[RESOLVED] <? filetyp('$How do you tell php to load a [specific] filetyp?'); ?>

ManWithNoName

My knowledge of php: COLOR=Red blissfully ignorant [/COLOR] --> GURU :evilgrin:
My learning curve: Diety
Environment: win2k, apache, php5 --> the Code adapts itself to my needs.

Question: how do you tell php to load a [specific] file?

E.g. In the code below how do you tell the variable to just load whatever xml is in its path (filename shouldn't be an issue)

<?php

/* load the xml file and stylesheet as domdocuments */
$xsl = new DomDocument();
$xsl->load("stylesheet.xsl");
$inputdom = new DomDocument();
$inputdom->load("document.xml");

/* create the processor and import the stylesheet */
$proc = new XsltProcessor();
$xsl = $proc->importStylesheet($xsl);
$proc->setParameter(null, "titles", "Titles");

/* transform and output the xml document */
$newdom = $proc->transformToDoc($inputdom);
print $newdom->saveXML();

?>

(Code from: http://www.zend.com/php5/articles/php5-xmlphp.php)

I tried *.xml (yes... I really tried that—what did you expect?)

Now I'm fresh out of ideas...

I browsed php.net, the closest thing I found was something called ”filetyp”, but that’s just crazy.

SOLUTION: just change the "document.xml" with a optional variable, e.g., "$file". Thats it. No need to define the variable.

<?php 
{ 
/* load the xml file and stylesheet as domdocuments */ 
$xsl = new DomDocument(); 
$xsl->load("stylesheet.xsl"); 
$inputdom = new DomDocument(); 
$inputdom->load($file); 

/* create the processor and import the stylesheet */ 
$proc = new XsltProcessor(); 
$xsl = $proc->importStylesheet($xsl); 
$proc->setParameter(null, "titles", "Titles"); 

/* transform and output the xml document */ 
$newdom = $proc->transformToDoc($inputdom); 
print $newdom->saveXML(); 
} 
?>

That's really all.

bpat1434

<?php
/* Get all XML files into an array... */
$files = array();
$dir = 'path/to/xml/files';
$d = dir($dir);
while(FALSE != ($entry=$d->read()))
{
  if($entry != '.' && $entry != '..')
  {
    if(substr($entry, strripos($entry, '.'))=='.xml')
    {
      // If the file extension is .xml
      $files[] = $entry;
    }
  }
}

foreach($files as $file)
{
/* load the xml file and stylesheet as domdocuments */
$xsl = new DomDocument();
$xsl->load("stylesheet.xsl");
$inputdom = new DomDocument();
$inputdom->load($file);

/* create the processor and import the stylesheet */
$proc = new XsltProcessor();
$xsl = $proc->importStylesheet($xsl);
$proc->setParameter(null, "titles", "Titles");

/* transform and output the xml document */
$newdom = $proc->transformToDoc($inputdom);
print $newdom->saveXML();
}
?>

strripos() is a PHP 5 function, but you said you were using PHP 5.

~Brett

ManWithNoName

Hi again Brett, and once again, thanks for your input

I gave your code a try and some interesting things happened.

Settings: I have created a test folder which contains 3 xml files (page0.xml, page1.xml, page2.xml) that are processed with one stylesheet and the needed php files (nothing new: besides this code, the ones from my earlier post). One ordinary index.html file that linked to respective xml file.

I needed to enter a dot as directory path before the script reacted ($dir = '.' ; ).

I tried out the settings in three different scenarios.

A) In the primary scenario, all files were put in one directory.

Result: only one file (page0.xml) was parsed (can I say that? Parsed.) correctly regardless on which url you click; all links loaded the page0.xml. I took a look at outputted html code and it showed that ALL three pages had been transformed after each other:

<html>
Page 0
</html>
<html>
Page 1
</html>
<html>
Page 2
</html>

But only page0 being shown and correctly transformed.

😎 In the second scenario I moved page1 and page2 to separate folders and kept the rest of the files in place.

Result: all pages are rendered correctly, and their file name is showing their correct path and name, but the data is that of page0.xml.

Page1.xml and page2.xml have disappeared (not been rendered at all). The html output is correct, one html start/end tag, but the data is, as mentioned, the page0.xml data.

C) In the last try all pages have their own folder.

Result: there is no output from any of the pages. They all load empty pages, no html, but correct file names.

Conclusion: I'm not quite sure what to make. Someone has set me up the bomb.

What you say? I'm really tired now, but I believe it has something to do with the way I have handled the $dir variable. I took a look at php.net. But there seems to be very few alternatives for making an error with that variable; you choose either a dot or a path. That's it.

I will give this a fresh try tomorrow.

bpat1434

Well the error isn't in my code entirely. I don't know what your Object code outputs and what not, so if it has multiple HTML tags, it should really screw up.

You might try at the beginning of the foreach loop either emptying all the vars used by the object class, so as all new data is being created. I think that might be what's causing your trouble: cached data in the class.

~Brett

ManWithNoName

Gee Brett, why did you jus not say that? All you needed to do was to enter the variable $file.

That's totally uncool dude.

All I have left now is a cookie code. Want to see the code?

$when the settings the user want you to remember is inputted: remember them. That is all. For now.  Be gone.

Cool huh? PHP is kick-ass simple. 🆒

[...] Object code outputs [...]

[...] your trouble: cached data in the class.

Psst. Totally noob. :rolleyes:

And to think I wasted time looking up those stuff.

(I'm just kidding with you (duh). Once again thanks for your help and time. Good luck in the future.)

dalecosp

ManWithNoName wrote:
My knowledge of php: COLOR=Red blissfully ignorant [/COLOR] --> GURU :evilgrin:
My learning curve: Diety
Environment: win2k, apache, php5 --> the Code adapts itself to my needs.

Diety? You mean you live, sleep, and eat PHP????? 😉

Anybody else see some issues here? W isn't the first letter of LAMP, BTW!!!!

😃

SOLUTION: just change the "document.xml" with a optional variable, e.g., "$file". Thats it. No need to define the variable.

Ah. So, it's fixed? Can you mark the thread 'resolved' now, then?

ManWithNoName

Deity, more like: I tell PHP to script and it echoes: "how much?".

Anybody else see some issues here? W isn't the first letter of LAMP, BTW!!!!

W? LAMP? First letter? You got some explaining to do.

Ah. So, it's fixed? Can you mark the thread 'resolved' now, then?

I'm going to wait a little while before marking this one as "resolved". I want to see if anybody has a comment about why this worked. Because this doesn't make any sense to me.

And if this is a bug, then maybe I should continue looking for a REAL answer.

bpat1434

I'm going to go out on a limb here and say:
Linux Apache MySQL PHP (LAMP)
Windoze Apache MySQL PHP (WAMP)
Just my guess though...

What do you want explained? How what works? In my script, the $file variable is actually populated with a value (whatever the current value of the key in the $files array is).

~Brett

ManWithNoName

... Windows Apache XML PHP 🆒

Hi Brett,

No no there was no problems with your code; I understood your code perfectly.

And I found some good tutorial about caching (if you want to know: http://www.phpguru.org/static/Caching.html). The Object code output was a bit harder to narrow down (the closest I came: http://www.bigbold.com/snippets/tag/object (It was way to advanced for me).

But all that doesn't matter.

If you looked at my SOULTION you will see that to make the php process code load a "*.xml" file I put "$file" instead of "the-document-name.xml".

But the thing is: that variable isn't defined anywhere in the code. I was just curious in how the code knows how to load any XML file it is assigned to do (the prevous setting wanted a specific file name).

Your code was logical, the solution I found is not.

I was just curious as if this is a bug or not? If it is, this could give me problems later.

But this is not that important, it works, I'm happy 🙂. But I thought that for future use (mine as others) this issue should be raised within the post.