[ale] vim search/replace syntax request

[Byron Jeff]

On Thu, Mar 10, 2011 at 10:36:26AM -0500, Dow Hurst wrote:
> In vim what you sent translates to my needs as listed below and works like a charm!  

Cool Beans...

> Thanks for the information!

No problem. Glad to help.

BAJ

> 
> :6,14 s/^\(.\{84\}\)1\(.*\)$/\10\2/
> 
> Best wishes,
> Dow
> ________________________________________________
> Dow Hurst, Research Scientist
> 340 Sullivan Science Bldg., Dept. of Chem. and Biochem.
> University of North Carolina at Greensboro
> PO Box 26170 Greensboro, NC 27402-6170
> 
> 
> 
> 
> On Thu, Mar 10, 2011 at 7:28 AM, Byron Jeff <byronjeff at mail.clayton.edu<mailto:byronjeff at mail.clayton.edu>> wrote:
> On Thu, Mar 10, 2011 at 02:04:15AM -0500, Dow Hurst wrote:
> 
> > I'd like to construct a vim based search and replace statement.  I want
> > to search between a set of line numbers for the 85th character on a line
> > and if it contains a "1" then turn it into a "0".  I just don't know how
> > to set up that particular search pattern and replace pattern.
> 
> The following should work in vim. Not sure about raw vi:
> 
> :%s/^\(.\{84\}\)1\(.*\)$/\10\2/
> 
> Here is an expanded version with some backslashes eliminated for explanation:
> 
> :%s/^ ( .{84} ) 1 ( .* ) $ / \1 0 \2 /
> 
> The :%s/^ is the standard substitute all lines and mark the beginning of the line.
> 
> The parens in the expression are grouping operators. Needs backslashes to
> mark as special operators. Each group captures and saves the match so that
> it can be reinserted in the substitution. The groups are numbered starting
> with 1.
> 
> So the first group captures .{84} which matches exactly 84 charaters.
> 
> The 1 matches the 1 in the 85th column.
> 
> The .* is the second group which captures the rest of the line after the
> 85th character. Probably could get away without it (*). The $ is the end of the
> line.
> 
> Once a line is matched then the substitution is pretty simple:
> 
> The \1 is the first group which is the first 84 characters.
> the 0 substitutes for the 1.
> the \2 is the group for the rest of the line.
> 
> (*) This should also work:
> 
> :%s/^\(.\{84\}\)1/\10/
> 
> Good Luck.
> 
> BAJ
> 
> >
> > It's been a long time since I've posted to the list.  It is nice to see the ALE group is flourishing.  I miss going to the meetings...
> >
> > Best wishes,
> > Dow
> > ________________________________________________
> > Dow Hurst, Research Scientist
> > 340 Sullivan Science Bldg., Dept. of Chem. and Biochem.
> > University of North Carolina at Greensboro
> > PO Box 26170 Greensboro, NC 27402-6170
> >
> >
> 
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> 
> 
> --
> Byron A. Jeff
> Department Chair: IT/CS/CNET
> College of Information and Mathematical Sciences
> Clayton State University
> http://cims.clayton.edu/bjeff
> _______________________________________________
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> http://mail.ale.org/mailman/listinfo/ale
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> 

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-- 
Byron A. Jeff
Department Chair: IT/CS/CNET
College of Information and Mathematical Sciences
Clayton State University
http://cims.clayton.edu/bjeff