CODEWITHSUNDEEP
phpregexpreg-replace
Learning
Learning

Reputation: 868

Regex pattern extract string between curly braces and exclude curly braces

$str='\add[sometext]{\begin{equation}\label{eqn:3}
f_{1} =
\begin{cases}}
\beta_{1} + \beta_{2}f_{2} & f_{2}\leq \gamma\\
\beta_{1} + \beta_{2}\gamma + \beta_{4}(f_{2}-\gamma) & f_{2} >\gamma
\end{cases}sdsdssd,
\end{equation}}
 it may have some extra code here with {}
 \end{equation}}'

I need to extract the string between \add[sometext]{ and }(i.e.till \add tag end curly braces)The string between \add[sometext]{ and } may varies so I can't specify these string in regex pattern I should only consider starting and ending curly braces of \add[sometext]

Expected output:

\begin{equation}\label{eqn:3}
    f_{1} =
    \begin{cases}
    \beta_{1} + \beta_{2}f_{2} & f_{2}\leq \gamma\\
    \beta_{1} + \beta_{2}\gamma + \beta_{4}(f_{2}-\gamma) & f_{2} >\gamma
    \end{cases}sdsdssd,
    \end{equation}

I tried:

$str=preg_replace('/\\\\add\s*\[\s*\w*\]\s*{(.*?)}/s,$1,$match)

I don't know how to get related curly braces (i.e.\add tag start { till end })

Upvotes: 0

Views: 480

Answers (3)

Learning
Learning

Reputation: 868

I have got a regex for my requirement.

$str = preg_replace('/\\\\add\s*\[.*]\s*{(.*?)\\\\end{(.[^\s]*?)}}/s', "$1\\end{\$2}", $str);

Working demo

Upvotes: 0

Federico Piazza
Federico Piazza

Reputation: 30985

You can use a simple regex like this:

\{([\s\S]*)\}

Regular expression visualization

Working demo

enter image description here

Match information

MATCH 1
1.  [21-226]    `\begin{equation}\label{eqn:3}
f_{1} =
\begin{cases}}
\beta_{1} + \beta_{2}f_{2} & f_{2}\leq \gamma\\
\beta_{1} + \beta_{2}\gamma + \beta_{4}(f_{2}-\gamma) & f_{2} >\gamma
\end{cases}sdsdssd,
\end{equation}`

As you can see in the match information, the captured content is what you need.

The idea behind this regex is

\{([\s\S]*)\}
      ^--- Capture everything in a greedy way from the first `{` to the last `}`

But also you can do the same thing if you use the s flag (single line):

(?s)\{(.*)\} --> using inline `s` flag
    \{(.*)\} --> using external `s` flag

For PHP code you can have:

$re = "@\\{(.*)\\}@s"; 
$str = "\$str='\add[sometext]{\begin{equation}\label{eqn:3}\nf_{1} =\n\begin{cases}}\n\beta_{1} + \beta_{2}f_{2} & f_{2}\leq \gamma\\\n\beta_{1} + \beta_{2}\gamma + \beta_{4}(f_{2}-\gamma) & f_{2} >\gamma\n\end{cases}sdsdssd,\n\end{equation}}'"; 

preg_match($re, $str, $matches);

Update:

You can use this regex for your updated comments in the question:

\{([\s\S]*equation\})\}

Working demo

Upvotes: 1

benny
benny

Reputation: 315

how about this:

$str='\add[sometext]{\begin{equation}\label{eqn:3}
f_{1} =
\begin{cases}
\beta_{1} + \beta_{2}f_{2} & f_{2}\leq \gamma\\
\beta_{1} + \beta_{2}\gamma + \beta_{4}(f_{2}-\gamma) & f_{2} >\gamma
\end{cases}sdsdssd,
\end{equation}}';

$str= preg_match('/\\\\add\s*\[\s*\w*\]\s*{(.*?)}$/s',$str,$match);

var_dump($match[1]);

Upvotes: 1

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