Ignore percent sign in batch file
batch-file, escaping
Solution
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`.
Problem
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.