Escaping exclamation marks with delayed expansion - batch-file

I have a batch file I'm using to search for a piece of text and replace it with another. It works, but what's happening is that when the 'new' file is created, anything after an exclamation mark is deleted.
So original file
Hello! I have some cheese
Just becomes
Hello
Although the text I am trying to replace is fine.
I understand that since I'm using delayed expansion I need to somehow escape the exclamation marks with ^^! but can't figure out where to do this. Adding it at the echo just echoes the ^^! literally.
Any help would be appreciated.
set "rootname=Common Filename"
set "replace="old">"
set "replaced="new">"
set "source="%rootname%_unqiue_suffix.txt""
set "target=Fixed\%SOURCE%"
setlocal enableDelayedExpansion
(
for /F "tokens=1* delims=:" %%a in ('findstr /N "^" %source%') do (
set "line=%%b"
if defined line set "line=!line:%replace%=%replaced%!"
echo(!line!
)
) > %target%
endlocal

To avoid loss of exclamation marks, enable delayed expansion only when it is really needed and expand normal % variables and for meta-variables only when delayed expansion is disabled:
#echo off
setlocal EnableExtensions DisableDelayedExpansion
set "rootname=Common Filename"
set "replace="old">"
set "replaced="new">"
set "source=%rootname%_unqiue_suffix.txt"
set "target=Fixed\%source%"
(
for /F "tokens=1* delims=:" %%a in ('findstr /N "^" "%source%"') do (
set "line=%%b"
setlocal EnableDelayedExpansion
if defined line set "line=!line:%replace%=%replaced%!"
echo(!line!
endlocal
)
) > "%target%"
endlocal
This code still causes trouble in case the variables replace and replaced contain !-signs, because they are percent-expanded.

Related

Semicolons at beginning of the line, delims and FOR cycle

I cannot figure out how to correctly identify an empty/undefined variable having two semicolon one after one other or the line that starts with it.
This is the cycle:
for /F "delims=; tokens=1-7" %%m IN (testlist.txt) DO echo FUNCGROUP=%%r a=%%m b=%%n c=%%o d=%%p e=%%q f=%%s
I also tried adding "eol=;" and "......eol=" without success.
This is content of the first line of the file testlist.txt:
;xxxxxx;Active;;FALSE;con ter - dong;HWID000001;Item;sites/coll-
The result I need is, for the first cycle:
a=
b=xxxxxx
c=Active
d=
e=FALSE
f=con ter - dong
g=HWID000001
Thanks for any help.
To achieve that, you need to put a "NULL value" for empty fields in the lines of the file.
As there is no direct string substitution utilities with batch, you have to make substitutions beforehand to create "empty" fields. I suggest you to use a "NULL" character for empty fields like unbreakable space (Alt+0160).
In your case, this gives :
#echo off
SETLOCAL ENABLEDELAYEDEXPANSION
for /F "eol= tokens=*" %%l in (file.txt) DO (
SET "LINE=%%l"
SET "LINE=###!LINE:;;=; ;!###"
SET "LINE=!LINE:###;= ;!"
SET "LINE=!LINE:;###=; !"
SET "LINE=!LINE:###=!"
for /F "delims=; tokens=1-7" %%m in ("!LINE!") DO (
SET "RES=FUNCGROUP=%%r a=%%m b=%%n c=%%o d=%%p e=%%q f=%%s"
echo !RES: =!
)
)
Note that the SET "LINE=###!LINE:;;=; ;!###", SET "LINE=!LINE:###;= ;!" and the SET "LINE=!LINE:;###=; !" sections use the unbreakable space (Alt+0160) and replace beginning ";" with "Alt+0160;", the ending ";" with ";Alt+0160" and any following ";;" with ";Alt+0160;". The for loop parses then correctly the line, and next you just have to remove the unbreakable space to get "empty" variables.
EDIT: As brightly suggested by #jeb in the comments, you can also use quotes to handle empty fields. Each for loop variables can be then directly and simply unquoted.
#echo off
SETLOCAL ENABLEDELAYEDEXPANSION
for /F "eol= tokens=*" %%l in (file.txt) DO (
SET "LINE=%%l"
SET "LINE=###^"!LINE:;=^";^"!^"###"
SET "LINE=!LINE:###^";=^"^";!"
SET "LINE=!LINE:;^"###=;^"^"!"
SET "LINE=!LINE:###=!"
for /F "delims=; tokens=1-7" %%m in ("!LINE!") DO (
echo FUNCGROUP=%%~r a=%%~m b=%%~n c=%%~o d=%%~p e=%%~q f=%%~s
)
)
Unfortunately, you have not provided sufficient information for what you intend to do with your empty field, so this is only a demonstration to provide you with output similar to that which you've indicated in your question:
#Echo Off
SetLocal EnableExtensions DisableDelayedExpansion
For /F "Delims==" %%G In ("(Set Field[) 2>NUL") Do Set "%%G="
Set "i=0"
For /F UseBackQ^ Delims^=^ EOL^= %%G In ("testlist.txt") Do Call :GetFields "%%G"
Pause
GoTo :EOF
:GetFields
Set "Record=%~1"
Set /A i += 1
SetLocal EnableDelayedExpansion
Set "Index=1"
Set "Field[!Index!]=%Record:;=" & Set /A Index +=1 & Set "Field[!Index!]=%"
Echo(&Echo Record !i!
For /L %%G In (1,1,!Index!) Do If Not Defined Field[%%G] (Echo Field[%%G]=) Else Set Field[%%G]
Exit /B

How to get piped input in windows batch file?

I want to use command like xxx yyy | deal_everyline.bat, where the command xxx yyy will generate some output. When I use deal_everyline.bat in console, everything runs normally. However, when I test it with cat A.txt | deal_everyline.bat, it only output the first line. What should I do to get all lines?
#echo off
:loop
set input=
set /p input=
if "%input%" neq "" (
echo %input%
goto loop
)
penknife suggests using MORE with FOR /F, but that will corrupt tabs, and it hangs at 64K lines of input. Better to use FINDSTR "^". The odd FOR /F options syntax is a way to disable both the EOL and DELIMS options so that non-blank lines are preserved exactly, as long as they are < ~8191 bytes long.
#echo off
for /f delims^=^ eol^= %%A in ('findstr "^"') do (
echo(%%A
)
The above code will skip empty lines. If you want to preserve them, then use the FINDSTR /N option to prefix each line with the line number followed by a colon, save the value in an environment variable, and then use variable expansion to strip the prefix. Variables must be expanded with delayed expansion within the FOR loop, but delayed expansion cannot be enabled when %%A is expanded, else it will corrupt ! characters. So the delayed expansion is toggled on and off within the loop
#echo off
setlocal disableDelayedExpansion
for /f delims^=^ eol^= %%A in ('findstr /n "^"') do (
set "ln=%%A"
setlocal enableDelayedExpansion
set "ln=!ln:*:=!"
echo(!ln!
endlocal
)
If you want to preserve environment variable changes across iterations, then CALL out to a subroutine with delayed expansion constantly off. But beware that normal percent expansion of variables can fail depending on the content. For example unquoted <, >, &, and | will all cause problems.
#echo off
setlocal disableDelayedExpansion
for /f delims^=^ eol^= %%A in ('findstr /n "^"') do (
set "ln=%%A"
call :processLine
)
exit /b
:processLine
set "ln=%ln:*:=%"
echo(%ln%
exit /b

Batch Script - Findstr "/n" flag?

I found a example of what I wanted to do online, which was search a text file for a string and replace only that string and write out to a file. Which sort-of works...
The script does replace the correct text. But, its printing line numbers, which I don't want. However, if I remove the \n flag from findstr, it only prints lines containing data, and lines that aren't comments *(i.e. beginning with ";;").
How do I use findstr to print all the lines without the line numbers?
#echo off &setlocal
set "search=string to replace"
set "replace=replace string with me"
set "textfile=input.ini"
set "newfile=output.ini"
(for /f "delims=" %%i in ('findstr /n "^" "%textfile%"') do (
set "line=%%i"
setlocal enabledelayedexpansion
set "line=!line:%search%=%replace%!"
echo(!line!
endlocal
))>"%newfile%"
type "%newfile%"
The /n switch is required to keep the empty lines, so if want the empty lines in input be copied to the output you should not remove /n switch from findstr otherwise you do not need findstr and can directly read file with FOR. instead you can remove the line numbers by this modified code of yours:
#echo off
setlocal
set "search=string to replace"
set "replace=replace string with me"
set "textfile=input.ini"
set "newfile=output.ini"
(for /f "tokens=1* delims=:" %%i in ('findstr /n "^" "%textfile%"') do (
if "%%j" NEQ "" (
set "line=%%j"
setlocal enabledelayedexpansion
echo(!line:%search%=%replace%!
endlocal
) else (
echo(
)
))>"%newfile%"
type "%newfile%"
Also there is no need for set "line=!line:%search%=%replace%!" you can directly pass !line:%search%=%replace%! to echo
Or you can get ride of findstr altogether and simply write the FOR loop as below but you will loos empty lines
(for /f "usebackq delims="eol^= %%i in ("%textfile%") do (
set "line=%%i"
setlocal enabledelayedexpansion
echo(!line:%search%=%replace%!
endlocal
))>"%newfile%"
Please note that in "usebackq delims="eol^= there must be no spaces between the quotation mark(") and eol. It is not a typo.
This is to disable the default eol character(;) so the lines that beginning semicolon(;) will not be ignored by FOR /F command

I am writing a .bat program to find and replace text in a file without changing its position

I am writing a .bat program that will find and replace text in a file. The problem that I am having is that it is removing blank lines and left justifying the other lines. I need the blank lines to remain and the new text to remain in the same location. Here is what I have wrote, and also the result. Can anybody please help.
program:
#ECHO OFF
cls
cd\
c:
setLocal EnableDelayedExpansion
For /f "tokens=* delims= " %%a in (samplefile.tx) do (
Set str=%%a
set str=!str:day=night!
set str=!str:winter=summer!
echo !str!>>samplefile2.txt)
ENDLOCAL
cls
exit
samle File:
this line is the first line in my file that I am using as an example.This is made up text
the cat in the hat
day
winter
below is the result:
this line is the first line in my file that I am using as an example.This is made up text
the cat in the hat
night
summer
I need the lines, spaces and new text to remain in the same position while making the text replacement. Please help
Your use of "tokens=* delims= " will trim leading spaces. Instead, use "delims=" to preserve leading spaces.
FOR /F always skips empty lines. The trick is to insert something before each line. Typically FIND or FINDSTR is used to insert the line number at the front of each line.
You can use !var:*:=! to delete the the line number prefix from FINDSTR.
Use echo(!str! to prevent ECHO is off message when line is empty
It is more efficient (faster) to redirect only once.
#echo off
setlocal enableDelayedExpansion
>samplefile2.txt (
for /f "delims=" %%A in ('findstr /n "^" samplefile.txt') do (
set "str=%%A"
set "str=!str:*:=!"
set "str=!str:day=night!"
set "str=!str:winter=summer!"
echo(!str!
)
)
This still has a potential problem. It will corrupt lines that contain ! when %%A is expanded because of the delayed expansion. The trick is to toggle delayed expansion on and off within the loop.
#echo off
setlocal disableDelayedExpansion
>samplefile2.txt (
for /f "delims=" %%A in ('findstr /n "^" samplefile.txt') do (
set "str=%%A"
setlocal enableDelayedExpansion
set "str=!str:*:=!"
set "str=!str:day=night!"
set "str=!str:winter=summer!"
echo(!str!
endlocal
)
)
Or you could forget custom batch entirely and get a much simpler and faster solution using my JREPL.BAT utility that performs regular expression search and replace on text. There are options to specify multiple literal search/replace pairs.
jrepl "day winter" "night summer" /t " " /l /i /f sampleFile.txt /o sampleFile2.txt
I used the /I option to make the search case insensitive. But you can drop that option to make it case sensitive if you prefer. That cannot be done easily using pure batch.
#ECHO Off
SETLOCAL
(
FOR /f "tokens=1*delims=]" %%a IN ('find /n /v "" q27459813.txt') DO (
SET "line=%%b"
IF DEFINED line (CALL :subs) ELSE (ECHO()
)
)>newfile.txt
GOTO :EOF
:subs
SET "line=%line:day=night%"
SET "line=%line:winter=summer%"
ECHO(%line%
GOTO :eof
Thi should work for you. I used a file named q27459813.txt containing your data for my testing.
Produces newfile.txt
Will not work correctly if the datafile lines start ].
Revised to allow leading ]
#ECHO Off
SETLOCAL
(
FOR /f "delims=" %%a IN ('type q27459813.txt^|find /n /v "" ') DO (
SET "line=%%a"
CALL :subs
)
)>newfile.txt
GOTO :EOF
:subs
SET "line=%line:*]=%"
IF NOT DEFINED line ECHO(&GOTO :EOF
SET "line=%line:day=night%"
SET "line=%line:winter=summer%"
ECHO(%line%
GOTO :eof

skip=!count! is not working in windows(7) batch file

In for loop i have checked condition, if condition is true i have set the count value and skipped some lines in file,
#echo off
SetLocal EnableDelayedExpansion
set /a count=0
set for_parameters="skip=!count! delims="
for /f %for_parameters% %%a in ('list.txt') do (
echo %%a
if %%a==Exception: (
set /a count+=2
)
)
Endlocal
Its shows delims=" was unexpected at this time. Error
Can anyone help me to fix this problem .....
What jeb said about delayed expansion and FOR options is true - you can't use it - but that has nothing to do with the problem in your code.
You are using delayed expansion before you use the expression in your FOR statement, so there is no problem there.
Your problem is that you are attempting to set SKIP=0. The SKIP value must be >0 to be valid. The fix is simple: don't include the SKIP option if you don't want to skip any lines :-)
#echo off
SetLocal EnableDelayedExpansion
set /a count=0
set "skip="
if !count! gtr 0 set "skip=skip=!count!"
set for_parameters="!skip! delims="
for /f %for_parameters% %%a in ('list.txt') do (
echo %%a
if %%a==Exception: (
set /a count+=2
)
)
Endlocal
Expanding on jeb's point: you cannot do the following
for /f !for_parameters! %%a in ...
because FOR has special parsing rules. Most commands parse their options after delayed expansion. But FOR and IF parse their options before FOR variable expansion and delayed expansion take place. Neither FOR variables nor delayed expansion can be used to specify FOR or IF options.
In the for-options you can only use percent expansion, but not delayed expansion.
If your skip value itself is calculated in a block, then you need to extract the for loop into a function
You can use the More command to skip lines.
#echo off
:: By Elektro H#cker
SetLocal EnableDelayedExpansion
set /a count=2
for /F %%a in ('Type "list.txt" ^| MORE +!COUNT!') do (echo %%a)
Pause&exit

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