calling a PHP script within <img src> tag

jmpalmer

I have an HTML file (temp.html) that has the following line of HTML code:

where picscript.php is a file in the same directory that contains a query.

However, when temp.html is loaded into the browser it doesn't "call" picscript.php.

I'm new to PHP, but not programming in general.

I'm not using an INCLUDE statment... should I? If so, what should it look like?

What the heck am I doing wrong?

John

bpat1434

You do not need to use an include statement. What you do need to do is actually have the picscript.php send the appropriate image headers so that the output is read by the browser as an image, and not text:

<?php

$image = $_GET['imgnum'];



$query = "SELECT `photo` FROM `photos` WHERE `id`=" . intval($image);
$result = mysql_query($query);

// Send the content-type header for JPG
header('Content-Type:image/jpg');
echo mysql_result($result, 0, 'photo');

That's a basic script, and it has no error checking, but it gives you an idea.

bradgrafelman

Also make sure you're using valid HTML (e.g. quoting all values for attributes).

jmpalmer

I'm sure I missing something simple so ... here are the two files:

Temp.html

<html>
<body>
Here is your picture: 
<img src=temp2.php?imgnum=1> 
</body>
</html>

Temp2.php

<?php

$image = $_GET['imgnum'];

$query = "SELECT filename FROM img_tbl WHERE index_fn=" . intval($image);
$result = mysql_query($query);

// Send the content-type header for JPG
header('Content-Type:image/jpg');
echo mysql_result($result, 0, 'photo');

It doesn't even seem to be "calling" Temp2. How can I verify that temp2.php is even be fired off?

I get the broken image icon when temp.html is loaded into the browser.
When I look at the properties of the of the broken image icon I see:
http://www.mywebsite.com/temp2.php?imgnum=1

not the name of the image.
So this makes me think that temp2.php isn't being executed.

Thanks for the help in advance.

John

leatherback

Have you tried using valid html, as Brad commented..

Most importantly:
<img src="temp2.php?imgnum=1">

But obviously you need to specify which HTML version en encodings you are using too.

NogDog

The correct content-type should be "image/jpeg".

But more importantly, your SQL query is only retrieving the file name from the database, but you are trying to output the 'photo' column. Assuming that is correct, you need to select photo instead of filename in the query.

jmpalmer

I made the corrections per the last two postings, but still nothing but a broken link for the image.

If I just run temp2.php (I set $image=1 as if it were being passed from temp.html) it works fine.

This makes me think that when I run temp.html (and it is supposed to call temp2.php) it isn't really doing it. I'll mess up the SQL code on purpose to see if I get an error... and I don't get any errors.

The two files look like this:

temp.html

<html>
<body>
Here is your picture: 
<img src="temp2.php?imgnum=1"> 
</body>
</html>

temp2.php

<?php

$image = $_GET['imgnum'];

$host="x.x.x.192";
$user="xxxxxx";
$password="xxxxxxx";
$db="spjp";

$cxn = mysql_connect($host,$user,$password)
or die ("Couldn't connect to server");

$dbselected = mysql_select_db($db)
or die("Couldn't select db");

$query = "SELECT filename FROM img_tbl WHERE index_fn=" . intval($image);
$result = mysql_query($query);

// Send the content-type header for JPG
header('Content-Type:image/jpeg');
mysql_result($result, 0,'filename');

Thanks for the help

bpat1434

If you're reading the filename, use [man]file_get_contents[/man](mysql_result($result, 0, 'filename')); output the actual file:

header('Content-Type: image/jpeg');
echo file_get_contents(mysql_result($result, 0, 'filename'));

Note that you also need to echo or print the data to the browser

Tegaman

Very, very frustrating. This thread is perfect description of my problem that I cannot get resolved. Can someone please focus on this one issue; the php page is NOT being opened/entered/executed as expected from the HTML page. Can anyone please help with this problem? Lets assume that all the php page has in it is an ECHO statement.

bradgrafelman

@: You might want to start your own thread and give some more details about your problem, including some actual code.

The concept is pretty simple... setting the source of an <img> element to a PHP script:

<img src="http://mysite.com/foobar.php">

will cause the user's browser to make a separate HTTP request for 'foobar.php' after it parses the HTML code and sees that it needs additional resources to render the page.

Derokorian

If you are not getting an image to show with this method try accessing the script directly to see if an image is shown. If not the problem is the script. Also, the script in the src CANNOT output anything other than the associated headers and the binary data making the image, if it outputs anything else the image won't show! This is why I suggest accessing the script directly, because if an error is being output, then the image will not work.