I've a problem with some lines of code. I try to generate a picture with php and the gd2 librarie.
<?php
header('Content-type: image/jpeg');
$im = imagecreatefromjpeg('profil.jpg');
imagejpeg($im);
imagedestroy($im);?>
This code doesn't work. There's no image displayed, just a message : "this image can't be displayed because there's error" displayed in white on grey background.
The picture is well loaded and "$im" is a resource created with "imagecreatefromjpeg"...
If i remove all lines except header('Content-type: image/jpeg');, I've got the same problem.
Who can give me the right way ??
Thanks by advance !
Mickael
Nmae that php file image.php and call it as source of image tag in another php file. Then that will show the image.
<img src="image.php" height="250px" width="162px">
Click Here To Download Example From Google
Working good after creating a new php file where I pasted my code.
I think that my file was corrupt.
I just tried your code.
The only error I get is when profil.jpg is not present in the same directory.
is profil.jpg in the same directory as your .php file ?
Related
I want my PHP to output a simple image but when I do it gives me a white square, so I thought the IMG couldn't be found but when I put a simple HTML IMG tag it can be found the code is very simple. also if I try an image from the internet it also won't work.
<!--testing if image can be found-->
<!--<img src="img.jpeg">-->
<?php
header("Content-type: image/jpeg");
readfile("img.jpeg");
exit(0);
?>
Basically what's happening here is that you output some HTML comments before you output the image which means your image file is invalid and so your internet browser doesn't know what to display and defaults to what you're seeing (a white box in the middle of your screen).
Additionally when you use header you should ensure that you haven't already returned data!
If you check your source in your browser you'll see something like...
<!--testing if image can be found-->
<!--<img src="img.jpeg">-->
ÿØÿàJFIFÿá‹rExifII*ÀÐ ¶..................................
...with the additioonal data continuing for some time!
If you remove the comments from your PHP file then it will work as intended; your (entire) file should look something like:
<?php
header("Content-type: image/jpeg");
readfile("img.jpeg");
Note: You may want to add additional headers RE content length etc. but these aren't strictly necessary for most browsers.
To display an image using content type header, you must put only php on the page. If you add any other content it'll display a blank square
ex.
<?php header("Content-type: image/png"); readfile("./resources/img/favicon.png"); ?>
I uploaded the .doc file using php code and saved it in a folder ,the path is stored in database.
When i tried to make a view of the doc file in a div ...a dialog box appears asking whether to save or openwith ..
I'm using wampserver ....I just tried like this
<iframe name="awindow" frameborder=2 width=580 height=440 src="www/siva_example/pdf/1_siva.doc"></iframe>
Any help regarding..
Is it possible to view .doc file in browser or i have to convert it to pdf format...
It's not possible to display a .doc[x] in HTML. You can try to convert the .doc[x] to HTML or to an image.
you can view doc file by adding this to your header at request time
<?php
header('Content-disposition: inline');
header('Content-type: application/msword');
// not sure if this is the correct MIME type
readfile('MyWordDocument.doc');
exit;
You could try using google docs for this, use the below code, it works perfectly for ppts, have not been able to test it for a doc yet
<iframe src="http://docs.google.com/gview?url=http://yourdomain.com/document.doc&embedded=true" style="width:550px; height:450px;" frameborder="0"></iframe>
Ofcourse you would have to update the url as per your program using PHP
image displays in the browser whrn the following code is used.But when the code is used inside html,the image does not appear instead a small box appear.
<?php
header('Content-Type: image/jpeg');
readfile('http://localhost/picture013.jpg');
?>
for example when i use no image is displayed.
<html>
<body>
<?php
header('Content-Type: image/jpeg');
readfile('http://localhost/picture013.jpg');
?>
</body>
</html>
in short i wanted to display all the images from a folder.
Maybe you can try this one
<?php
$dir = "localhost/image"; // your folder name ex: image
$imgs = glob($dir ."/*.jpg"); // get your image files with .jpg
foreach ($imgs AS $i) {
echo "<img src='$i'>"; //
}
?>
Think about what this is doing for a moment. That file already exists, you're not inventing image data from nothing, so why are you not just creating an element and link to that image?
Put an <img src="some/location/picture013.jsp"> on the page with an echo/print instead, and whatever page needs to show that image will work just fine. In this case it's already a legal URL, so you can just add <img src="http://localhost/picture013.jpg"> to whatever page needs it.
If it wasn't, you could make the server simply resolve the location you're pointing to with an .htaccess rule, or simply have the file live in an already web-visible directory. There is no reason to make PHP proxy the image data when it's already a normal web-visible resource; in fact, you're just over-complicating your code with unnecessary steps.
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.
After using the PHP QrCode lib I discovered that for some reason when using a dynamic page with scripts and dialog boxes (JQuery) that when trying to output a QR code in a .png format I get weird symbols instead of the actual generated .png file.
Heres what I have tried:
Created a seperate file with just:
<?php
//include only that one, rest required files will be included from it
include "phpqrcode/qrlib.php";
QRcode::png('barrda554');
?>
Works great,
Attempt 2:
File opened within a dialog box type using JQuery UI:
<?php
header stuff...
include "phpqrcode/qrlib.php";
...
?>
<html>
...
<?php
QRcode::png('barrda554');
?>
..
</html>
In this attempt I get multiple funky symbols for some reason:
�PNG IHDRWWKK/PLTE���U��~�IDAT8��ѱ � P# �c��n :V�L�#�k
y��)�|F��5`ڸzHF|l���
%Z"e�Ы�D{\�ގ����p`�f�eh�������k�[BJeJ�c����,�^�gu�m|Q��o��W����g�
�#�s�<�y��k�m��!v�.��(+�u���$s�-�n$߫>�gR�`IEND�B`�
This has stumped me and I am unsure on how I should approach this to fix it.
Let me know your ideas,
David
UPDATE:
After putting header('Content-Type: image/png'); within the file that JQuery opens, no cigar.
Here is the actual file:
http://jsfiddle.net/T4nEP/
The problem is here:
<html>
<?php
QRcode::png('barrda554');
?>
</html>
To understand what this is doing, imagine that you open a regular PNG file in a text editor, and just copy/paste the contents directly into your HTML file. It's not going to show an image - it'll just be garbage, like you're seeing.
To include an image in an HTML file, you need to use the <img> tag, and point to the URL of the image. In this case, the URL of the image would be a PHP script that outputs nothing except the PNG contents - like this:
<img src="qrcode.php">
And then in qrcode.php, generate the image:
<?php
include "phpqrcode/qrlib.php";
QRcode::png('barrda554');
?>
If you need some information from the HTML page in order to generate the image, you can include it as query parameters in the URL, like this:
<img src="qrcode.php?product=1&format=2">
And then get those values in your PHP like this:
<?php
include "phpqrcode/qrlib.php";
$product = $_GET['product'];
$format = $_GET['format']
// ...
// whatever you need to do to generate the proper code
// ...
QRcode::png('barrda554');
?>
And finally - there are ways to include the image data directly into the HTML page, but it's not supported by all browsers, and is not recommended because it makes the page size much larger and prevents the browser from being able to cache the image separately.
You can see more about base64 data URLs here and here.
Make sure you have the following php code to generate the correct Content-Type header so the browser knows how to render the data its receiving.
header("Content-Type: image/png");