Regular expressions - Ignore phrase in brackets

ladovnik

Hello! I have a problem of using regular expressions. I want to use preg_replace to replace all word's book in the variable $str which aren't in quotes to <B>book</B>. I make this pattern:

$str='I have super book! This book is cool! I like fantastic books but i dont like political "book"...';
$searchFor='book';
$new_str=preg_replace('#([^"]|^)('.$searchFor.')([^"]|$)#', '\\1<b>\\2</b>\\3', $str);

Result for this code is:

I have super book! This book is cool! I like fantastic books but i dont like political "book"...

But... if I replace variable $str to:

$str='I have super book! This book is cool! I like fantastic books but i dont like political "bad book, bad"...';

I have result:

I have super book! This book is cool! I like fantastic books but i dont like political "bad book, bad"...

How I must replace this code, to word book in quote (in that form: "[some_text]book[some_text2]") be ignore?

Please help me and I'm salute
ladovnik

PS I'm sorry for my English 😉

NogDog

There may be a better way, but this seems to work:

<?php
$str='"Hello, book." I like my book. I do not like "this book". Books are great.';
$searchFor = 'book';
$parts = explode('"', $str);
$counter = 1;
foreach($parts as $ix => $val)
{
   if($counter++ % 2)
   {
      $parts[$ix] = preg_replace('/'.preg_quote($searchFor).'/i', "<b>$0</b>", $val);
   }
}
$str = implode('"', $parts);
echo $str;

ladovnik

It's working! Thank You! Who have any else method for this problem?(maybe better?)

nrg_alpha

Another method may be using negative look behind / ahead assertions:

$str='"Hello, book." I like my book. I do not like "this book". Books are great.';
$str = preg_replace('#(?!<".*?)book(?!.*?")#i', "<b>$0</b>", $str);
echo $str;

Output:

"Hello, book." I like my book. I do not like "this book". [B]Book[/B]s are great.

Granted, if you did want to make an exception and bold Books (plural as well as single), you could use this pattern:

$str = preg_replace('#(?!<".*?)books?(?!.*?")#i', "<b>$0</b>", $str);

EDIT -Oops.. hold on.. just noticed the other word book wasn't bold.. I know why.. I'll revise...

nrg_alpha

Ok.. perhaps I was on the wrong course. Going in a route more simular to Nog's...

$str='"Hello, book." I like my book. I do not like "this book". Books are great.';
$arr = preg_split('#(".*?book.*?")#', $str, -1, PREG_SPLIT_DELIM_CAPTURE);
foreach($arr as $val){
   if(!strstr($val, '"')){
      $val = str_ireplace('book', '<b>book</b>', $val);
   }
   $newStr .= $val;
}
echo $newStr;

Output:

"Hello, book." I like my [B]book[/B]. I do not like "this book". [B]Book[/B]s are great.

If you don't want to bold any instance of the word book that has any of its letters capitalized, then simply use str_replace instead of str_ireplace.

EDIT - I am treating this like a search engine that highlights even part of words.. hence Books get the treatment.. if this is not desireable, you can simply use the following line in place of str_replace:

$val = preg_replace('#\bbook\b#', "<b>book</b>", $val);

ladovnik

nrg_alpha;10895417 wrote:

$str='"Hello, book." I like my book. I do not like "this book". Books are great.';
$arr = preg_split('#(".*?book.*?")#', $str, -1, PREG_SPLIT_DELIM_CAPTURE);
foreach($arr as $val){
   if(!strstr($val, '"')){
      $val = str_ireplace('book', '<b>book</b>', $val);
   }
   $newStr .= $val;
}
echo $newStr;

Yes! This is this! This fix my problem! Thank You very, very much!🙂

I'm salute,
ladovnik