Replacing all latex occurences $$ ... $$ to inline $ ... $ by modifying the string in PHP

Question

I have tried a couple of solutions regarding "replace second occurence in a string" (like here) but could not get the solution for my problem.

String example:

$string = "$$ f'(x) = 12x^2 + 4x $$ $$ f'(x) = 12x+2x-12 $$ $$ f'(x) = 12x^3 +4x^2 $$";

Goal:

$string = "$ f'(x) = 12x^2 + 4x $ $ f'(x) = 12x+2x-12 $ $ f'(x) = 12x^3 +4x^2 $";

I found this gist snippet:

function str_replace_n($search, $replace, $subject, $occurrence)
{
    $search = preg_quote($search);
    return preg_replace("/^((?:(?:.*?$search){".--$occurrence."}.*?))$search/", "$1$replace", $subject);
}

And tried to carefully call it on the second instance by:

$string = str_replace_n('$$', '\)', $string, 2);

So that I can then go over the remaining $$ and replace them by \(.

However, the function does nothing when I call it like that.

I thought it is an escape issue since $$ are rexeg operators, but even escaping them did not change the outcome (the preg_quote in the function should do the magic, I assume): $string = str_replace_n('\$\$', '\\)', $string, 2); - nope …

What am I doing wrong?

anubhava · Accepted Answer · 2018-02-06T11:05:34.307

1

You can search using this regex:

\$\$((?:(?!\$\$).)*)\$\$

and replace using:

\\($1\\)

RegEx Demo

Lookahead expression (?:(?!\$\$).) matches any character that is not a $$

Code:

$re = '/\$\$((?:(?!\$\$).)*)\$\$/';
$str = '$$ f\'(x) = 12x^2 + 4x $$  $$ f\'(x) = 12x+2x-12 $$  $$ f\'(x) = 12x^3 +4x^2 $$';
$subst = '\\\\($1\\\\)';

$result = preg_replace($re, $subst, $str);

As @Wiktor suggested below a simple non-greedy regex pattern will also work here:

\$\$(.*?)\$\$

edited Feb 06 '18 at 11:05

answered Feb 06 '18 at 11:04

anubhava

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Did you just answer within 60 seconds after I have asked?! ... and it works... hallelujah, outstanding. – Avatar Feb 06 '18 at 11:05
1

It is much easier to use `\$\$(.*?)\$\$`. Using a tempered greedy token here is not justified. – Wiktor Stribiżew Feb 06 '18 at 11:05
@WiktorStribiżew: Yes right, in this case `\$\$(.*?)\$\$` will also work – anubhava Feb 06 '18 at 11:06
Anyway, the most efficient pattern variation for this approach will be [`\$\$([^$]*(?:\$(?!\$)[^$]*)*)\$\$`](https://regex101.com/r/GCoQcS/1). – Wiktor Stribiżew Feb 06 '18 at 11:12
Feel free to add it as an answer or we are going to see this comment section overfilled. @WiktorStribiżew – revo Feb 06 '18 at 11:14

Srdjan M. · Answer 2 · 2018-02-06T11:22:01.090

Regex: \${2}([^$]+)\${2} or \${2}(.+?)\${2} Substitution: ($1)

Details:

{n} Matches exactly n times
[^] Match a single character not present in the list
+ Matches between one and unlimited times

PHP code:

$string = "$$ f'(x) = 12x^2 + 4x $$  $$ f'(x) = 12x+2x-12 $$  $$ f'(x) = 12x^3 +4x^2 $$";

$string = preg_replace('~\${2}([^$]+)\${2}~', '($1)', $string);
print_r($string);

Output:

( f'(x) = 12x^2 + 4x )  ( f'(x) = 12x+2x-12 )  ( f'(x) = 12x^3 +4x^2 )

Replacing all latex occurences $$ ... $$ to inline \( ... \) by modifying the string in PHP

2 Answers2