displaying an image

Anon

I want to diplay a random image in a html document, this is what the html looks like:

script.php3 is the script that select the image and send it back to the html

script.php3

<?
header("Content-type: image/gif");
include("ttt.gif");
?>

ttt is a .gif file in the same directory that .html and .php3 are, what I get as output is the picture holder, you know the square border where the pic was supposed to showed up, but the real pic never gets display, could anyone has another way to do that, or can give me some hint to get it fix.

Thanks for you help guys, I've been stuck in this for almost a week, have tried so many different ways, none of them worked.
Thanks again. sanddy

brandonschnell

include("ttt.gif"); tries to open the gif as a php script. how bout this:

<?
header("Content-type: image/gif");

$filename = "ttt.gif";
$fp = fopen($filename, "r");
$data = fread($fp,filesize($filename));
fclose($fp);
echo $data;

lordstjohn

???? This is the first time I can see this kind of syntax... This could not working !

If you know the name of picture :

in HTML : <img src="image.gif">

If youd'like to search the name in a DB, or anywhere else, from a form...

<?
//---- Do what I want to do ----//
// In this part $picture will be assigner to a name //
$picture = "ttt.gif";
print("<img src=\"$picture\">\n");
?>

brandonschnell

problem is often you won't know the filename till the script runs. you could echo the filename to a img src tag but then it doesn't play nice when other sites need to link to your cool pict of day (i assume hes trying to show a pict and use the php code to decide which one). setting the headers and echoing the pict data will make php return a pict just as if the php script itself were a pict. theres a number of cool trick you can do with this.

Anon

Thanks Brandon, but that code didnt work either, I'm still getting the same effect, the pic holder only , nothing else, I made sure that the pic is in the same directory. As you say, my script will do some processing before displaying the pic.

As andre said( thanks for your comments) you can insert the pic really easy following the format:

$picture = "ttt.gif";
print("<img src=\"$picture\">\n");

but:
"print <img....." but this will force you to name the page .php3 (i'm trying to avoid this), if you need an image (e.g. a banner) display in every single page of your site, all your pages will be .php3, what means that all of them will be parsed by php.exe, even though you are trying to just displayed a small pic, this will slow down you server a lot, calling the .php file from your pic tag in your html will force the .php file and only the .php file to be parsed, so you are free to have all your web documents in plain .html format.

thanks guys for your help, hope someone else comes out with any other idea how to do this, thanks

brandonschnell

whats the source of your script? the problem is from something thats happening before the image is being shown. maybe a php warning, maybe an error. have you tried disable showing the pict and just output the results of the page as txt looking for an error?

Anon

this is exactly what I have I'm just trying to get this part to work to keep building on top of it, but not even this is working:

script.php3

<?
header("Content-type: image/gif");

$filename = "ttt.gif";
$fp = fopen($filename, "r");
$data = fread($fp,filesize($filename));
fclose($fp);
echo $data;
?>

test.html

i tried this to make sure the pic can be found:

$filename =http://127.0.0.1/ttt.gif";
$fp = @fopen($filename, "r");
if ($fp)
{ print"The file exists!";}
else
{ print"The file does not exist";exit(); }

when I called this from a html document I get the string The file exists.

hey , thank you very much for you help.

brandonschnell

when that happens to me it typically means php is finding an error or a warning. i'd check to be sure fread is actually returning data and don't suppress warnings/errors ("@"). also comment out the header and see if any errors are returned. with the header set to gif any warnings errors will prevent the image from showing up.

Anon

I also tried this to make sure that the pic is been loaded into $data.

script.php3
<?
if (!isset($id)){header("Content-type: image/gif");}
else{
$filename = "ttt.gif";
$fp = fopen($filename, "r");
$data = fread($fp,filesize($filename));
fclose($fp);
print "$data";
}
?>

test.html

of course, when you click the anchor "pic" you wont get the pic, but you get the binary information (bunch of "@#$%&%%&$!") that are stored in the file, this proved that the variable $data is been loaded with whatever is contain in the .gif file. However the pic from the <img> tag still not shown. really weird.

"if (!isset($id))" this statement is to avoid sending the header more than one time, since the header gets sent for the first time when the browser make the request
<img src="script.php3>.

thanks

brandonschnell

when the browser makes the request to img src no image header is sent automatically (well, the text/html header is sent but that won't work with images)

the header has to be sent before the data regardless of how the script was called. the code you posted above would only send the header or the data (both have to be sent). if the data was sent but no header and the script was called from a <img src> tag nothing would be shown but the missing pict icon.

Anon

thanks for your help: this seems to work

script.php3
<?
$image = fread(fopen("ttt.gif", "r"), 1000000);
Header("Content-type:image/gif");
echo $image;
?>

test.html

now I realized that even though I'm able to display the image, I have no way to associate the image to an anchor since the image in randomly picked from a mysql data base, since I'm calling the .php3 script from an image tag from the html, the only thing I get is the binary information of the pic, so I have no way to figure out which pic was actually picked from the data base,
for this reason there is no way I can wrap the image inside a <a></a> statement.

I wanted to display the image(banner) so user will be able to click on them and be taken to a given place, since I have no information about which image has been displayed I'm not able to do that.

Any ideas?

Thanks for you help Brandom.

brandonschnell

there isn't going to be an easy way to do the anchor with out using php at some point to echo the img tag.

what i use is a page to show pics and another page to choose which one to use. so pick_a_pict.php would choose the pict to show (not selecting image data to speed up the query, select * is a bad bad thing with images) then would echo "<img src=\"pict.php?pict_id=1\"> click me </a> with pict.php doing the image select from db and header & data output.

Anon

this takes me to the begining again, if I

echo "<img src=\"pict.php?pict_id=1\">

from a .php3 file I have to make the whole page .php, I dont know if I'm making myself clear,
to do this you have to do something like this
<?
.....
echo "<img src=\"pict.php?pict_id=1\">
.....
?>

then this page will have to be named .php and be parsed completely to output a single image, if I need the image in all the pages, then every single page in the site will have to be .php, I think this will affect the performance since all the pages will have to be parse by php.exe, if you are getting a considerable numbers of hits, this solution does not seem to be the best one.
that was the reson I wanted to call the .php from the image tag so only that file will be parse while the html document will be trasnferred from the server without getting thru php.exe
I was looking for a way to insert the image from the data base into the page but keeping the document .html or .htm, so no parcing would be necessary.

Thanks again for your ideas, thank you very much

brandonschnell

sounds like you want html to handle dynamic content (change the img src info for every picture). dynamic content requires something like php to handle it. =)

About the only thing that would kinda let you do this would be to create php pages that are rendered to html and copied to a html page via crontab on a regular basis, but this method has its own set of issues.

for what php offers, the peformance hit isn't huge. Most sites will likely run out of bandwidth before running out of system resources.