filepath

问题 I'm importing a CSV file to MySQL database. This can be done using java.mysql support for forward slash in file path. If user gives the path String filepath=" c:\upload\date\csv\sample.csv"; we can do it by replacing '\' with '/' using following line of code. String filepath2=filepath.replace("\\","/"); but when user enters path as follows String filepath=" c:\test.csv"; the replace function unable to replace "\\" with "/" as '\' in the filepath does not remain '\' as '\' is followed by 't'

问题 I'm importing a CSV file to MySQL database. This can be done using java.mysql support for forward slash in file path. If user gives the path String filepath=" c:\upload\date\csv\sample.csv"; we can do it by replacing '\' with '/' using following line of code. String filepath2=filepath.replace("\\","/"); but when user enters path as follows String filepath=" c:\test.csv"; the replace function unable to replace "\\" with "/" as '\' in the filepath does not remain '\' as '\' is followed by 't'

问题 I'm developing an application for iPad. I want to access a text file inside the Resources folder! I'm doing that in following way: NSString* filePath = @"/Users/net4uonline/Desktop/slots2/paylines.txt"; Now, if I move my whole project from Desktop to somewhere else I know this won't work. So, is there any way to give a relative path for this file instead of the current path. Maybe like the following? NSString* filePath = @"Resources/paylines.txt"; I know this won't work but as my file is

问题 My grunt file is shown below: module.exports = function(grunt) { // Project configuration. grunt.initConfig({ pkg: grunt.file.readJSON('package.json'), uglify: { options: { banner: '/*! <%= pkg.name %> <%= grunt.template.today("yyyy-mm-dd") %> */\n' }, build: { src: 'src/**/*.js', dest: 'dist/<%= pkg.name %>.min.js' } }, watch: { js: { files: ['src/**/*.js'], options: { livereload: '<%= connect.options.livereload %>' } }, livereload: { options: { livereload: '<%= connect.options.livereload %>

问题 I was wondering if anyone here knows how to find the complete path (from the drive letter onwards) of a ContentManager instance. using this I could create a string with the right number of "..\" to append to the file path when I want to load a file from anywhere else on the computer (eg. from a registry key). So basically I'm asking if there is a way. 回答1: You may just want to use System.GetFolderPath with one of these locations, most likely Program Files. From there, you can navigate to your

问题 I tried to track the file with server in the filename and i can print all the file in directory with server** but when I try to read the file it gives me error" saying: Traceback (most recent call last): File "view_log_packetloss.sh", line 27, in <module> with open(filename,'rb') as files: IOError: [Errno 2] No such file or directory: 'pcoip_server_2014_05_19_00000560.txt' I have seen similar question being asked but I could not fix mine, some error were fixed using chdir to change the

问题 There are two directories on my desktop, DIR1 and DIR2 which contain the following files: DIR1: file1.py DIR2: file2.py myfile.txt The files contain the following: file1.py import sys sys.path.append('.') sys.path.append('../DIR2') import file2 file2.py import sys sys.path.append( '.' ) sys.path.append( '../DIR2' ) MY_FILE = "myfile.txt" myfile = open(MY_FILE) myfile.txt some text Now, there are two scenarios. The first works, the second gives an error. Scenario 1 I cd into DIR2 and run file2

问题 There are two directories on my desktop, DIR1 and DIR2 which contain the following files: DIR1: file1.py DIR2: file2.py myfile.txt The files contain the following: file1.py import sys sys.path.append('.') sys.path.append('../DIR2') import file2 file2.py import sys sys.path.append( '.' ) sys.path.append( '../DIR2' ) MY_FILE = "myfile.txt" myfile = open(MY_FILE) myfile.txt some text Now, there are two scenarios. The first works, the second gives an error. Scenario 1 I cd into DIR2 and run file2

问题 There are two directories on my desktop, DIR1 and DIR2 which contain the following files: DIR1: file1.py DIR2: file2.py myfile.txt The files contain the following: file1.py import sys sys.path.append('.') sys.path.append('../DIR2') import file2 file2.py import sys sys.path.append( '.' ) sys.path.append( '../DIR2' ) MY_FILE = "myfile.txt" myfile = open(MY_FILE) myfile.txt some text Now, there are two scenarios. The first works, the second gives an error. Scenario 1 I cd into DIR2 and run file2

问题 I have a folder which contains few jar files. I am referring to those jar files from another jar file which is in some other location. My problem is, when I give the path of the jar folder like this C:\Trial Library\jar Folder\ ie. with space in the folder names (Trial Library) then it is unable to locate this folder. If I give without space ie C:\Trial_Library\jar_Folder\ then it works fine. Please help me to fix this issue ASAP. Here is my Batch File set CURRENT_DIRECTORY=%~dp0 set ANT_HOME