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POVariable number of matches / back references in Regex?
 

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  1. POVariable number of matches / back references in Regex?
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    <p>I spent some time recently converting some SCSS code to Stylus, and one of my regex steps was to strip out the <code>$</code> character in variable names, while formatting those names to camelCase rather than hyphenated-case. </p> <p>For example, <code>$long-hyphenated-variable-name: value;</code> became <code>longHyphenatedVariableName = value;</code> but parameter names like <code>text-size</code> did not become <code>textSize</code>.</p> <p>anyway, my Regex skills are pretty limited, so I could only figure out how to do it as a series of passes, looking first for variable names that might have as much as 6 parts (5 hyphens) and then for shorter ones on subsequent passes and passing each part of the string as a separate backreference so I could capitalize the first letter. But I wondered if there was a better way to do it: i.e. is there a way to get a variable number of sub-string matches within the string and supply back-references in accordance with how many you got? I just couldn't find any information on this.</p> <p>This is my code, in six steps, which I'd like to condense to one if it's possible (I'm thinking of writing a multi-step regex routine that can convert SCSS to Stylus for future use). In each case the search expression is followed by the replacement expression:</p> <pre><code>\$(\S*?)\-(\S*?)\-(\S*?)\-(\S*?)\-(\S*?)\-(.) $1\u$2\u$3\u$4\u$5\u$6 \$(\S*?)\-(\S*?)\-(\S*?)\-(\S*?)\-(.) $1\u$2\u$3\u$4\u$5 \$(\S*?)\-(\S*?)\-(\S*?)\-(.) $1\u$2\u$3\u$4 \$(\S*?)\-(\S*?)\-(.) $1\u$2\u$3 \$(\S*?)\-(.) $1\u$2 \$(.) $1 </code></pre>
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    1. USAlan Moore
    1. USlunelson
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    1. COJust FYI, I added the `boost-regex` tag because that seems to be the flavor Sublime Text uses.
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      1. USAlan Moore
 

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