So in batch there are a lot of special characters that are reserved and cause issues if you use them, if I want to set up a password like so:
SET PASSWORD="123%6!###$^&*_-"
Some of these characters will be stripped after assignment for example the percent '%' character is interpreted as a parameter, is there a way to make batch see that as a string instead of interpreting special characters? no I cannot escape those special characters, this is basically a file that we tell the user to edit teh batch file and set the password.
You could use some magic batch functions to handle with any string, like magic echo.
But I suppose it's better that the user doesn't edit the batch file itself, instead edit a password file, like pwd.txt.
Then you can read this file simply with:
#echo off
setlocal EnableDelayedExpansion
set /p password= < pwd.txt
echo !password!
This read the first line of the file and has no problems with any character.
Related
I have a variable with html code (having major, minor symbols)
and I need it to be exported and appended to a txt
set WORD1=^<p^>^<strong^>PROBLEM^</strong^> with something;n^</p^>
I can't echo the variable like this
echo %WORD1%
And I need to export it/append it to a file. I used:
echo %WORD1% >body.txt
But this generates an error as the variable has minor/Major symbol
If I double quote the variable, the exported text is exported with double quotes (and obviusly this is not what i need)
To define the variable, use:
set "WORD1=<p><strong>PROBLEM</strong> with something;n</p>"
To "export" the variable, use:
(
set /P "=%WORD1%"
echo/
) > body.txt < NUL
To define a variable in a safe way you need to enclose the whole assignment expression in quotation marks:
set "WORD1=<p><strong>PROBLEM</strong> with something;n</p>"
This avoids the need of escaping, unless the string itself contains quotation marks on its own.
Note that this syntax only works with command extensions enabled, but this is the default in Command Prompt anyway.
To return/expand an arbitrary string in a safe manner, even when it contains quotation marks on its own, is to use delayed variable expansion:
echo(!WORD1!
To safely write the output to a file, place the redirection expression at the front:
> "body.txt" echo(!WORD1!
You can also do this on one line:
set "WORD1=<p><strong>PROBLEM</strong> with something;n</p>"&&>body.txt cmd/v/cecho.!WORD1!
rem :: Or, without defining a previous variable, if it is not necessary:
>body.txt <nul set/P "=<p><strong>PROBLEM</strong> with something;n</p>"
I define variable like:
set LOGGING_MANAGER=-Djuli-logback.configurationFile=file:"%CATALINA_HOME%\conf\logback.xml"
Notice, that I wrap %CATALINA_HOME%\conf\logback.xml in double quotes, because path may contain spaces. And at execution step this variable will substitute to java program.
Example of substitution:
path_to_jre/java.exe %LOGGING_MANAGER%
Next code I can't change: (it's from catalina.bat file from Apache Tomcat)
if not "%LOGGING_MANAGER%" == "" ...
This if statement will fail, because variable contains quotes.
There I have one of 2 problems:
if statement fail (error occured).
without quotes when substitute to java program have problems (space is delimiter).
How can I change variable definition so that the path may contain spaces and if statement will work fine?
note: Not tested, but after reading the catalina.bat this is a way to deal with the problem.
If you are running this from a batch file, you can use
set LOGGING_MANAGER_FILE="%CATALINA_HOME%\conf\logback.xml"
set LOGGING_MANAGER=-Djuli-logback.configurationFile=file:%%LOGGING_MANAGER_FILE%%
set _RUNJAVA=CALL "%JRE_HOME%\bin\java.exe"
call catalina.bat
What it does is
Save the file name with quotes in a separate variable
Set the LOGGING_MANAGER variable using a escaped reference of the previous variable. As it will not be expanded, the final content of the LOGGING_MANAGER variable is -Djuli-logback.configurationFile=file:%LOGGING_MANAGER_FILE%
That way, when the if "%LOGGING_MANAGER%"=="" ... is executed, there will be not any problematic quote.
And now the set _RUNJAVA line. Inside catalina.bat, the contents of the _RUNJAVA are used to initialize the _EXECJAVA variable that will launch the server. _RUNJAVA is initialized in setclasspath.bat IF it has not been initialized by the user (code from setclasspath.bat).
rem Don't override _RUNJAVA if the user has set it previously
if not "%_RUNJAVA%" == "" goto gotRunJava
rem Set standard command for invoking Java.
rem Also note the quoting as JRE_HOME may contain spaces.
set _RUNJAVA="%JRE_HOME%\bin\java.exe"
The default value is "%JRE_HOME%\bin\java.exe", we just add a CALL batch command before to force a second iteration of the batch parser that will expand the quoted %LOGGING_MANAGER_FILE% variable in the final command line.
for %%a in ("%CATALINA_HOME%\conf\logback.xml") do set "LOGGING_MANAGER=-Djuli-logback.configurationFile=file:%%~sa"
should set logging_manager appropriately so that you don't have to "quote the filename"
for %%a in ("%CATALINA_HOME%\conf") do net use o: "\\%userdomain%\sharenameforthedrivewherecatalina_homeisresident%%~pa"
set "LOGGING_MANAGER=-Djuli-logback.configurationFile=file:o:\conf\logback.xml"
may also work - if you know the share name for the drive where catalina_home is resident and assuming o: is an unused drive.
Special characters in batch files are a pain, but I haven't found the right workaround for properly escaping the first two characters of this particular string I'm trying to pass the application.
SET pass=^&AntiBatchfileString
A_Program.exe /pass=%pass%
Things I have tried:
:: Escaping the escape twice, first for ^, second for &.
SET pass=^^^^&AntiBatchfileString
echo %pass%
:: Combining escapes.
SET first=^^
SET second=^^&AntiBatchfileString
SET pass=%first%%second%
echo %pass%
:: Preventing expansion
SET first=^^
SET second=^^&AntiBatchfileString
SET pass=!first!%second%
echo %pass%
:: I got this to print correctly
SET "pass=^&AntiBatchfileString"
echo ^^%pass%
Still when passing the last one it doesn't accept the login, I don't know what the final output is. That got me thinking maybe it was trying to do another expansion when passing the parameter to the application, so I quoted that as well.
SET "pass=^&AntiBatchfileString"
A_Program.exe "/pass=^^%pass%"
It's still not working, I'm not sure what I'm missing at this point.
Supposing you want the string ^&AntiBatchfileString literally, this is the best set syntax, as most special characters (^ & ( ) < > | and also the standard delimiters , ; = SPACE TAB) lose their particular meaning as soon as ther are placed in between "", and the "" themselves do not become part of the variable value:
set "pass=^&AntiBatchfileString"
This works only as long as the command extensions are on, which is the Windows default anyway (type cmd /? and see the /E option).
When expanding (reading) a variable like "%pass%" (with enclosing ""), special characters are still treated literally.
However, as soon as you expand it like %pass% (no ""), they get back their special meaning. So you have the following options:
Use set "pass=^^^&AntiBatchfileString", where ^^ escapes the literal ^ and ^& the literal & when reading like %pass%.
Enable delayed expansion (see set /? about how it works and setlocal /? or cmd /? about how to enable it), where the variable value is expanded (read) at a point of time where parsing of special characters has already been completed.
I prefer the latter approach, because no special escaping is necessary, and it can also deal with " appearing in the string value (even if unsymmetrically present).
By the way, " can also be escaped by ^", as long as this does not appear within unescaped "".
Nevertheless, % signs cannot be escaped like ^% in a batch file, because percent expansion happens before escaping, but you need to double them like %% to get one literal one each, independent whether or not the string is in between "".
Note that on the console, %% does not work.
Finally, literal ! are consumed by the delayed expansion feature when enabled, therefore you need to pay particular attention to those in case, by escaping them like ^!, or also by intelligently toggling delayed expansion (hence to enable it only when it is actually needed and to disable it otherwise, when a literal string is provided, like in a set command line, for instance, when expanding a standard variable like %pass% and when reading a for variable like %%I (batch file) or %I (console), for example). Of course this is also not the ultimate solution, because you need setlocal and endlocal to enable/disable delayed expansion, which are intended to localise environment changes, so any variable changes since the most recent setlocal command are lost as soon as endlocal is executed (there are some tricks for passing a variable value over the endlocal barrier though).
If you want to use % as a string without escaping in a batch file:
Like %20, you can use %%%20.
git clone "https:// abc.com /D%%%220an"
I'm having a problem where I can't get all characters (especially special characters) to show when I try to echo them
echo 1234567890q!w#e#r$t%y^u&io()pa;s/d.f,ghjklzxcvb nmQAZXSWEDCVFRTGBNHYUJMKIOLP
When using this code in batch, it doesn't show the characters (especially the special characters) but instead say "This program is not recognizable as an internal or external command, operable program or batch file.". The reason to why this is saying this is because the lines have special characters but how do I show the special characters in echo on? Thanks people.
Neither of those lines results in that particular error message when I test them from the command line (although the first results in "Environment variable 123456etc. not defined"). You have properly quoted your "variable=value" pair, and there is no problem setting CHARSET to that value.
I suspect the problem is that when you echo %charset% the & is being evaluated as a command separator, and evaluating the string thereafter as a new command. Try retrieving the value of CHARSET in the delayed expansion style.
set "charset=1234567890q!w#e#r$t%%y^u&io()pa;s/d.f,ghjklzxcvb nmQAZXSWEDCVFRTGBNHYUJMKIOLP"
setlocal enabledelayedexpansion
echo !charset!
endlocal
By the way, if you want to include a literal percent, you should do it double (%%).
For example I have the file sample.txt. This file contains:
1111101
2222203
3333303
44444A1
55555A1
66666A1
Now, I want to replace user defined specific pattern. For example I have other file where use defines what he want to replace with. Example the file name is replace.txt. This file contains 2 Columns, first column for the pattern and the 2nd column for the text to be replace.
Example:
replace.txt
2222203 2222203ADD
55555A1 55555A1SUB
Now, when the batch file has been executed, I would like the file sample.txt to have a contents like this:
1111101
2222203ADD
3333303
44444A1
55555A1SUB
66666A1
Also is it possible to have a "space" as part of the text to be replace(column 2?
You may use FindRepl.bat program that is a Batch-JScript hybrid application that perform these replacements in a very efficient way via regular expressions; it uses JScript language that is standard in all Windows versions from XP on. In the basic use of FindRepl.bat you redirect the input file to it and place two strings as parameters, a "search" string and a "replacement" string. For example:
< sample.txt FindRepl.bat "2222203" "2222203ADD"
Previous command will replace all 2222203 strings in the file by 2222203ADD. In order to perform the replacement of several strings, you may include several alternatives in both the search and replacement strings separated by a pipe character (this is called alternation), and include the /A switch to select this feature; for example:
< sample.txt FindRepl.bat "2222203|55555A1" /A "2222203ADD|55555A1SUB"
If you want to define the set of replacements in a separated file, you just need to load the strings from the file, assemble the alternations in two variables and use they in FindRepl preceded by an equal-sign to indicate that they are variables, not literal strings. If you want that the strings may have spaces, then you must use a different character to separate the search and replace parts in the file. For example, if you use a colon in replace.txt file this way:
2222203:2222203 ADD
55555A1:55555A1 SUB
Then the Batch file below solve your problem:
#echo off
setlocal EnableDelayedExpansion
set "search="
set "replace="
for /F "tokens=1,2 delims=:" %%a in (replace.txt) do (
set "search=!search!|%%a"
set "replace=!replace!|%%b"
)
set "search=!search:~1!"
set "replace=!replace:~1!"
< sample.txt FindRepl.bat =search /A =replace
You may download FindRepl.bat and review an explanation of its use from this site; you must place it in the same folder of previous program or, better yet, in a folder included in PATH variable.