Counting Replacements in Strings

[gerkintrigg]

Hi everyone!

 

I wondered whether there's a way of counting the number of replacements made from a string replacement.

 

Example:

$string="I like strings. I think they are brilliant. I will get to the bottom of this issue.";
$word_to_replace="I";
$replacement="We";
string=preg_replace($word_to_replace, $replacement, $string);
echo $string;

 

That would just output:

We like strings. We think they are brilliant. We will get to the bottom of this issue.

 

But what I need (strange as it might seem) is this:

We1 like strings. We2 think they are brilliant. We3 will get to the bottom of this issue.

 

I understand loops and how to increase a counting variable, but not how to do this with a preg_replace function.

 

Any suggestions?

Thanks.

[taquitosensei]

just add $count as the 5th

string=preg_replace($word_to_replace, $replacement, $string,-1, $count); // -1 is no limit 

 

[rajivgonsalves]

I am not sure if this is the optimum way to do it but it kinda works

 

<?php
$counter = 1;
$string="I like strings. I think they are brilliant. I will get to the bottom of this issue.";
$string=preg_replace("#I#e", 'W.$counter++' , $string);
echo $string;
?>

[rajivgonsalves]

actually the code should have been, this will not throw a notice

 

<?php
$counter = 1;
$string="I like strings. I think they are brilliant. I will get to the bottom of this issue.";
$string=preg_replace("#I#e", '"W".$counter++' , $string);
echo $string;
?>

[gerkintrigg]

just add $count as the 5th

string=preg_replace($word_to_replace, $replacement, $string,-1, $count); // -1 is no limit 

 

That didn't work at all...

[cags]

What about it didn't work?

[taquitosensei]

that's straight from phps documentation.

mixed preg_replace  ( mixed $pattern  , mixed $replacement  , mixed $subject  [, int $limit = -1  [, int &$count  ]] )

If specified, this variable will be filled with the number of replacements done.

 

nm I just read your post all the way through.  You'll probably have to use a callback function to increment your $replacement.

 

[Andy-H]

$string="I like strings. I think they are brilliant. I will get to the bottom of this issue.";
$word_to_replace="I";
$replacement="We";
string=preg_replace($word_to_replace, $replacement, $string);
echo $string;

 

It didn't work because you didn't test your example code, you missed a $ from before "string" and and "I" is not a regular expression...

 

$string="I like strings. I think they are brilliant. I will get to the bottom of this issue.";
$word_to_replace="#I#";
$replacement="We";
$string=preg_replace($word_to_replace, $replacement, $string, -1, $count);
echo '<pre>' . "\r\n";
echo 'Output:' . "\t\t\t" . $string . "\r\n";
echo 'Replacements made:' . "\t" . number_format($count, 0) . "\r\n";
echo '</pre>';

 

Output:			We like strings. We think they are brilliant. We will get to the bottom of this issue.
Replacements made:	3

 

[gerkintrigg]

Baaa. I feel sheepish.

 

Eventually I did this as a 2 stage approach like this:

#original (case insensitive) word from database:
$uc_word=ucwords($word);
$upper_word=strtoupper($word);
$pattern = array(
'~\b'.$word.'\b(?![^<]*?>)~',
'~\b'.$uc_word.'\b(?![^<]*?>)~',
'~\b'.$upper_word.'\b(?![^<]*?>)~'
);

#Replacement highlighted phrase:	
$new_word = array(
'<span id="'.$word_all.'_nst_id_007" class="'.$style.'">'.$word.'</span>',
'<span id="'.$word_all.'_nst_id_007" class="'.$style.'">'.$uc_word.'</span>',
'<span id="'.$word_all.'_nst_id_007" class="'.$style.'">'.$upper_word.'</span>'
);
#now highlight the words:
$page = preg_replace($pattern, $new_word, $page);

Then:

$replacement='nst_id_';
$counter=1;
$page=preg_replace("#nst_id_007#e", '"$replacement".$counter++', $page);

[Alex]

You could've used substr_count