filepath

问题 i have a problem with create path for rsync. x := filepath.Join("home", "my_name", "need_folder", ".") fmt.Println(x) I get "home/my_name/need_folder" , but need "home/my_name/need_folder/." , how fix without concat? In linux folder with name "." not impossible. Thank you! 回答1: You can't do that with the filepath.Join() as its documentation states: Join calls Clean on the result... And since . denotes the "current" directory, it will be removed by filepath.Clean(): It applies the following

问题 I am trying to enter a file path into a function where part of the function reads data from .csv files from a folder specified in the function. This part of the script: prep_data <- function("filepath", Year_First, Year_Last) { ... FileName <- paste0(filepath,"/details", Year_Index, "moredetails", Year_Index, ".csv") Tbl_Year <- read.csv(FileName) Where prep_data("/users/me/etc", 1980, 2014) Is giving me this error: In file(file, "rt") : cannot open file 'filepath/details1980moredetails1980

问题 I am trying to enter a file path into a function where part of the function reads data from .csv files from a folder specified in the function. This part of the script: prep_data <- function("filepath", Year_First, Year_Last) { ... FileName <- paste0(filepath,"/details", Year_Index, "moredetails", Year_Index, ".csv") Tbl_Year <- read.csv(FileName) Where prep_data("/users/me/etc", 1980, 2014) Is giving me this error: In file(file, "rt") : cannot open file 'filepath/details1980moredetails1980

问题 I have three questions: If I choose to store image.png in file system where should I store it res/drawable/image_1.png or res/drawable/images/image_1.png And I'm going to store path of image in database. What should I put in image_path field ex. image_1 or images/image_1 or etc. How can I get path of image from database and set image to view follow my code in the bottom? Could you guys change it for me? I already have answer in case storing image file in file system in assets/images/pic_1.png

问题 This is a seemingly simple problem but I am having trouble doing it in a clean manner. I have a file path as follows: /this/is/an/absolute/path/to/the/location/of/my/file What I need is to extract /of/my/file from the above given path since that is my relative path. The way I am thinking of doing it is as follows: String absolutePath = "/this/is/an/absolute/path/to/the/location/of/my/file"; String[] tokenizedPaths = absolutePath.split("/"); int strLength = tokenizedPaths.length; String

问题 I'm porting a Linux script to Windows & MinGW, which accesses the Android phone through ADB. Sometime I need to pass the Android's file path as ADB command line option. However, when invoking the ADB.exe, MinGW translates it to Windows' path. For example, adb shell cat /proc/version Is translated as follows, resulting "No such file or directory" error in Android. adb shell cat C:/Program Files (x86)/Git/proc/version I found double-quotation helps to prevent that. adb shell "cat /proc/version"

问题 How can I escape the backslashes in the string: 'pictures\12761_1.jpg' ? I know about raw string. How can I convert str to raw if I take 'pictures\12761_1.jpg' value from xml file for example? 回答1: You can use the string .replace() method. >>> print r'pictures\12761_1.jpg'.replace("\\", "/") pictures/12761_1.jpg 回答2: You can also use split/join: print "/".join(r'pictures\12761_1.jpg'.split("\\")) EDITED: The other way you may use is to prepare data during it's retrieving(e.g. the idea is to

问题 My project package doesn't show in DDMS > File Explorer (I had selected my phone in window devices). I found that an error is shown in logcat, which says can't open file for reading. Is it possible that I didn't install Eclipse properly? Before that, my project ran well and I could see the project path in the file explorer (under data>data>). After I had make few changes (I don't recall what changes I had made, too many), when doing another project, I uninstalled and reinstalled Eclipse, then

问题 I have a path (including directory and file name). I need to test if the file-name is a valid, e.g. if the file-system will allow me to create a file with such a name. The file-name has some unicode characters in it. It's safe to assume the directory segment of the path is valid and accessible ( I was trying to make the question more gnerally applicable, and apparently I wen too far ). I very much do not want to have to escape anything unless I have to. I'd post some of the example characters

问题 Given a folder path (like C:\Random Folder ), how can I find a file in it that holds a certain extension, like txt ? I assume I'll have to do a search for *.txt in the directory, but I'm not sure how I'm supposed to start this search in the first place. 回答1: Look at the System.IO.Directory class and the static method GetFiles . It has an overload that accepts a path and a search pattern. Example: string[] files = System.IO.Directory.GetFiles(path, "*.txt"); 回答2: You could use the Directory