using preg_replace()

[dsaba]

trying to understand how to use preg_replace()

I read this in the manual:

<?php
$string = 'April 15, 2003';
$pattern = '/(\w+) (\d+), (\d+)/i';
$replacement = '${1}1,$3';
echo preg_replace($pattern, $replacement, $string);
?>

 

the output of that sample regex is:

Array
(
    [0] => Array
        (
            [0] => April 15, 2003
        )

    [1] => Array
        (
            [0] => April
        )

    [2] => Array
        (
            [0] => 15
        )

    [3] => Array
        (
            [0] => 2003
        )

)

 

 

now seeing the example I want to do something similar with this code I made up:

<?php
$str = '$var = -34;
if ($this == -134) {
$x = 5 - -9;
$b = 7 + -9832;
$notThis = 8 - 10;
$again = 1234-1234;
$here = -9 + 78;
}';
$pat = '~[-+=](| )(-)[0-9]+~';
preg_match_all($pat, $str, $out);
preg_replace($pat, '${2}negative', $str);
output($out,100,100);
output($str, 100,100);
?>

 

I get this output for the $out:

Array
(
    [0] => Array
        (
            [0] => = -34
            [1] => = -134
            [2] => - -9
            [3] => + -9832
            [4] => = -9
        )

    [1] => Array
        (
            [0] =>  
            [1] =>  
            [2] =>  
            [3] =>  
            [4] =>  
        )

    [2] => Array
        (
            [0] => -
            [1] => -
            [2] => -
            [3] => -
            [4] => -
        )

)

 

I get this output for the changed $str with preg_replace

$var = -34;
if ($this == -134) {
$x = 5 - -9;
$b = 7 + -9832;
$notThis = 8 - 10;
$again = 1234-1234;
$here = -9 + 78;
}

 

 

So you see preg_replace() did not change anything! Apparently I'm not using it correctly. My regex grabs all negative signs, take note I said negative signs and not subtraction signs. I simply want to replace all negative signs in the $str string with the string 'negative'

 

The captured negative signs with my regex can be seen in $out[2]

So how can I replace these matched negative signs with the string 'negative' with the preg_replace() function?

Also if there is a better way to write my regex pattern for matching negative signs, let me know. I couldn't find a way to say '' or ' ' , (space or null) I used (| ), if i didnt' use the parenthesis it wouldn't catch it, however by using parenthesis it also thinks its a substring, and spits it out in $out[1] which is just a waste, because i'm not really looking for that substring, just the negative

 

Thanks for reading

[BenInBlack]

haystack

$var = -34;

if ($this == -134) {

$x = 5 - -9;

$b = 7 + -9832;

$notThis = 8 - 10;

$again = 1234-1234;

$here = -9 + 78;

}

 

regex

/([ ]\-)(?!([ ]\-|[ ][0-9]))/

 

replace with (note there is a space in front of negative)

negative

 

Result

$var = negative34;

if ($this == negative134) {

$x = 5 - negative9;

$b = 7 + negative9832;

$notThis = 8 - 10;

$again = 1234-1234;

$here = negative9 + 78;

}

 

Here is the command

$output = preg_replace(/([ ]\-)(?!([ ]\-|[ ][0-9]))/, negative,$var = -34; if ($this == -134) { $x = 5 - -9; $b = 7 + -9832; $notThis = 8 - 10; $again = 1234-1234; $here = -9 + 78; });

[dsaba]

your command came up with a parse error

but the format was like this:

preg_replace(PATTERN, 'negative', HAYSTACK);

                                ^^^

                              here is my source of confusion

 

in the manual example they do something like this:

$replacement = '${1}1,$3';

 

how this special format of  { } brackets and $ dollar sign works I don't know

you did none of this for the replacement, and it came up with a parse error anyways..

so...?

 

 

from my original post you can see my attempt:

${2}negative

why I chose 2 ? because in $out[2] are the captured substrings of negative signs in my original regex

 

[BenInBlack]

in the manual example they do something like this:

$replacement = '${1}1,$3';

 

ok, up in the Replacement notes of the of doc for preg_replace you will see that in this case they want a litteral 1 after the replace $1

so then needed to set the $1 apart from being look at as $11

 

The $1 and $3 match the parens in the PATTERN '/(\w+) (\d+), (\d+)/i', so $1 = what is returned with (\w+) and $3 = (\d+)/i

 

Ok here is a revised, the code using the $# and this I wrapped the first space with parens

so

$1 = ([ ])

$2 = (\-)

the rest is look ahead code (?!([ ]\-|[ ][0-9]))

 

so using this REPLACE - $1$2~negative~

I am restoring the (space) and the (-) in front of string ~negative~

 

Haystack

$var = -34;
if ($this == -134) {
   $x = 5 - -9;
   $b = 7 + -9832;
   $notThis = 8 - 10;
   $again = 1234-1234;
   $here = -9 + 78;
}

 

Patern

/([ ])(\-)(?!([ ]\-|[ ][0-9]))/

 

Replace

$1$2~negative~

 

Result

$var = -~negative~34;
if ($this == -~negative~134) {
   $x = 5 - -~negative~9;
   $b = 7 + -~negative~9832;
   $notThis = 8 - 10;
   $again = 1234-1234;
   $here = -~negative~9 + 78;
}

 

Code (and this time ill add the quotes ;-))

$output = preg_replace('/([ ])(\-)(?!([ ]\-|[ ][0-9]))/','$1$2~negative~','$var = -34; if ($this == -134) { $x = 5 - -9; $b = 7 + -9832; $notThis = 8 - 10; $again = 1234-1234; $here = -9 + 78; }');