Quotes

pbbruce has asked for the wisdom of the Perl Monks concerning the following question:

Replies are listed 'Best First'.
Re: Quotes by dvergin (Monsignor) on Apr 21, 2001 at 05:20 UTC
Your presentation of the problem is a bit puzzling because `$dummp = "Phillip "B" Bruce";` is not a valid statement... Is this what you are after? `my $dummp = 'Phillip "B" Bruce'; print $dummp; # prints: Phillip "B" Bruce $dummp =~ s/"//g; print $dummp; # prints: Phillip B Bruce` [download]	[reply] [d/l] [select]
Re: Re: Quotes by pbbruce (Initiate) on Apr 21, 2001 at 05:48 UTC
That exactly what I thought I typed but I left out the ~ in my assignment statement. Thanks.	[reply]
Re: Re: Re: Quotes by Rudif (Hermit) on Apr 21, 2001 at 16:33 UTC
but I left out the ~ in my assignment statement. The perlman:perlop defines =~ as a binding operator, not an assignment: Binary ``=~'' binds a scalar expression to a pattern match. Rudif	[reply]
Re: Quotes by Beatnik (Parson) on Apr 21, 2001 at 13:52 UTC
Valid as well... `$dummp = qq/Phillip "B" Bruce/;` and `$dummp = q/Phillip "B" Bruce/;` and ofcourse `$dummp = "Phillip \"B\" Bruce";` Greetz Beatnik ... Quidquid perl dictum sit, altum viditur.	[reply] [d/l] [select]
Re: Quotes by rchiav (Deacon) on Apr 21, 2001 at 05:27 UTC
could you explain what you mean a little better? Your statement `$dummp = "Phillip "B" Bruce";` [download] really isn't a valid Perl statement because "Phillip" and "Bruce" are quoted and the B is not. Perl isn't going to know what you mean by that. Does your string have quotes around the B? If so, the above answer will do what you want it to. If not, we're going to need more info. -Rich	[reply] [d/l]