PHP, RegEx and Objects

[Alexandre Lahure]

Hi all

I'd like to execute a piece of code when I find a particular string. So I
used preg_replace('/my_regex/e', 'my_piece_of_code', $my_string)

Actually, I'd like to convert something like this :
[img-32]
into something like that :
<img alt="" height="600" src="image.jpg" width="800" />

This wouldn't be very complicated if I hadn't to initialize an Image
object, then use a method called display(). My piece of code looks like
that :
$img = new Image(32); $img->display();

But instead of writing my tag (and displaying the image), PHP writes
'Object'
I'm looking for answers to this fundamental question : "Why ?" and of
course if somebody can solve my problem...

Thanks to all that are going to help me
--
Alexandre Lahure
Point 52, Solutions Internet "Ready to Start"
http://www.point52.com/

"Computers are like air conditioners,
They don't work when you open windows"

[Alvaro G Vicario]

*** Alexandre Lahure wrote/escribió (Wed, 05 Nov 2003 20:06:26 +0100):

$img = new Image(32); $img->display();

But instead of writing my tag (and displaying the image), PHP writes
'Object'

<IMG> tag expects text (the name of the file). What does display() method
return?

--
--
-- Álvaro G. Vicario - Burgos, Spain
--

[Alexandre Lahure]

> <IMG> tag expects text (the name of the file). What does display() method

return?

I know that. Don't worry about what the display() method returns (but, if
you really want to know, it displays a fully functionnal <img> tag, with
all necessary attributes), it works just fine (actually, it was tested
under different conditions without problems).

The following piece of code works well in every case tested but the
preg_replace() case (and that's what gives me headache)
$img = new Image(32);
$img->display();

PHP seems to fail at Object initialization, or preg_replace() doesn't
consider Object initialization as valid PHP code (???)
--
Alexandre Lahure
Point 52, Solutions Internet "Ready to Start"
http://www.point52.com/

"Computers are like air conditioners,
They don't work when you open windows"

[Alvaro G Vicario]

*** Alexandre Lahure wrote/escribió (Thu, 06 Nov 2003 14:04:56 +0100):

The following piece of code works well in every case tested but the
preg_replace() case (and that's what gives me headache)
$img = new Image(32);
$img->display();

PHP seems to fail at Object initialization, or preg_replace() doesn't
consider Object initialization as valid PHP code (???)

<?
class Image{
function display(){
return 'foo boo foo';
}
}
$img = new Image();
echo preg_replace('/boo/', 'this works for me', $img->display());
?>

--
--
-- Álvaro G. Vicario - Burgos, Spain
--

[Alexandre Lahure]

> <?

class Image{
function display(){
return 'foo boo foo';
}
}
$img = new Image();
echo preg_replace('/boo/', 'this works for me', $img->display());
?>

<?php
class Image {
function Image($image_id)
{
$this->image_id = $image_id;
$query = "SELECT width, height, src FROM images WHERE
img_id = '$this->image_id';";
$result = mysql_query($query);
if ($row = mysql_fetch_row($result)) {
$this->width = $row[0];
$this->height = $row[1];
$this->src = $row[2];
}
mysql_free_result($result);
}

function display() {
$image_tag = "<img src=\"$this->src\" width=\"$this->width\"
height=\"$this->height\" />";
return $image_tag;
}
}

$image_tag = preg_replace('/\[img-([0-9]+)\]/e', '\$img = new Image($1);
\$img->display();', '[img-32]');

print($image_tag); /* displays 'Object'
* instead of '<img src="image.jpg" width="800"
height="600" />' */
I simplified the code but this would be OK.

--
Alexandre Lahure
Point 52, Solutions Internet "Ready to Start"
http://www.point52.com/

"Computers are like air conditioners,
They don't work when you open windows"

[Alvaro G Vicario]

*** Alexandre Lahure wrote/escribió (Fri, 07 Nov 2003 09:46:25 +0100):

$image_tag = preg_replace('/\[img-([0-9]+)\]/e', '\$img = new Image($1);
\$img->display();', '[img-32]');

print($image_tag); /* displays 'Object'
* instead of '<img src="image.jpg" width="800"
height="600" />' */

The '=' operator has a return value: the value itself you're assigning. For
example:

$age=33; // returns 33
echo ($age=33) // prints 33

Since new Image() returns an object, you are inserting an object within a
string. PHP does its best to handle that: it makes a string representation
of the object: "Object".
--
--
-- Álvaro G. Vicario - Burgos, Spain
--

[Alexandre Lahure]

> The '=' operator has a return value: the value itself you're assigning.

For
example:

$age=33; // returns 33
echo ($age=33) // prints 33

Since new Image() returns an object, you are inserting an object within a
string. PHP does its best to handle that: it makes a string
representation
of the object: "Object".

I agree, but if you look closer at the PHP manual, here what you will see
(PCRE / preg_replace section) :

"/e modifier makes preg_replace() treat the replacement parameter as PHP
code after the appropriate references substitution is done. Tip: make sure
that replacement constitutes a valid PHP code string, otherwise PHP will
complain about a parse error at the line containing preg_replace()."

As long as I know, my PHP code is valid since it works well when executed
separately. And admitting that PHP outputs 'Object', why does it not
output the second part of the code ?

--
Alexandre Lahure
Point 52, Solutions Internet "Ready to Start"
http://www.point52.com/

"Computers are like air conditioners,
They don't work when you open windows"

[Stefan Farnik]

Alexandre Lahure <ad***@point52.com> wrote in message news:<op**************@news.wanadoo.fr>...

<?php
class Image {
function Image($image_id)
{
$this->image_id = $image_id;
$query = "SELECT width, height, src FROM images WHERE
img_id = '$this->image_id';";
$result = mysql_query($query);
if ($row = mysql_fetch_row($result)) {
$this->width = $row[0];
$this->height = $row[1];
$this->src = $row[2];
}
mysql_free_result($result);
}

function display() {
$image_tag = "<img src=\"$this->src\" width=\"$this->width\"
height=\"$this->height\" />";
return $image_tag;
}
}

$image_tag = preg_replace('/\[img-([0-9]+)\]/e', '\$img = new Image($1);
\$img->display();', '[img-32]');

print($image_tag); /* displays 'Object'
* instead of '<img src="image.jpg" width="800"
height="600" />' */

try something like this:

<?php
class Image {
function Image($image_id)
{
$this->image_id = $image_id;
$query = "SELECT width, height, src FROM images
WHERE
img_id = '$this->image_id';";
$result = mysql_query($query);
if ($row = mysql_fetch_row($result)) {
$this->width = $row[0];
$this->height = $row[1];
$this->src = $row[2];
}
mysql_free_result($result);
}

function display($image_id = null)
{
if ($image_id !== null) {
$img = new Image($image_id);
return $img->display();
} else {
$image_tag = "<img src=\"$this->src\"
width=\"$this->width\" height=\"$this->height\" />";
return $image_tag;
}
}
}

$image_tag = preg_replace('/\[img-([0-9]+)\]/e',
'Image::display($1);', '[img-32]');

?>

[Alexandre Lahure]

> try something like this:

<?php
class Image {
function Image($image_id)
{
$this->image_id = $image_id;
$query = "SELECT width, height, src FROM images
WHERE
img_id = '$this->image_id';";
$result = mysql_query($query);
if ($row = mysql_fetch_row($result)) {
$this->width = $row[0];
$this->height = $row[1];
$this->src = $row[2];
}
mysql_free_result($result);
}

function display($image_id = null)
{
if ($image_id !== null) {
$img = new Image($image_id);
return $img->display();
} else {
$image_tag = "<img src=\"$this->src\"
width=\"$this->width\" height=\"$this->height\" />";
return $image_tag;
}
}
}

$image_tag = preg_replace('/\[img-([0-9]+)\]/e',
'Image::display($1);', '[img-32]');

?>

Yes, it works, thank you, but I can't stop thinking it's cheating. I'm
sure there is a way to do this "fairly", or the PHP manual is lying about
the power of the 'e' modifier of preg_replace().

One day, truth will be mine...
--
Alexandre Lahure
Point 52, Solutions Internet "Ready to Start"
http://www.point52.com/

"Computers are like air conditioners,
They don't work when you open windows"

[Alvaro G Vicario]

*** Alexandre Lahure wrote/escribió (Fri, 07 Nov 2003 13:54:47 +0100):

"/e modifier makes preg_replace() treat the replacement parameter as PHP
code after the appropriate references substitution is done. Tip: make sure
that replacement constitutes a valid PHP code string, otherwise PHP will
complain about a parse error at the line containing preg_replace()."

I've been playing around with my code:

<pre><?
class Image{
function display(){
return '[Here goes image tag]';
}
}
$img1=new Image();
echo preg_replace('/i/', $img1->display(), "This is a test\n");
echo preg_replace('/i/e', '$img1->display()', "This is a test\n");
echo preg_replace('/i/e', '$img2=new Image(); $img2->display();',
"This is a test\n")
?>

This prints:

Th[Here goes image tag]s [Here goes image tag]s a test
Th[Here goes image tag]s [Here goes image tag]s a test
ThObjects Objects a test
Could it be a variable scope issue?

--
--
-- Álvaro G. Vicario - Burgos, Spain
--

[Alexandre Lahure]

> I've been playing around with my code:

<pre><?
class Image{
function display(){
return '[Here goes image tag]';
}
}
$img1=new Image();
echo preg_replace('/i/', $img1->display(), "This is a test\n");
echo preg_replace('/i/e', '$img1->display()', "This is a test\n");
echo preg_replace('/i/e', '$img2=new Image(); $img2->display();',
"This is a test\n")
?>

This prints:

Th[Here goes image tag]s [Here goes image tag]s a test
Th[Here goes image tag]s [Here goes image tag]s a test
ThObjects Objects a test
Could it be a variable scope issue?

I don't think so. Actually, variable initialization and method call must
be in the same replace pattern.

--
Alexandre Lahure
Point 52, Solutions Internet "Ready to Start"
http://www.point52.com/

"Computers are like air conditioners,
They don't work when you open windows"

[Stefan Farnik]

Alexandre Lahure <ad***@point52.com> wrote in message news:<op**************@news.wanadoo.fr>...

Yes, it works, thank you, but I can't stop thinking it's cheating. I'm
sure there is a way to do this "fairly", or the PHP manual is lying about
the power of the 'e' modifier of preg_replace().

One day, truth will be mine...

I don't think so.
You pass two calls of functions to preg_replace, since your first
function returns something (object), preg_replace will use this for
replacement.
You also can write a wrapper function:

<?php
imgDisplay($img_id) {
$img = new Image($img_id);
return $img->display();
}
?>

This will work, too.