one inside strtoupper

[hisoka]

In this code :

<?php
$string = "this is my lower sting";
print preg_replace('/(.*)/e', 'strtoupper("\\1")', '$string');
?>

I do not understand what does

\\1 

do inside

strtoupper 

I read here :

 

http://php.net/manual/en/function.strtoupper.php

 

that 

strtoupper

 

takes a string as a parameter not \\1 ??

Second question : What does regex (.*) do ? I know that .* means any character, any number of repetitions but not (.*)

 

??

Edited April 17, 2016 by hisoka

[Ingolme]

That "strtoupper" is not a function, it's just a string. It's not actually PHP code.

 

As for the \1, it's called a backreference. It's explained in the PHP manual for preg_replace():

http://php.net/manual/en/function.preg-replace.php

 

replacement may contain references of the form \\n or (since PHP 4.0.4) $n, with the latter form being the preferred one. Every such reference will be replaced by the text captured by the n'th parenthesized pattern. n can be from 0 to 99, and \\0 or $0 refers to the text matched by the whole pattern. Opening parentheses are counted from left to right (starting from 1) to obtain the number of the capturing subpattern. To use backslash in replacement, it must be doubled ("\\\\" PHP string).

 

You should do more research on regular expressions if you really want to understand all of this. The information is out there if you look for it.

[hisoka]

Hello FoxyMod I hope you got a good weekend

 

I still do not know what the regex (.*) does ? I looked for it here for example http://www.regular-expressions.info/refcapture.html

 

in all the site but I could not find what does (.*) do ?

[Ingolme]

It's a combination of (), . and *. Each of those does a separate thing. Look for the explanations of each of those three rules on the page you just linked to.

[hisoka]

After some researches , I find that the parentheses are used to group the . and * together , I find , too, that .* does the same thing as (.*) . So I deleted the parentheses and run the code again like this :

<?php
$string = "this is my lower sting";
print preg_replace('/.*/e', 'strtoupper("\\1")', '$string');
?>

but nothing happened . If (.*) does the same thing as .* so why I got a blank page and nothing happened ?

Edited April 17, 2016 by hisoka

[Ingolme]

Don't wrap $string in quotation marks. It's a variable, you don't put variables in quotation marks.

[hisoka]

Don't wrap $string in quotation marks. It's a variable, you don't put variables in quotation marks

 

 

I took the code from this page :

 

http://www.madirish.net/402

 

and the $string is wrapped in quotation marks . Why is $string wrapped in quotation marks ?

[Ingolme]

The code on that page is wrong. The person who wrote it made a mistake.

 

$string should not be wrapped in quotation marks.

[hisoka]

The code on that page is wrong. The person who wrote it made a mistake

 

 

but this code :

<?php
$string = "this is my lower sting";
print preg_replace('/(.*)/e', 'strtoupper("\\1")', '$string');
?>

gives exactly the same result as this code :

<?php
$string = "this is my lower sting";
print preg_replace('/(.*)/e', 'strtoupper("\\1")', $string);
?>

which is : THIS IS MY LOWER STING

 

So if

'$string'

is wrong why it did not engender any error and gave a normal result as $string without quotation marks ? that is what I wanted to know

[Ingolme]

The first block of code you provided would actually output $STRING, the second one would output THIS IS MY LOWER STING

 

There would be no error message, it just would not work correctly.

[hisoka]

The first block of code you provided would actually output $STRING

 

No my friend the first block , I provided , outputs THIS IS MY LOWER STING and not $STRING . Try to run it yourself and you will see that it outputs THIS IS MY LOWER STING

Edited April 18, 2016 by hisoka

[justsomeguy]

I believe that is a side-effect of using the /e modifier, although I don't see it documented. It's also worth pointing out that the /e modifier is deprecated in PHP 5.5 and removed in PHP 7, as it can be a security risk. The preg_replace_callback function is the safe alternative.

 

edit: it's because this is what gets eval'd:

 

strtoupper("$string")

[hisoka]

I am still confused concerning this:

 

After some researches , I find that the parentheses are used to group the . and * together , I find , too, that .* does the same thing as (.*) . So I deleted the parentheses and run the code again like this :

<?php
$string = "this is my lower sting";
print preg_replace('/.*/e', 'strtoupper("\\1")', '$string');
?>

but nothing happened . If (.*) does the same thing as .* so why I got a blank page and nothing happened ?

[justsomeguy]

If (.*) does the same thing as .* so why I got a blank page and nothing happened ?

It doesn't do the same thing. If it did the same thing, why would we need another way to do it? The parentheses indicate that it should match that part. That's what the \1 reference is referring to, the match of subpattern #1, which is inside those parentheses. If you had a pattern like this:

 

/(\d)(\w)/

Then using \1 as a back-reference would refer to the \d match, and using \2 would refer to the \w match. The parentheses create a subpattern.

[hisoka]

understand . You really save my life

 

back references confused me a lot and now after your explanation I get it