[VELOCITY-395] Semantics of escaping unclear

• Type: Improvement
• Status: Resolved
• Priority: Minor
• Resolution: Fixed
• Affects Version/s: 1.3.1
• Fix Version/s: 1.6
• Component/s: Documentation
• Environment: Operating System: Windows 2000; Platform: PC
• Bugzilla Id: 35738
• Assignee: Unassigned
• Reporter: Frank Fischer
• Created: 14/Jul/05 23:11
• Updated: 02/Sep/08 23:12
• Resolved: 02/Sep/08 23:12

Description

I think the semantics of the escaping operator \ (backslash) are not documented
clearly. Especially I miss this ones:

Why generates
\$foo ==> $foo
but
\${foo} ==> \${foo}
this seems inconsistent to me. I expect
\${foo} ==> ${foo}

I understand that
${foo:bar}
raises an error since there are no : (colons)
allowed in a variable-name.
But why raises
\${foo:bar}
also an error? I would expect that \$ disables the function of $
as an operator, so that the following

{..}

will not be parsed as
a variable name.

There should be a simple way to generate something like
${foo:bar}
or even
${not a variable name at all &%$}
The workaround
#set($D='$')
${D}

{foo:bar}

is undocumented and REALLY ugly.