Trivial string substitution/parser

[Graham Breed]

Duncan Booth wote:

> If you must insist on using backslash escapes (which introduces the
> question of how you get backslashes into the output: do they have to be
> escaped as well?) then use string.Template with a custom pattern.

If anybody wants this, I worked out the following regular expression
which seems to work:

(?P<escaped>\\)\$ | #    backslash escape pattern
\$(?:
  (?P<named>[_a-z][_a-z0-9]*)    | # delimiter and Python identifier
  {(?P<braced>[_a-z][_a-z0-9]*)} | # delimiter and braced identifier
  (?P<invalid>)              # Other ill-formed delimiter exprs
)

The clue is string.Template.pattern.pattern

So you compile that with verbose and case-insensitive flags and set it
to "pattern" in a string.Template subclass.  (In fact you don't have
to compile it, but that behaviour's undocumented.)  Something like

>>> regexp = """
... (?P<escaped>\\\\)\\$ | # backslash escape pattern
... \$(?:
...   (?P<named>[_a-z][_a-z0-9]*)    | # delimiter and identifier
...   {(?P<braced>[_a-z][_a-z0-9]*)} | # ... and braced identifier
...   (?P<invalid>)              # Other ill-formed delimiter exprs
... )
... """
>>> class BackslashEscape(Template):
...     pattern = re.compile(regexp, re.I | re.X)
...


                   Graham