Displaying images using php3

Anon

How do I display an array of images using php3. The images I'm trying to display are in a subdirectory directory which can only be accessed by using ftp functions. My question is considering the above scenario how do write the IMG SRC statement and include php scripting within it. I've tried using echo $image in the IMG SRC statement, but it doesn't. Would someone who has more experience with php 3 answer this question.

Anon

This question is a bit clearer than your first one (other topic...)

try this:

<img src="script.php"> (I know, that .php is not a typical image type 🙂

In your script.php you add the following command as first statement:

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

Whatever your output is of this script is tried to be interpreted as gif. You so have to ftp to whereever you want, copy your gif and flush it as output in your script (might be done using echo, but actually I assume there are better methods to flush a file as output...)

Hope this helped

Anon

I'm assuming that all code related to displaying the images goes in the script.php3
file and not the main file.

Anon

right 🙂 The main file can even be a "normal" HTML-File

Anon

How do I call the script file from the main file? <img src = "script.php3"> by itself doesn't work.

Anon

... just insert an image-Tag and set (instead of "somewhat.gif") your script-name.

i.e. (in your html-file where you want to display the gif):

In this case, script.php has to be placed in the same directory as your main (html) file.

With the header-Command (as described before) in script.php the script "acts" like a picture, altough its PHP-code is parsed and executed by the server....

Anon

I don't understand what you mean by image tag. Where do I put this image tag?

Anon

I already did what you told me to do. Is image-Tag a variable like $image-Tag. Would it help to tell you that I also opened up a database connection in the script.php3 file?
Your answer is too vague. Please be more specific. I've only been doing php for a month.

Brian wrote:

I don't understand what you mean by image tag. Where do I put this image tag?

Anon

Dear Brian

It seems that you are not really familiar with HTML... I try to explain:

a tag is "<whatever>"... so a correct image-tag is: "<img src="images/anything.gif">"

This is plain HTML so far....

You have to put this img-tag into your so-called main-file -> there where you want to display all the informations AND the picture.

As you need a script to grab the correct picture you can't just use "<img src="path/to/picture.gif">. To solve your problem, use: <img src="script.php">.

You even can give any parameters along the scriptname. i.e. <img src="script.php?person_id=<? echo $person_id;?>">

(which would result in <img src="script.php?person_id=23"> assuming 23 is the current value when the main file was executed)

Let's talk about basic HTML and server processing...

Let's assume you point your browser to a webpage which contains 1 picture. That's what's going on until you see the whole page including the picture:

browsers searches page on specific server
server sends file (HTML) to browser
browser parses HTML-Code and displays Page
browser recognized the img-tag and sends another request to server:
browsers sends request to transfer picture
server sends picture
browser implements picture into page.

Everything ok so far?

The second request (picture) in detail:

The browser parses the HTML-Code which he gets from the server (first request). When the browser finds a <img src="path/to/whatever.gif"> the request is: Server, send me "path/to/whateverfile.ext". No matter if whateverfile.ext is a valid picture-File.
If the File transmitted by the second request IS a valid picture File (see below), the browser places it into the page where the img-tag was found.

(here is below) 🙂
It is actually NOT the extension itself that makes a file to a valid picture-File! Every file transmitted has got it's own so-called "header". This header information says what the content of the file should be treated as - NOT THE EXTENSION.
With the header-command in the script.php (described approx. 1 hour ago...) you make the output of that script to be interpreted as a (valid) gif File, altough it has the extension .php.

So assuming you want to display some data about a person AND its picture, you have to add the code for retrieving all the information (except the picture) in your MAIN-File.
To retreive the picture you have to have your script.php executed (which ftp's to whereever and streams the picture as output). This is easily done by adding the image-tag "<img src="script.php"> in the MAIN-File whereever the picture should be placed.

I hope I cleared any missunderstandings...
anyway.. I'm off now (midnight here).

Good look

Anon

... your whole issue might be a bit too tricky programming only for about a month...

anyway.. good night and good look

Thomas

Anon

In what file should I open the database connection and the ftp connection.

Anon

... in the scripts.php which you put into the img-tag

Anon

I did all that but all I get is the little box with the red x in it. The script.php is in the same directory so I know I don't have to change my path. What am I missing?

Anon

i guess the best thing is you send me both scripts (mainfile and scripts.php) and I will have a look on it. Send it to jeger@inetconsult.ch

The red cross says "picture faulty or missing".. probably there is an error in your script.php

Anon

This is the main file.

This the script.php3

<?php header("content-Type: image/gif");

$conn = mysql_connect($host,$user,$password);
mysql_select_db($database);
$sql = "Select * from artsub where medium = $medium";
$result = mysql_query($sql);
if(!$result)
{
echo "SELECT query failed";
}

$firsttime = "true";
while($row = mysql_fetch_array($result))
{

$art = $row["photo"];







$hostname = "hidden";
     $port = 21;



      $ftp = ftp_connect($hostname,$port);
      $login = ftp_login($ftp, $user,$password);
      if(!$login)
      {
        echo "login failed/n";
        exit;
      }




      if($firsttime == "true")
      {

      	$i=0;
      }
      $list = Array();
      $list = ftp_nlist($ftp,"uploads/images");
    do
    {

      if($lis[$i== "uploads/images$art") 
        {
           $slash = "/";


           $path = $slash.$list[$i];

           $size = ftp_size($ftp, $path);

           ImageGIF($path);
           ImageDestroy($path);

        }
        $i++;
      }while($list[$i]);

}

ftp_quit($ftp);
mysql_close();

Anon

I found out that my problem is with open the file. The files are there, but that can't be opened using the ImageCreateFromGif command.
If I use a filesystem command like fopen, it doesn't even recognize the directory because the directory is password protected. I'm using FTP functions to access the files, but I don't see any FTP functions that actually open files.