The CreateProcess documentation states that in order to run batch file one has to use cmd.exe /C <path to bat>. In fact it can run batch files just fine unless path to a script contains spaces (Microsoft confirms that in the description of the MS14-019 vulnerability). I wonder if it's possible to escape spaces in the path to batch file to make CreateProcess work. Adding quotes doesn't help, CreateProcess fails with the error:
cannot spawn "<path to batch>": No such file or directory
Update
A workaround is to use short file names as pointed by #jac in the comments. I still wonder why enclosing a path in double quotes works for normal executables but doesn't work for batch files.
There is no escape character for CreateProcess.
Since the code is broken and you can't fix it, you'll have to work around the problem. For example, create a junction point to the target directory and launch the batch file via the junction point path, or use short paths as jac suggested. (Do note that not all volumes will necessarily have short paths enabled, but if you are dealing with the system volume it is probably safe to assume that they will be.)
I am trying to write a batch file in windows, where "%0" needs to be understood as "%0" instead of the name of the batch file. How can I achieve this?
Thanks!
You may use the "escape" trickery, whereby %, preceded by another %, will be treated like a literal:
%%0
Sorry... Just googled it.
Use double percent signs.
"%%0" will be fine.
I think this is a simple batch programming question, but after much searching, I can't find the answer.
I'm trying to figure out how to refer to files whose names contain difficult characters.
It seems that double quotes have the effect of treating most enclosed characters literally. For example, for a file named ^^.txt, dir "^^.txt" will find the file, while dir ^^.txt won't.
However, I don't know how to escape %. For example, for a file named %ERRORLEVEL%.txt, none of these find the file:
dir "%ERRORLEVEL%.txt"
dir "%%ERRORLEVEL%%.txt"
dir "^%ERRORLEVEL^%.txt"
Any suggestions would be appreciated.
The escape character for the Windows command prompt is ^. It works without the quotes.
dir ^%errorlevel^%.txt
For your other example ^^.txt, use:
dir ^^^^.txt
Always avoid % and ^ in filenames. Rename % to 'percent' in a global file manager.
They are next to impossible to handle properly in many batch commands.
In a batch script you must double the %, eg.:
#echo off
dir %%errorlevel%%.txt
Btw. you should not use cmd default variable or command names for other things (errorlevel).
I receive files which names contain spaces and change every week (the name contains the week number)
IE, the file for this week looks like This is the file - w37.csv
I have to write a script to take this file into account.
I didn't succeed in writing this script.
If I do :
$FILE="This is the file - w*.csv"
I don't find /dir/${FILE}
I tried "This\ is\ the\ file - w*.csv"
I tried /dir/"${FILE}" and "/dir/${FILE}"
But I still can't find my file
It looks like the space in the name needs the variable to be double-quoted but, then, the wildcard is not analysed.
Do you have an idea (or THE answer)?
Regards,
Olivier
echo /dir/"This is the file - w"*.csv
or — you almost tried that —
echo /dir/This\ is\ the\ file\ -\ w*.csv
Use a bash array
v=( /dir/This\ is\ the\ file - w*.csv )
If there is guaranteed to be only one matching file, you can just expand $v. Otherwise, you can get the full list of matching files by expanding as
"${v[#]}"
or individual matches using
"${v[0]", "${v[1]}", etc
First of all, you should not use the dollar sign in an assignment.
Moreover, wildcard expansion is not called in an assignment. You can use process substitution for example, though:
FILE=$(echo 'This is the file - w'*.csv)
Note that the wildcard itself is not included in the quotes. Quotes prevent wildcard expansion.
I have a batch file which moves files from one folder to another. The batch file is generated by another process.
Some of the files I need to move have the string "%20" in them:
move /y "\\myserver\myfolder\file%20name.txt" "\\myserver\otherfolder"
This fails as it tries to find a file with the name:
\\myserver\myfolder\file0name.txt
Is there any way to ignore %? I'm not able to alter the file generated to escape this, such as by doubling percent signs (%%), escaping with / or ^ (caret), etc.
You need to use %% in this case. Normally using a ^ (caret) would work, but for % signs you need to double up.
In the case of %%1 or %%i or echo.%%~dp1, because % indicates input either from a command or from a variable (when surrounded with %; %variable%)
To achieve what you need:
move /y "\\myserver\myfolder\file%%20name.txt" "\\myserver\otherfolder"
I hope this helps!
The question's title is very generic, which inevitably draws many readers looking for a generic solution.
By contrast, the OP's problem is exotic: needing to deal with an auto-generated batch file that is ill-formed and cannot be modified: % signs are not properly escaped in it.
The accepted answer provides a clever solution to the specific - and exotic - problem, but is bound to create confusion with respect to the generic question.
If we focus on the generic question:
How do you use % as a literal character in a batch file / on the command line?
Inside a batch file, always escape % as %%, whether in unquoted strings or not; the following yields My %USERNAME% is jdoe, for instance:
echo My %%USERNAME%% is %USERNAME%
echo "My %%USERNAME%% is %USERNAME%"
On the command line (interactively) - as well as when using the shell-invoking functions of scripting languages - the behavior fundamentally differs from that inside batch files: technically, % cannot be escaped there and there is no single workaround that works in all situations:
In unquoted strings, you can use the "^ name-disrupter" trick: for simplicity, place a ^ before every % char, but note that you're not technically escaping % that way (see below for more); e.g., the following again yields something like My %USERNAME% is jdoe:
echo My ^%USERNAME^% is %USERNAME%
In double-quoted strings, you cannot escape % at all, but there are workarounds:
You can use unquoted strings as above, which then requires you to additionally ^-escape all other shell metacharacters, which is cumbersome; these metacharacters are: <space> & | < > "
Alternatively, unless you're invoking a batch file, , you can individually double-quote % chars as part of a compound argument (most external programs and scripting engines parse a compound argument such as "%"USERNAME"%" as verbatim string %USERNAME%):
some_exe My "%"USERNAME"%" is %USERNAME%
From scripting languages, if you know you're calling a binary executable, you may be able to avoid the whole problem by forgoing the shell-invoking functions in favor of the "shell-free" variants, such as using execFileSync instead of execSync in Node.js.
Optional background information re command-line (interactive) use:
Tip of the hat to jeb for his help with this section.
On the command line (interactively), % can technically not be escaped at all; while ^ is generally cmd.exe's escape character, it does not apply to %.
As stated, there is no solution for double-quoted strings, but there are workarounds for unquoted strings:
The reason that "^ name-disrupter" trick (something like ^%USERNAME^%) works is:
It "disrupts" the variable name; that is, in the example above cmd.exe looks for a variable named USERNAME^, which (hopefully) doesn't exist.
On the command line - unlike in batch files - references to undefined variables are retained as-is.
Technically, a single ^ inside the variable name - anywhere inside it, as long as it's not next to another ^ - is sufficient, so that %USERNAME^%, for instance, would be sufficient, but I suggest adopting the convention of methodically placing ^ before each and every % for simplicity, because it also works for cases such as up 20^%, where the disruption isn't even necessary, but is benign, so you can apply it methodically, without having to think about the specifics of the input string.
A ^ before an opening %, while not necessary, is benign, because ^ escapes the very next character, whether that character needs escaping - or, in this case, can be escaped - or not. The net effect is that such ^ instances are ultimately removed from unquoted strings.
Largely hypothetical caveat: ^ is actually a legal character in variable names (see jeb's example in the comments); if your variable name ends with ^, simply place the "disruptive" ^ somewhere else in the variable name, as long as it's not directly next to another ^ (as that would cause a ^ to appear in the resulting string).
That said, in the (very unlikely) event that your variable has a name such as ^b^, you're out of luck.
In batch files, the percent sign may be "escaped" by using a double percent sign ( %% ).
That way, a single percent sign will be used within the command line. from http://www.robvanderwoude.com/escapechars.php
I think I've got a partial solution working. If you're only looking to transfer files that have the "%20" string in their name and not looking for a broader solution, you can make a second batch file call the first with %%2 as the second parameter. This way, when your program tries to fetch the second parameter when it hits the %2 in the text name, it will replace the %2 with an escaped %2, leaving the file name unchanged.
Hope this works!
How to "escape" inside a batch file withoput modify the file**
The original question is about a generated file, that can't be modified, but contains lines like:
move /y "\\myserver\myfolder\file%20name.txt" "\\myserver\otherfolder"
That can be partly solved by calling the script with proper arguments (%1, %2, ...)
#echo off
set "per=%%"
call generated_file.bat %%per%%1 %%per%%2 %%per%%3 %%per%%4
This simply sets the arguments to:
arg1="%1"
arg2="%2"
...
How to add a literal percent sign on the command line
mklement0 describes the problem, that escaping the percent sign on the command line is tricky, and inside quotes it seems to be impossible.
But as always it can be solved with a little trick.
for %Q in ("%") do echo "file%~Q20name.txt"
%Q contains "%" and %~Q expands to only %, independent of quotes.
Or to avoid the %~ use
for /F %Q in ("%") do echo "file%Q20name.txt"
You should be able to use a caret (^) to escape a percent sign.
Editor's note: The link is dead now; either way: It is % itself that escapes %, but only in batch files, not at the command prompt; ^ never escapes %, but at the command prompt it can be used indirectly to prevent variable expansion, in unquoted strings only.
The reason %2 is disappearing is that the batch file is substituting the second argument passed in, and your seem to not have a second argument. One way to work around that would be to actually try foo.bat ^%1 ^%2... so that when a %2 is encountered in a command, it is actually substituted with a literal %2.