail);
die;
This technique has the advantage of being.. the normal way of doing things.
OR
Output the image inline
Using data uris outputting the contents as a base64 encoded string
<img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAUA
AAAFCAYAAACNbyblAAAAHElEQVQI12P4//8/w38GIAXDIBKE0DHxgljNBAAO
9TXL0Y4OHwAAAABJRU5ErkJggg==" alt="Red dot">
This technique is most appropriate with small images.
It has the advantage of meaning all the images are in the main http request - at the (possibly sizable) disadvantage of making the page harder to edit/build and possibly negating the benefits of browser caching (unless the html page is itself cached).
Being normal is easier
Regarding this statement in the question:
However, this just expects the full page to be an image
That's right - if you do it the normal way you want to point at your php script with the src attribute of an image tag, and server only an image - i.e. the exact same response as if you were pointing at an image file with a browser.
Unless you have a good reason to do things a different way - the "normal" way is a good place to start.
You can point an html img tag to an php file.
<img src='thumbnail.php?file=<?php echo $item['filename']; ?>' />
Then on your php file you display the image and change the headers since all it is doing is displaying an image.
$thumbnail = thumbnail($_GET['filename'], 209, 137);
imagejpeg($thumbnail);
header("Content-type: image/jpeg");
You need to insert the image like you would a normal image in HTML and create the image in a separate PHP file:
image.php
<?php
$img = imagecreate(100,100); //Create an image 100px x 100px
imagecolorallocate($img, 255,0,0); //Fill the image red
header('Content-type: image/jpeg'); //Set the content type to image/jpg
imagejpeg($img); //Output the iamge
imagedestroy($img); //Destroy the image
?>
myWebpage.html
<img src="image.php" />
render.php
$form = '<b>Hi</b><br /><img src="test.jpg" /><br />New File';
echo $form;
header("Content-Type: image/png");
imagejpeg($form);
But this code not work
I want this page render to img file :
<img src="render.php" />
you are trying to output both HTML and image-data - that doesnt make any sense.
Instead you should put an HTML-img-tag into your website, the src-attribute being render.php which sets the header via header() and directly outputs image data - without any additional HTML.
https://stackoverflow.com/a/1851856/351861
You should use extra php file for images, something like:
<img src="render.php?id=15" />
Because if your php script returns only one picture it is used wrong. so you pass it what image to show.
And render.php might be something like
$pic_id = (int)$_GET['id'];// cast to int because only numbers are allowed here
$pic_path = 'assets/img/';
$f = $pic_path.$pic_id.".jpg";// it's just an example so it's simplyfied
header("Content-Type: image/png");
imagejpeg($form);
The point here is that you can check user's permissions to see the picture very flexible and log some data about downloader.
I have this problem, i have this script in php that creates a image on the fly, the problem is that the outputted image on the browser is allright, but i need to change it's name.
Ex: Index.php
<?php $url = "http://www.somesite.com/cls/image_scrc?_aaa=qJ7VgSxWBLG3FfNQVB%2BKI58kfHQulPHZLuLYaZAG6Tk%3D&_ab=3ctGTf547wdsAGjY%2F5FASE%2BpBnbQEAcrhbJzCHQ7mGs%3D&_acb=89e62acf3b4d254abf0c3ab30d6ebb33" ?>
<img src="<?php echo $url ?>" />
The image_scrc.php is the file that creates the image, and as you can see i have several data that is passed by the get method.
In the image_scrc.php i have tryed
header('Content-type: image/jpg');
header('Content-Disposition:inline; filename="'.$random_name_jpeg.'"');
but the html link is is always appearing like this
http://www.somesite.com/cls/image_scrc?_aaa=qJ7VgSxWBLG3FfNQVB%2BKI58kfHQulPHZLuLYaZAG6Tk%3D&_ab=3ctGTf547wdsAGjY%2F5FASE%2BpBnbQEAcrhbJzCHQ7mGs%3D&_acb=89e62acf3b4d254abf0c3ab30d6ebb33
if i select the image on browser and then select copy image link it copies just like this also.
however, when I save the image it assumes the random_name.jpg, but only on save!
i've tried everything, even htaccess rules but nothing seems to work !!
it's this possible to acomplish? transform this
http://www.somesite.com/cls/image_scrc?_aaa=qJ7VgSxWBLG3FfNQVB%2BKI58kfHQulPHZLuLYaZAG6Tk%3D&_ab=3ctGTf547wdsAGjY%2F5FASE%2BpBnbQEAcrhbJzCHQ7mGs%3D&_acb=89e62acf3b4d254abf0c3ab30d6ebb33
to this
http://www.somesite.com/cls/random_name.jpg
i cant have the image on the server side! and must be displayed on the fly
Thanks in advance.
I have a PHP function that does on-the-fly image resizing for thumbnail creation.
I am having trouble as it's just displaying raw image stream instead of the actual image.
My code is using a function called thumbnail:
$thumbnail = thumbnail($item['filename'], 209, 137);
imagejpeg($thumbnail);
I've tried putting in:
header("Content-type: image/jpeg");
However, this just expects the full page to be an image. I have absolutely no idea where to go from here, been working at it for a while. I'd rather not save the image to disk although it's looking like this might be necessary.
You either
Do it the normal way
This mean you point at one url, and serve the contents of one image:
<img src="myimage.php">
and myimage.php is a script that looks like:
header('Content-type: image/jpeg');
imagejpeg($thumbnail);
die;
This technique has the advantage of being.. the normal way of doing things.
OR
Output the image inline
Using data uris outputting the contents as a base64 encoded string
<img src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAUA
AAAFCAYAAACNbyblAAAAHElEQVQI12P4//8/w38GIAXDIBKE0DHxgljNBAAO
9TXL0Y4OHwAAAABJRU5ErkJggg==" alt="Red dot">
This technique is most appropriate with small images.
It has the advantage of meaning all the images are in the main http request - at the (possibly sizable) disadvantage of making the page harder to edit/build and possibly negating the benefits of browser caching (unless the html page is itself cached).
Being normal is easier
Regarding this statement in the question:
However, this just expects the full page to be an image
That's right - if you do it the normal way you want to point at your php script with the src attribute of an image tag, and server only an image - i.e. the exact same response as if you were pointing at an image file with a browser.
Unless you have a good reason to do things a different way - the "normal" way is a good place to start.
You can point an html img tag to an php file.
<img src='thumbnail.php?file=<?php echo $item['filename']; ?>' />
Then on your php file you display the image and change the headers since all it is doing is displaying an image.
$thumbnail = thumbnail($_GET['filename'], 209, 137);
imagejpeg($thumbnail);
header("Content-type: image/jpeg");
You need to insert the image like you would a normal image in HTML and create the image in a separate PHP file:
image.php
<?php
$img = imagecreate(100,100); //Create an image 100px x 100px
imagecolorallocate($img, 255,0,0); //Fill the image red
header('Content-type: image/jpeg'); //Set the content type to image/jpg
imagejpeg($img); //Output the iamge
imagedestroy($img); //Destroy the image
?>
myWebpage.html
<img src="image.php" />
All,
I am executing a PHP script through CURL which returns a PNG file as an output. How can I show the png file in a php web page?
echo '<img src="urltotheimage.php" alt="Alt text" />';
First of all, why do you need to fetch a PNG through CURL? There's better ways to get a image from server, for example using the <img /> tag..
Anyways, I assume you are getting the binary data in a variable, you can output the image by setting appropriate headers and echoing the data:
header('Content-type: image/png');
echo $image;
as you are receiving the string representing the png and if that is base64 encoded then you can embed directly on the image tag as follows:
<img src="data:image/png;base64,aAbBcCdDeEfFgGhH..." />
where aAbBcCdDeEfFgGhH... would be the image string.
see more on data uri's here: http://en.wikipedia.org/wiki/Data_URI_scheme#Inclusion_in_HTML_or_CSS_using_PHP