Find and replace a string with special characters using a batch file - file

Question

So I'm having some trouble getting this batch file to work correctly. I've tried several different ways to find and replace a string.
Problem: We have to update a database ever so often and in doing so, a file gets changed. I have to change the one file back to get things to work. This line:
<network ipAdress="172.24.55.32" networkPort="9100" />
gets changed to this line:
<file filename="fffff" />
and needs to be changed back to the first line with the ip address.
I have tried to use a find and replace subroutine and can get it to work on the first part of changing "file filename" to "network ipAdress" but the quotes in the second section keep the script from working correctly. This is the code as I have it.
#echo off
setlocal
call :FindReplace "file filename" "network ipAdress" BCPrint.XML
Timeout 2
call :FindReplace "fffff" "172.24.55.32" networkPort="9100" BCPrint.XML
exit /b
:FindReplace <findstr> <replstr> <file>
set tmp="%temp%\tmp.txt"
If not exist %temp%\_.vbs call :MakeReplace
for /f "tokens=*" %%a in ('dir "%3" /s /b /a-d /on') do (
for /f "usebackq" %%b in (`Findstr /mic:"%~1" "%%a"`) do (
echo(&Echo Replacing "%~1" with "%~2" in file %%~nxa
<%%a cscript //nologo %temp%\_.vbs "%~1" "%~2">%tmp%
if exist %tmp% move /Y %tmp% "%%~dpnxa">nul
)
)
del %temp%\_.vbs
exit /b
:MakeReplace
>%temp%\_.vbs echo with Wscript
>>%temp%\_.vbs echo set args=.arguments
>>%temp%\_.vbs echo .StdOut.Write _
>>%temp%\_.vbs echo Replace(.StdIn.ReadAll,args(0),args(1),1,-1,1)
>>%temp%\_.vbs echo end with
I have also tried setting the second part as a variable like:
set R2=172.24.55.32" networkPort="9100
call :FindReplace "file filename" "network ipAdress" BCPrint.XML
Timeout 2
call :FindReplace "fffff" "%R2%" BCPrint.XML
With escape characters as well:
set ^R2=172.24.55.32" networkPort="9100^
I am sure there are a couple of different ways that I have tried that I haven't listed like escape characters without using a variable. I've come to the end of what I can figure out.
Is there a way to find and replace a line that has special characters with a line that has special characters? Any advice on how to do either the entire line or that second section with the quotes and space would be great. Thanks for your time.

Answer 1

Download JREPL.BAT written by Dave Benham which is a batch file / JScript hybrid to run a regular expression replace on a file using JScript and store it in same directory as the batch file below.
#echo off
call "%~dp0jrepl.bat" "file filename=\x22fffff" "network ipAdress=\x22172.24.55.32\x22 networkPort=\x229100" /XSEQ /F BCPrint.XML /O -
\x22 is " specified in hexadecimal form as an argument string can't contain a double quote character.

Answer 2

Passing arguments containing quotation marks is very difficult, because there are several places where those characters are recognised, so you had to build up individual escape sequences (^) every time.
The following example script fails for every call attempt:
#echo off
setlocal EnableExtensions DisableDelayedExpansion
rem // Define argument here:
set "_ARG=R2=172.24.55.32" networkPort="9100"
setlocal EnableDelayedExpansion
rem // Echo argument string to prove that variable is set:
echo(!_ARG!
echo/
rem // Call sub-routine with delayed expansion:
call :SUB "!_ARG!"
endlocal
rem // Call sub-routine with normal (immediate) expansion:
call :SUB "%_ARG%"
rem // Call sub-routine with double normal expansion (`call`):
call :SUB "%%_ARG%%"
endlocal
exit /B
:SUB
setlocal DisableDelayedExpansion
rem // Delayed expansion disabled during argument expansion:
set "ARG=%~1"
setlocal EnableDelayedExpansion
rem // Delayed expansion enabled for argument display (`echo`):
echo(!ARG!
endlocal
endlocal
exit /B
The easiest way to avoid such problems is to pass the variable name to the sub-routine rather than its value:
#echo off
setlocal EnableExtensions DisableDelayedExpansion
rem // Define argument here:
set "_ARG=R2=172.24.55.32" networkPort="9100"
setlocal EnableDelayedExpansion
rem // Echo argument string to prove that variable is set:
echo(!_ARG!
echo/
endlocal
rem // Call sub-routine and pass variable name:
call :SUB _ARG
endlocal
exit /B
:SUB
setlocal EnableDelayedExpansion
set "#ARG=%~1"
rem /* Normal expansion is used to get the variable name;
rem delayed expansion is used to read its value: */
echo(!%#ARG%!
endlocal
exit /B

Answer 3

To update a string in a file that contains "quotes", the following solution will work.
Create a file named replace_quot.ps1 in the same directory as your bat file. Place the following 2 lines in this script.
param([string]$Name_of_File)
(Get-Content $Name_of_File) -replace '"', '"' | set-content $Name_of_File
In your bat file enter the below lines
set File_Name=File Name.txt
set "Find_String=<file filename="fffff" />"
set "Replace_String=<network ipAdress="172.24.55.32" networkPort="9100" />"
set "Find_String=%Find_String:"="%"
set "Replace_String=%Replace_String:"="%"
powershell -command "(Get-Content '%File_Name%') -replace '[\x22]', '"' | Set-Content '%File_Name%'"
powershell -command "(Get-Content '%File_Name%') -replace '%Find_String%', '%Replace_String%' | Set-Content '%File_Name%'"
Powershell.exe -executionpolicy remotesigned -File replace_quot.ps1 -Name_of_File "%File_Name%"
Update the File_Name value to reflect your file name. The bat script will replace the quotes with their hex value in both the strings and the file. Next it will find the string that needs to be replaced and restore it to the desired value. Lastly it will call the replace_quot.ps1 script to update the file's quote hex values to the actual quote character.