wscrsurfdude wrote:
> Ik have an uml file
What is an "uml file" ?
> I want to read with readlines. I have the following
> code:
> infile = open("out2.txt","r")
> for line in infile.readlines():
> print line
>
> The print statement just gives the data,
You get what can be read from the file when opened as a text file and
read line by line. Be sure that neither file.readlines() nor print won't
invent data that are not in the file.
> not the uml headings. Is there
> a solution which also gives the uml headings.
If they're not in the file, obviously, no - whatever "uml heading" might
be.
--
bruno desthuilliers
python -c "print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for
p in '[EMAIL PROTECTED]'.split('@')])"
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