Perl Programming Language :: converting code via regex


	converting code via regex HI! I am converting some old code: $_ = " if (a.weight = '') and (a.w <> 'N') then" to the following $_ = "if ( a.weight[ a.L_NO_ ].ToString() == 0 ) and ( a.w <> 'N' ) then using s/(\w)\.\b$WordList{ $Word }\b/$1\.$WordList{ $Word }\[ $1.L_NO_ \].ToString/g; Why does this not change the"a.w". I have a hash of key words that I am looping through I am looking for a finished product of _$ = "if( ( a.weight[ a.L_NO_ ].ToString() == 0 ) && ( a.w[ a.L_NO_ ].ToString() != 'N' ) )" Thanks, Jerry On Tue, 22 May 2007 08:21:23 -0500, Jerry <jpres@general-steel.com> wrote: > I am converting some old code: > $_ = " if (a.weight = '') and (a.w <> 'N') then" > to the following > $_ = "if ( a.weight[ a.L_NO_ ].ToString() == 0 ) and ( a.w <> 'N' ) then ^^^^^ How did '' become 0 ? > using > s/(\w)\.\b$WordList{ $Word }\b/$1\.$WordList{ $Word }\[ $1.L_NO_ > \].ToString/g; No need to escape most of that in the replacement part, because those characters are only special in a regular expression (the left-hand part of the s/// operator), not a plain string (the right-hand part). Although I had to leave in the escaped [ otherwise perl assumes it's part of the $WordList{} expression and tries to interpolate it. > Why does this not change the"a.w". I have a hash of key words that I am > looping through It works for me. If you'd shown a short but complete program that illustrates your problem (as the posting guidelines for this group suggest) we could have shown you what you did wrong. #!/usr/bin/perl use strict; use warnings; $_ = " if ( a.weight == 0 ) and ( a.w <> 'N' ) then"; my %WordList = ( foo => 'weight', bar => 'w' ); for my $Word ( qw(foo bar) ) { s{(\w)\.\b$WordList{ $Word }\b} {$1.$WordList{ $Word }\[ $1.L_NO_ ].ToString()}g; } print "$_\n"; Produces: if ( a.weight[ a.L_NO_ ].ToString() == 0 ) and ( a.w[ a.L_NO_ ].ToString() <> 'N' ) then Apart from the removal of the unneccessary back-whacks and the use of {} as the delimiter (for better wrapping in this post), the s/// remains the same as your example. > I am looking for a finished product of > _$ = "if( ( a.weight[ a.L_NO_ ].ToString() == 0 ) && ( a.w[ > a.L_NO_ ].ToString() != 'N' ) )" What happened to the " then" at the end? ok! s/then//; my $wordCNT = 0; my ( @Word ) = /\.(\w+)/g; foreach my $Word ( @Word) { my %WordList = ( "footage" => "footage", "qty" => "qty", "total" => "total", "totalcost" => "totalcost", "count" => "count", "weight" => "weight", "estcost" => "estcost", "cost" => "cost", "totwgt" => "totwgt", "wt_SLASH_ft" => "wt_SLASH_ft", "tothigh" => "tothigh", "wt_SLASH_each" => "wt_SLASH_each", "wt_SLASH_sht" => "wt_SLASH_sht", "c" => "c", "w" => "w", "dimension" => "dimension", "freight" => "freight", "onhand" => "onhand", "ft_oh" => "ft_oh", "vendor" => "vendor", "ordered" => "ordered", "recd" => "recd", "ln_NO_" => "ln_NO_", "item_NO_" => "item_NO_", "amount" => "amount", "price" => "price", "wgt_ea" => "wgt_ea", "priceea" => "priceea", "unit" => "unit", "um" => "um", "u_SLASH_m" => "u_SLASH_m", "dl_NO_" => "dl_NO_", "pt_NO_" => "pt_NO_", "d" => "d", "p" => "p", "sku_NO_" => "sku_NO_", "invitem_NO_" => "invitem_NO_", "i" => "i", "pt_NO_" => "pt_NO_", "sell" => "sell", "invqty" => "invqty", "description" => "description", "t" => "t", "qtyout" => "qtyout", "qtyout" => "qtyout", "wgtout" => "wgtout", "pc" => "pc", "uofm" => "uofm", "baldue" => "baldue", "ptyp" => "ptyp", "po_NO_" => "po_NO_", "ap_NO_" => "ap_NO_", "fob" => "fob", "product" => "product", "onorder" => "onorder", "qtyin" => "qtyin", "wgtin" => "wgtin", "invdate" => "invdate", "sr" => "sr", "city" => "city", "name" => "name", "pt" => "pt", ); if( $WordList{ $Word } ) { #print "$File, $Word$#WordList$WordList{ $Word }\n"; #print $_ if /\"N\" \) \)/i; s/(\w)\.\b$WordList{ $Word }\b \);/$1\.$WordList{ $Word }\[ $1.L_NO_ \] \n/g if /get/; s/(\w)\.\b$WordList{ $Word }\b/$1\.$WordList{ $Word }\[ $1.L_NO_ \]/g if /^\s+(\w)\./ && $wordCNT == 0 && !/$WordList{ $Word }\[/i; s/(\w)\.\b$WordList{ $Word }\b/$1\.$WordList{ $Word }\[ $1.L_NO_ \].ToString/g if /sorts/ && $wordCNT > 0; s/(\w)\.\b$WordList{ $Word }\b/$1\.$WordList{ $Word }\[ $1.L_NO_ \].ToString/g if( /= \"/ \|\| /!= \"/ \|\| /== \"/ ) && !/\[/ && !/\* \w\./g; s/(\w)\.\b$WordList{ $Word }\b /$1\.$WordList{ $Word }\[ $1.L_NO_ \].ToString /g if( /!= \"n\"/i && $wordCNT > 0 ); #print $_ if /\"N\" \) \)/i; s/(\w)\.\b$WordList{ $Word }\b\s+==\s+\"(\w+)\"/utils\.IsMatch$ $1\.$WordList{ $Word }\[ $1.L_NO_ \].ToString\($, \"$2\" \)/g if !/\.ToString/ && !/\* \w\./g; s/(\w)\.\b$WordList{ $Word }\b\s+\!=\s+\"(\w+)\"/utils\.NotMatch$ $1\.$WordList{ $Word }\[ $1.L_NO_ \].ToString\($, \"$2\" \)/g if !/\.ToString/ && !/\* \w\./g; # s/(\w)\.\b$WordList{ $Word }\b\s+\)/$1\.$WordList{ $Word }\[ $1.L_NO_ \].ToString \)/g if !/\),/; s/(\w)\.\b$WordList{ $Word }\b\.ToString/$1\.$WordList{ $Word }\[ $1.L_NO_ \]\.ToString/g; s/(\w)\.\b$WordList{ $Word }\b\s+= /$1\.$WordList{ $Word }\[ $1.L_NO_ \] = /g; s/(\w)\.\b$WordList{ $Word }\b\s+\)/$1\.$WordList{ $Word }\[ $1.L_NO_ \] \)/g if !/\),/ && $wordCNT == 0; s/(\w)\.\b$WordList{ $Word }\b/float.Parse$ $1\.$WordList{ $Word }\[ $1.L_NO_ \].ToString\($ \)/g if !/(\w)\.$WordList{ $Word }\[/; # s/(\w)\.$WordList{ $Word };/float.Parse$ $1\.$WordList{ $Word }\[ $1.L_NO_ \].ToString\($ \);/g; # s/(\w)\.$WordList{ $Word } /float.Parse$ $1\.$WordList{ $Word }\[ $1.L_NO_ \].ToString\($ \) /g; # s/(\w)\.$WordList{ $Word }$/float.Parse$ $1\.$WordList{ $Word }\[ $1.L_NO_ \].ToString\($ \);/g; $wordCNT++; } } "Dave Weaver" <zen13@zen.co.uk> wrote in message news:465301cd$0$31824$db0fefd9@news.zen.co.uk... - Hide quoted text - > On Tue, 22 May 2007 08:21:23 -0500, Jerry <jpres@general-steel.com> > wrote: >> I am converting some old code: >> $_ = " if (a.weight = '') and (a.w <> 'N') then" >> to the following >> $_ = "if ( a.weight[ a.L_NO_ ].ToString() == 0 ) and ( a.w <> 'N' ) >> then > ^^^^^ > How did '' become 0 ? >> using >> s/(\w)\.\b$WordList{ $Word }\b/$1\.$WordList{ $Word }\[ $1.L_NO_ >> \].ToString/g; > No need to escape most of that in the replacement part, because those > characters are only special in a regular expression (the left-hand > part of the s/// operator), not a plain string (the right-hand part). > Although I had to leave in the escaped [ otherwise perl assumes it's > part of the $WordList{} expression and tries to interpolate it. >> Why does this not change the"a.w". I have a hash of key words that I >> am >> looping through > It works for me. If you'd shown a short but complete program that > illustrates your problem (as the posting guidelines for this group > suggest) we could have shown you what you did wrong. > #!/usr/bin/perl > use strict; > use warnings; > $_ = " if ( a.weight == 0 ) and ( a.w <> 'N' ) then"; > my %WordList = ( foo => 'weight', bar => 'w' ); > for my $Word ( qw(foo bar) ) { > s{(\w)\.\b$WordList{ $Word }\b} > {$1.$WordList{ $Word }\[ $1.L_NO_ ].ToString()}g; > } > print "$_\n"; > Produces: > if ( a.weight[ a.L_NO_ ].ToString() == 0 ) and > ( a.w[ a.L_NO_ ].ToString() <> 'N' ) then > Apart from the removal of the unneccessary back-whacks and the use of > {} as the delimiter (for better wrapping in this post), the s/// > remains the same as your example. >> I am looking for a finished product of >> _$ = "if( ( a.weight[ a.L_NO_ ].ToString() == 0 ) && ( a.w[ >> a.L_NO_ ].ToString() != 'N' ) )" > What happened to the " then" at the end? Jerry <jpres@general-steel.com> wrote: > s/then//; You want "authenticate" to become "auticate" ? -- Tad McClellan SGML consulting t@augustmail.com Perl programming Fort Worth, Texas Please don't top-post. If you're replying to someone, trim their message to something small enough to provide context for your post, and put your replies below the relevant parts of the post you're replying to. Jerry <jpres@general-steel.com> wrote: > my $wordCNT = 0; > my ( @Word ) = /\.(\w+)/g; ... That is not a short but complete program that exhibits your problem. When I run it it outputs nothing. What did you expect? > foreach my $Word ( @Word) { > my %WordList = ( "footage" => "footage", > "qty" => "qty", > "total" => "total", > # ... snip words ... > "pt" => "pt", > ); So the only point of %WordList is to see if a word in the original $_ is 'valid' ? There's no need to initialize %WordList each time through the loop, and there are less verbose ways to initialize it: my %WordList = map { $_ => 1 } qw(footage qty total ... pt); for my $Word ( /\.(\w+)/g ) { ... } > if( $WordList{ $Word } ) > { > s/(\w)\.\b$WordList{ $Word }\b \);/$1\.$WordList{ $Word }\[ $1.L_NO_ \] \n/g if /get/; By the very nature of your code, if you get here $WordList{$Word} will always be the same as $Word, so why use the more verbose form? s/(\w)\.\b$Word\b \);/$1\.$Word\[ $1.L_NO_ \] \n/g if /get/; Also (as I mentioned in my previous post) you don't need all those back-slashes in the right-hand side of the s///: So that line can become: s/(\w)\.\b$Word\b \);/$1.${Word}[ $1.L_NO_ ] \n/g if /get/; The \b's in the left-hand side are redundant too, given your word list above: s/(\w)\.$Word \);/$1.${Word}[ $1.L_NO_ ] \n/g if /get/; There - that's a little more readable than your original line. > s/(\w)\.\b$WordList{ $Word }\b/$1\.$WordList{ $Word }\[ $1.L_NO_ \] > ToString/g if( /= \"/ \|\| /!= \"/ \|\| /== \"/ ) && !/\[/ && !/\\w\./g; What's the "/g" on the last pattern match for? if( /= \"/ \|\| /!= \"/ \|\| /== \"/ ) is exactly the same as: if /= "/ because if your string contains '!= "' or '== "' it will still match the first pattern. And double quotes don't need to be escaped in a regex. I find the "and" "or" and "not" operators to be much more readable than "&&" "\|\|" and "!": ... if /= "/ and not /\[/ and not /\\w\./; > s/(\w)\.\b$WordList{ $Word }\b /$1\.$WordList{ $Word }\[ $1.L_NO_ \] > ToString /g if( /!= \"n\"/i && $wordCNT > 0 ); s/(\w)\.$Word /$1.${Word}[ $1.L_NO_ ].ToString() /g if /!= "n"/i and $wordCNT > 0; If you're still having problems, post a short but complete program (which uses 'strict' and 'warnings' and includes some sample data) that demonstrates your problem. Tell us what output you see, and what you're expecting to see.