path-separator | 易学教程

Why doesn't this path work to open a Windows file in PERL?

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问题 I tried to play with Strawberry Perl, and one of the things that stumped me was reading the files. I tried to do: open(FH, "D:\test\numbers.txt"); But it can not find the file (despite the file being there, and no permissions issues). An equivalent code (100% of the script other than the filename was identical) worked fine on Linux. 回答1: As per Perl FAQ 5, you should be using forward slashes in your DOS/Windows filenames (or, as an alternative, escaping the backslashes). Why can't I use "C:

Why doesn't this path work to open a Windows file in PERL?

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Why doesn't this path work to open a Windows file in PERL?

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Simple way of converting slashes in a Makefile?

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问题 I need to convert all paths with '\' in them to '/'. The makefile is quite long and doing this manually is impossible. Is there some way to quickly convert them? Keep in mind that a global replace is not possible because '\' is also used to denote that a command is continued on the following line. 回答1: It looks like you could do this with a sed command: sed -e 's/\\\(.\)/\/\1/g' This converts any backslash followed by some other character (which doesn't include newline) into a forward slash

Application.Executablepath in C# has mixed separator characters

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问题 I'm using someone else's code (licensed) on two different machines. On one machine, the Application.ExecutablePath returns the result the programmer must have expected, on the other it does not. Both are Windows 7 machines. On my machine, the Application.ExecutablePath returns something like: "C:\\Dir1\\Dir2\\Dir3/bin/Debug/APP.EXE" On the other machine, it returns "C:\\Dir1\\Dir2\\Dir3\\bin/Debug/APP.EXE" The programmer obviously expected the second return string, because the code does this:

Convert Windows style path into unix path in java code

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问题 I am working in a java code that was designed to run on windows and contains a lot of references to files using windows style paths "System.getProperty("user.dir")\trash\blah". I am in charge to adapt it and deploy in linux. Is there an efficient way to convert all those paths(\) to unix style (/) like in "System.getProperty("user.dir")/trash/blah". Maybe, some configuration in java or linux to use \ as /. 回答1: My approach is to use the Path object to hold the path information, handle

Is there QPath::Combine in QT4?

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问题 I need a similar to .NET method for safely combining path parts without worrying for platform specifics of the path separator. Is there such class and method in QT4? Something like: QPath::Combine 回答1: There is not any function that can be used as direct replacement for Path.Combine() so you have to write it by your own. You may do it in the hard way (handling everything by yourself) or simply use QDir::cleanPath() : QString pathAppend(const QString& path1, const QString& path2) { return QDir

Is there QPath::Combine in QT4?

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I need a similar to .NET method for safely combining path parts without worrying for platform specifics of the path separator. Is there such class and method in QT4? Something like: QPath::Combine Adriano Repetti There is not any function that can be used as direct replacement for Path.Combine() so you have to write it by your own. You may do it in the hard way (handling everything by yourself) or simply use QDir::cleanPath() : QString pathAppend(const QString& path1, const QString& path2) { return QDir::cleanPath(path1 + QDir::separator() + path2); } I used QDir::separator() but as pointed

Get directory separator char on Windows? ('\\', '/', etc.)

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tl;dr: How do I ask Windows what the current directory separator character on the system is? Different versions of Windows seem to behave differently (e.g. \ and / both work on the English versions, ¥ is apparently on the Japanese version, ₩ is apparently on the Korean version , etc... Is there any way to avoid hard-coding this, and instead ask Windows at run time? Note: Ideally, the solution should not depend on a high-level DLL like ShlWAPI.dll , because lower-level libraries also depend on this. So it should really either depend on kernel32.dll or ntdll.dll or the like... although I'm

File separators of Path name of ZipEntry?

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ZIP entries store the full path name of the entry because (I'm sure of the next part) the ZIP archive is not organized as directories. The metadata contains the info about how files are supposed to be stored (inside directories). If I create a ZIP file in Windows, when I unzip the data in another OS, e.g. Mac OS X, the file structure remains as it used to be in Windows. Is this because the unzipper is designed to handle this, or isit because the file separators inside the ZIP are standard? I'm asking this because I'm trying to find an entry inside a ZIP file using the name of the zipped file.