path

问题 Sorry for asking it as it may be answered many times before, but my question is little bit different I have tree like /var/www/path/to/my/app/ -- index.php -- b.php -- inc/ -- include.php (I'm accessing inc/include.php from index.php, include "inc/include.php"; ) but in include.php, I need to get absolute path to APPLICATION root, not DOCUMENT_ROOT so in result, I need to be able to use this command //inc/include.php include(APP_ROOT."/b.php"); repeating, I do NOT WANT TO CALL IT LIKE include

问题 Sorry for asking it as it may be answered many times before, but my question is little bit different I have tree like /var/www/path/to/my/app/ -- index.php -- b.php -- inc/ -- include.php (I'm accessing inc/include.php from index.php, include "inc/include.php"; ) but in include.php, I need to get absolute path to APPLICATION root, not DOCUMENT_ROOT so in result, I need to be able to use this command //inc/include.php include(APP_ROOT."/b.php"); repeating, I do NOT WANT TO CALL IT LIKE include

问题 I need to obtain the path to the directory created for a usb drive(I think it's something like /media/user/xxxxx) for a simple usb mass storage device browser that I am making. Can anyone suggest the best/simplest way to do this? I am using an Ubuntu 13.10 machine and will be using it on a linux device. Need this in python. 回答1: This should get you started: #!/usr/bin/env python import os from glob import glob from subprocess import check_output, CalledProcessError def get_usb_devices(): sdb

问题 I have two REST classes for a simple web service (Jersey and GlassFish) that involves user resources - one to operate on all users (e.g., a factory for @POSTing) and another on individual users (e.g., @GET, @PUT, @DELETE). They are at: @Stateless @Path("users") public class AllUsersResource {...} @Stateless @Path("user") public class OneUserResource {...} respectively. When POSTing to AllUsersResource I want to return the Location (via Response.created(uri).build() ) of the new User, e.g.,

问题 I'm on a Jupyter Notebook server (v4.2.2) with Python 3.4.2 and I want to use the global name __file__ , because the notebook will be cloned from other users and in one section I have to run: def __init__(self, trainingSamplesFolder='samples', maskFolder='masks'): self.trainingSamplesFolder = self.__getAbsPath(trainingSamplesFolder) self.maskFolder = self.__getAbsPath(maskFolder) def __getAbsPath(self, path): if os.path.isabs(path): return path else: return os.path.join(os.path.dirname(__file

问题 I think this is simply a general c++ question: I'm attempting to compile a local version of ffmpeg on Linux Fedora using the gnu c++ compiler. I have source code in a bunch of folders under: ~/<username>/Downloads/Code/ffmpeg_sources/ which is where I'm attempting to set the config flags to install the build to a target not under this tree but at a root level directory with local shared libraries: /usr/local/ There is this following section near the beginning of the configuration file:

问题 I've tried numerous methods now, including FilenameUtils.normalize() from commons IO, but I can't seem to be able to get a resource in another folder to get a Java FXML file. The code is the following try { root = FXMLLoader.load(getClass().getResource("../plugin/PluginSelection.fxml")); } catch (IOException ex) { Logger.getLogger(QueueOperationsController.class.getName()).log(Level.SEVERE, null, ex); } Where the desired FXML file is: gui dialogues plugins PluginSelection.fxml // desired file

问题 I've tried numerous methods now, including FilenameUtils.normalize() from commons IO, but I can't seem to be able to get a resource in another folder to get a Java FXML file. The code is the following try { root = FXMLLoader.load(getClass().getResource("../plugin/PluginSelection.fxml")); } catch (IOException ex) { Logger.getLogger(QueueOperationsController.class.getName()).log(Level.SEVERE, null, ex); } Where the desired FXML file is: gui dialogues plugins PluginSelection.fxml // desired file

问题 This question already has answers here : How to construct a relative path in Java from two absolute paths (or URLs)? (22 answers) Closed 3 years ago . I need to get the relative file path (relative to the program executable path) from a File object. What's the best way to do this? File offers only methods for the absolute path. Maybe getting execution path manually and then cut this path off from the absolute path to get a relative path? I am on Java 7, just in case java.nio has some helping

问题 This question already has answers here : How to construct a relative path in Java from two absolute paths (or URLs)? (22 answers) Closed 3 years ago . I need to get the relative file path (relative to the program executable path) from a File object. What's the best way to do this? File offers only methods for the absolute path. Maybe getting execution path manually and then cut this path off from the absolute path to get a relative path? I am on Java 7, just in case java.nio has some helping