You can pass a function to replace. Var r = a. Replace(/(f)/, function(v) { return v.toUpperCase(); }); Explanation a.
Replace( /(f)/, "$1".toUpperCase()) In this example you pass a string to the replace function. Since you are using the special replace syntax ($N grabs the Nth capture) you are simply giving the same value. The toUpperCase is actually deceiving because you are only making the replace string upper case (Although the return value will still be "$1").A.
Replace( /(f)/, String.prototype.toUpperCase. Apply("$1")) Believe it or not the semantics of this expression are exactly the same.
– Evan Carroll May 26 at 17:57 @Evan Carroll: Please see my answer. – kay May 26 at 17:58 Ah, I see what you mean, I'm upercasing "\$1". Not the result of the voodoo that replace will do that is apparently substituting $1 for the first capture group.
– Evan Carroll May 26 at 18:04.
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