Batch file edit text file

[msswamy]

Thanks for your solution and it solved problem with exclamation mark ('!').
Regarding my new line issue, below is my sample text file.

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[Components.Section]
 Element_1
 !Element_2
 Element_3

Below are the string variables,

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set "replace=[Components.Section]"
set "replaced=[Components.Section]\nMyElement_0"

Finally my text file should come like below,

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[Components.Section]
 MyElement_0
 Element_1
 !Element_2
 Element_3

But it is coming like below,

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[Components.Section]\nMyElement_0
 Element_1
 !Element_2
 Element_3

[T3RRY]

Thanks for your solution and it solved problem with exclamation mark ('!').
Regarding my new line issue, below is my sample text file.

\n is the literal string \n in batch scripting.

a linefeed can be defined as a variable using the following, however it requires delayed expansion to be enabled to use.

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@Echo Off
(Set LF=^


%= Linefeed. do not modify this or above three lines. =%)
Setlocal enabledelayedexpansion
>>"%~dp0outfile.txt" Echo/this is!LF!a demo
Type "%~dp0outfile.txt"

[penpen]

@msswamy: You didn't react to the post of T3RRY, so i don't know whether you found it solved your issue or not.

penpen