Need help in Python regular expression

[Jean-Michel Pichavant]

To the OP,

I suggest if you haven't yet Kodos, to get it here 
http://kodos.sourceforge.net/.
It's a python regexp debugger, a lifetime saver.

Jean-Michel

John S wrote:
> On Jun 11, 10:30 pm, meryl <silverburgh.me... at gmail.com> wrote:
>   
>> Hi,
>>
>> I have this regular expression
>> blockRE = re.compile(".*RenderBlock {\w+}")
>>
>> it works if my source is "RenderBlock {CENTER}".
>>
>> But I want it to work with
>> 1. RenderTable {TABLE}
>>
>> So i change the regexp to re.compile(".*Render[Block|Table] {\w+}"),
>> but that breaks everything
>>
>> 2. RenderBlock (CENTER)
>>
>> So I change the regexp to re.compile(".*RenderBlock {|\(\w+}|\)"),
>> that also breaks everything
>>
>> Can you please tell me how to change my reg exp so that I can support
>> all 3 cases:
>> RenderTable {TABLE}
>> RenderBlock (CENTER)
>> RenderBlock {CENTER}
>>
>> Thank you.
>>     
>
> Short answer:
>
> r = re.compile(r"Render(?:Block|Table)\s+[({](?:TABLE|CENTER)[})]")
>
> s = """
>     blah blah blah
>     blah blah blah RenderBlock {CENTER} blah blah RenderBlock {CENTER}
>     blah blah blah RenderTable {TABLE} blah blah RenderBlock (CENTER)
>     blah blah blah
> """
>
> print r.findall(s)
>
>
>
> output:
> ['RenderBlock {CENTER}', 'RenderBlock {CENTER}', 'RenderTable
> {TABLE}', 'RenderBlock (CENTER)']
>
>
>
> Note that [] only encloses characters, not strings; [foo|bar] matches
> 'f','o','|','b','a', or 'r', not "foo" or "bar".
> Use (foo|bar) to match "foo" or "bar"; (?xxx) matches xxx without
> making a backreference (i.e., without capturing text).
>
> HTH
>
> -- John Strickler
>