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    <pre><code>/ ^ # start of string ( # first group start (?: (?:[^?+*{}()[\]\\|]+ # literals and ^, $ | \\. # escaped characters | \[ (?: \^?\\. | \^[^\\] | [^\\^] ) # character classes (?: [^\]\\]+ | \\. )* \] | \( (?:\?[:=!]|\?&lt;[=!]|\?&gt;)? (?1)?? \) # parenthesis, with recursive content | \(\? (?:R|[+-]?\d+) \) # recursive matching ) (?: (?:[?+*]|\{\d+(?:,\d*)?\}) [?+]? )? # quantifiers | \| # alternative )* # repeat content ) # end first group $ # end of string / </code></pre> <p>This is a recursive regex, and is not supported by many regex engines. PCRE based ones should support it.</p> <p>Without whitespace and comments:</p> <pre><code>/^((?:(?:[^?+*{}()[\]\\|]+|\\.|\[(?:\^?\\.|\^[^\\]|[^\\^])(?:[^\]\\]+|\\.)*\]|\((?:\?[:=!]|\?&lt;[=!]|\?&gt;)?(?1)??\)|\(\?(?:R|[+-]?\d+)\))(?:(?:[?+*]|\{\d+(?:,\d*)?\})[?+]?)?|\|)*)$/ </code></pre> <hr> <p>.NET does not support recursion directly. (The <code>(?1)</code> and <code>(?R)</code> constructs.) The recursion would have to be converted to counting balanced groups:</p> <pre><code>^ # start of string (?: (?: [^?+*{}()[\]\\|]+ # literals and ^, $ | \\. # escaped characters | \[ (?: \^?\\. | \^[^\\] | [^\\^] ) # character classes (?: [^\]\\]+ | \\. )* \] | \( (?:\?[:=!] | \?&lt;[=!] | \?&gt; | \?&lt;[^\W\d]\w*&gt; | \?'[^\W\d]\w*' )? # opening of group (?&lt;N&gt;) # increment counter | \) # closing of group (?&lt;-N&gt;) # decrement counter ) (?: (?:[?+*]|\{\d+(?:,\d*)?\}) [?+]? )? # quantifiers | \| # alternative )* # repeat content $ # end of string (?(N)(?!)) # fail if counter is non-zero. </code></pre> <p>Compacted:</p> <pre><code>^(?:(?:[^?+*{}()[\]\\|]+|\\.|\[(?:\^?\\.|\^[^\\]|[^\\^])(?:[^\]\\]+|\\.)*\]|\((?:\?[:=!]|\?&lt;[=!]|\?&gt;|\?&lt;[^\W\d]\w*&gt;|\?'[^\W\d]\w*')?(?&lt;N&gt;)|\)(?&lt;-N&gt;))(?:(?:[?+*]|\{\d+(?:,\d*)?\})[?+]?)?|\|)*$(?(N)(?!)) </code></pre>
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    1. POIs there a regular expression to detect a valid regular expression?
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    1. USMarkus Jarderot
    1. USMarkus Jarderot
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    1. POIs there a regular expression to detect a valid regular expression?
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    3. VO
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    1. COVoting you up on the presumption that this actually works. Since I've only had two cups of coffee today, I'm not up to the task of verifying it.
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      1. USKirk Strauser
    2. COWill this validate substitutions and translations?
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      1. USslashmais
    3. COIt will validate just the regex part of substitutinos and translations. s/<this part>/.../
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      1. USMarkus Jarderot
 

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