I have a problem understanding why a submatch is performing the way it does. 

Start with: {FIELDA}.getTime()-{FIELDB}-{EXISTING_UNITS_LOST}

Apply this regex: (})([^\.]|$)

Replace using: $1.toDoubleValue()

Gives this: {FIELDA}.getTime()-{FIELDB}.toDoubleValue(){EXISTING_UNITS_LOST}.toDoubleValue()

I want to replace all closing braces that don't already have a '.' after them witha '.toDoubleValue()' so I don't replace the } that already have a method attached, such as '.getTime().

My expression tries to submatch on the '}' and then replace  just that using the '$1'. The problem is that the '-' after {FIELDB} disappears even though its outside the submatch.

I use the brilliant RegExp tester at http://www.gskinner.com/RegExr/ to try expressions and I just can't get it to only replace the '}' and not the following 'non-dot' character as well.

Any help much appreciated.

Raw Match Pattern:
\}(?!\.)

Raw Replace Pattern:
}.toDoubleValue()

$sourcestring after replacement:
{FIELDA}.getTime()-{FIELDB}.toDoubleValue()-{EXISTING_UNITS_LOST}.toDoubleValue()

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Two points.

1) You seem confused by the replacement operation.  You can replace matches, not sub-matches.   You can use backreferences in your replacement but the entire pattern match is what is replaced.

2) You don't need to escape a period in a character class.

Raw Match Pattern:
\}(?:(?!\.)|(?=$))

Raw Replace Pattern:
$0.toDoubleValue()

$sourcestring after replacement:
{FIELDA}.getTime()-{FIELDB}.toDoubleValue()-{EXISTING_UNITS_LOST}.toDoubleValue()

---

Michael

"In theory, theory and practice are the same. In practice, they are not."
Albert Einstein

Hi Mash,

Yes, I did think I could replace submatches as you suspected. 

You're correct regex is fantastic and works perfectly - I would never have come up with that but will try and understand it so as to improve my regex skills.

 Thank you very much,

 Mike