converting a php file to a xml feed

[arcanine]

hi there, I'm running a forum (smf) and decided I wanted to have a

coverflow application

http://dinogod.adrianhosting.com/smf/index.php?action=showcase

you can see in action above, although there is a problem with it the

links to the pictures are hardcoded so if I wanted a new image I have to recompile it in

flex builder 3 I've asked around in flex forums as to a function to scan the image directory

for new images but it has to be done in php then flex can consume it in xml format

 

So on to my php problem I'm using a code snippet from phparadise

which lists all images in my directory and echos each filename per line

you can see the script live here http://dinogod.adrianhosting.com/smf/coverflow/list.php

the problem is flex won't consume a standard php script so I need help in generating

a xml feed from script for reference here is the source of the script

 

<?php

$imgdir = 'img/'; // the directory, where your images are stored
$allowed_types = array('png','jpg','jpeg','gif'); // list of filetypes you want to show

$dimg = opendir($imgdir);
while($imgfile = readdir($dimg))
{
if(in_array(strtolower(substr($imgfile,-3)),$allowed_types))
{
  $a_img[] = $imgfile;
  sort($a_img);
  reset ($a_img);
} 
}

$totimg = count($a_img); // total image number

for($x=0; $x < $totimg; $x++)
{
$size = getimagesize($imgdir.'/'.$a_img[$x]);

// do whatever
$halfwidth = ceil($size[0]/2);
$halfheight = ceil($size[1]/2);


echo "$a_img[$x] <br />";


}

?>

I've tried learning how to do this myself (on wc3.org) but either nothing shows at all, or xml complains at me furiously

can someone take me through it, I'm not a php programmer by trade and need clear guidance, or an example

thanks for your time 

 

 

[RichardRotterdam]

I dont think feed is the correct term. All you need is to use php to create an xml. The file extention would still be php only it would have xml data in it so flex can read the file.

 

i dunno what that last forloop does but replace this

<?
for($x=0; $x < $totimg; $x++)
{
$size = getimagesize($imgdir.'/'.$a_img[$x]);

// do whatever
$halfwidth = ceil($size[0]/2);
$halfheight = ceil($size[1]/2);


echo "$a_img[$x] <br />";


}
?>

 

with

<gallery>
<?php
foreach($a_img as $imageUrl){
?>
    <image>
        <url><?php echo($imageUrl) ?></url>
    </image>
<?php
}
?>
</gallery>

well something like that it really depends on how you want your xml to look its best to use a static xml first.

[arcanine]

Thanks for your reply I tried your suggestion and the php file

becomes blank so I did a little testing it seems $x needs to be

echoed for the filename to come out I tried echoing

 

"$a_img"

on its own and as the output I got

 

Array

Array

 

but no file names just the word array, so $a_img and $x need to be merged or

something to make your script work how would I go about doing this

because I see you've changed $a_img to $imageUrl but it doesn't have all the information

in that string

 

also some of that code can be ignored the original script would output image size and other details

I didn't need so I changed the echo however some of the original scripting will be in there

 

thanks again for your help I appreciate it

[arcanine]

okay I've been working non stop on this but not really getting anywhere here is where I'm at thus far

 

<?php //define the xml standard
header("Content-type: text/xml");
echo "<?xml version='1.0' encoding='ISO-8859-1'?>";
?>
<?php //this section calculates the image names in the directory

$imgdir = 'img/'; // the directory, where your images are stored
$allowed_types = array('png','jpg','jpeg','gif'); // list of filetypes you want to show
$dimg = opendir($imgdir);
while($imgfile = readdir($dimg))
{
if(in_array(strtolower(substr($imgfile,-3)),$allowed_types))
{
  $a_img[] = $imgfile;
  sort($a_img);
  reset ($a_img);
} 
}

$totimg = count($a_img); 

for($x=0; $x < $totimg; $x++)
{
$size = getimagesize($imgdir.'/'.$a_img[$x]);
?>


<?php //echo xml tree
echo "<gallery>";
echo "<image>";
echo "<url>$a_img[$x]"; 
echo "</url>";
echo "</image>";
echo "</gallery>";


}

?>

 

whichoutputs

 

<?xml version='1.0' encoding='ISO-8859-1'?><gallery><image><url>11.jpg</url></image></gallery><gallery><image><url>12.jpg</url></image></gallery>

 

which is of course wrong I need the url tags to repeat not gallery and images

you can see the errors for yourself at

 

http://dinogod.adrianhosting.com/smf/coverflow/list.php

 

how can I say only print gallery and image tags once but the url tags as many times as the string needs

[wildteen88]

Try:

<?php
//define the xml standard
header("Content-type: text/xml");
echo "<?xml version='1.0' encoding='ISO-8859-1'?>";

//this section calculates the image names in the directory
$imgdir = 'img/'; // the directory, where your images are stored
$allowed_types = array('png','jpg','jpeg','gif'); // list of filetypes you want to show

$dimg = opendir($imgdir);

$url = 'http://' . $_SERVER['HTTP_HOST'] . '/' . $imgdir;

echo "<gallery>
  <image>\n";

while($imgfile = readdir($dimg))
{
    if(in_array(strtolower(substr($imgfile,-3)),$allowed_types))
    {
        echo '    <url>' $url . $imgfile . "</url>\n";
    }
}

echo '  </image>
<gallery>';

?>

[arcanine]

Thanks for your help but the script generates nothing not even in the source code :S

 

I've uploaded your script to

http://dinogod.adrianhosting.com/smf/coverflow/test.php

so you can see for yourself

 

I'm really confused as to why it won't echo anything :S

[BlueSkyIS]

you're probably not seeing the parse error:

 

parse error: syntax error, unexpected T_VARIABLE, expecting ',' or ';' in /Users/lesbrown/Sites/site1/test.php on line 21

 

do you have error_reporting turned on?

[arcanine]

I imagine I don't if I'm not seeing anything

 

I don't know how to find out whether error_reporting is on or off

as I've mentioned I don't know php  :-\

 

your error does mention line 21 though which is

 

        echo '    <url>' $url . $imgfile . "</url>\n";

does anything look wrong there?

[wildteen88]

I missed a period from my code, I forgot to correct it before I posted the code. Fixed code

<?php
//define the xml standard
header("Content-type: text/xml");
echo "<?xml version='1.0' encoding='ISO-8859-1'?>";

//this section calculates the image names in the directory
$imgdir = 'img/'; // the directory, where your images are stored
$allowed_types = array('png','jpg','jpeg','gif'); // list of filetypes you want to show

$dimg = opendir($imgdir);

$url = 'http://' . $_SERVER['HTTP_HOST'] . '/' . $imgdir;

echo "<gallery>
  <image>\n";

while($imgfile = readdir($dimg))
{
    if(in_array(strtolower(substr($imgfile,-3)),$allowed_types))
    {
        echo '    <url>' . $url . $imgfile . "</url>\n";
    }
}

echo '  </image>
<gallery>';

?>

If you still get a blank screen then you should turn on display_errors

[arcanine]

oh right, thank you for all this by the way

 

the corrected script is at

 

http://dinogod.adrianhosting.com/smf/coverflow/test.php

 

I get XML

Parsing Error: no element found

Location: http://dinogod.adrianhosting.com/smf/coverflow/test.php

Line Number 6, Column 10:

 

looking at the source the output is

 

<?xml version='1.0' encoding='ISO-8859-1'?><gallery>

  <image>

    <url>http://dinogod.adrianhosting.com/img/11.jpg</url>

    <url>http://dinogod.adrianhosting.com/img/12.jpg</url>

  </image>

<gallery>

which to me looks right, can I ignore that xml error and start trying to call the

xml file to my flex project? or is it not valid yet?

[wildteen88]

This:

echo "<gallery>
  <image>\n";

should be:

echo "\n<gallery>
  <image>\n";

 

Also this:

echo '  </image>
<gallery>';

should be:

echo '  </image>
</gallery>';

[arcanine]

Thanks a lot  just what I needed I'll be sure to bookmark this

forum in google bookmarks

 

Final script just in case anyone wanted something similar

 

<?php
//define the xml standard
header("Content-type: text/xml");
echo "<?xml version='1.0' encoding='ISO-8859-1'?>";

//this section calculates the image names in the directory
$imgdir = 'img/'; // the directory, where your images are stored
$allowed_types = array('png','jpg','jpeg','gif'); // list of filetypes you want to show

$dimg = opendir($imgdir);

$url = 'http://' . $_SERVER['HTTP_HOST'] . '/' . $imgdir;

echo "\n<gallery>
  <image>\n";

while($imgfile = readdir($dimg))
{
    if(in_array(strtolower(substr($imgfile,-3)),$allowed_types))
    {
        echo '    <url>' . $imgfile . "</url>\n";
    }
}

echo '  </image>
</gallery>';

?>