android get file path from content uri

Preface or Today write a self-starter packaging script. Used to get the last file name from the path name. Here is a record of the implementation process.Of course, in the end I will also give the official approach. (PS: Very embarrassing, realized that the original bash Shell has a ready-made function)Gets the file na

var filem=document.queryselector ("#fileUp");$ ("#fileUp"). On ("Change", function () { Gets the file object, files is a property of the file selection control, which stores the file object selected by the file selection control, and the type is an array var fileobj = filem.files[0]; Creates a Formdata object

interface3.3Value file: Used to store data used by the app, such as size, string4.Assets File: Resource folder. In addition to providing/res folder to store resource files, the/assets folder can also store resource files, pictures or data. However, in the/assets folder the resource file will no longer r.java itself to generate the ID, so read the/assets folder

String path=request.getsession (). Getservletcontext (). Getrealpath ("upload/img/product");// Binary upload multiparthttpservletrequest multipartrequest = (multiparthttpservletrequest) request;// Get the file Commonsmultipartfile fpic= (commonsmultipartfile) multipartrequest.getfile ("Fpic");//Infer if there is a file

Application. conf configuration file content # Custom config:Test. config= "Test. config. Value" Get in Scala Play. Current. configuration. getstring ("test. config ") View @ Play. Play. Application (). configuration (). getstring ("test. config ") Unit Test Package taskimport Org. specs2.mutable. _ import play. API. test. _ import p

#include Key Reference Classes Ishelllink:ishelllink The Main method:1, getarguments: Obtaining parameter information2, GetDescription: Obtaining Descriptive information3, GetHotKey: Get shortcut keys4, Geticonlocation: Get shortcut icon5, Getidlist: Gets the shortcut to the target object of item identifier list (each object in the Windows shell, such as files, directories, and printers, has a unique item I

Why can't I get the correct file content?

C #. net drag to get the file pathAuthor attilax, email: 1466519819@qq.com Idea: Use the dragenter event to obtain the "information" of the dragged window (several files, some text, and so on ),Parse "information" in the dragdrop event. The allowdrop attribute of the form must be set to true; The dragenter event must exist. (If you write a dragdrop event separately, the dragenter event will not be dra

File upload image Get path addressSimilar: Click button, select the image, in the corresponding input box to display the image path addressLike thisPrinciple: Opacity to hide the native input file and then use the. file-btn to sim

Use JS to get uploaded file name, not circulated online, with Path.substring (Path.lastindexof ('/') +1); There are a lot of problems in this kind of work. For example, the value value of the Firefox input form can be obtained by default, while IE displays the file path. So to differentiate, under Firefox lastindex

Android file operation-read assets and RAW file content 1. Create Assets folders and Res/raw folders individually: (Note that raw files are new in res and then create a folder with the name Raw) 2. Create two txt files and copy to asset and raw folders: 3. Effect of implementation: 4. Implementation code:

Os:windows7 x64 Jdk:jdk-8u131-windows-x64 Ide:eclipse Oxygen Release (4.7.0) CodePackage Jizuiku0;import java.io.file;/* * @version V17.09 */public class Getdemo {public static void main (string[] args) { System.out.println ("The current default working path is:" +new file (""). GetAbsolutePath ());//Get this path

Transferred from: http://blog.csdn.net/honker110/article/details/4355618@echo off setlocal enabledelayedexpansion echo the path of the currently running batch file:!cd!pause @echo off echo the current directory is:%cd%pause @echo off::Set "abc=%cd%"echo the path of the batch file that is currently running:%~dp0 pause @

1. BaseName ("/mnt/img/image01.jpg") function: Get the file name; the output is: image01.jpg.Using basename ($uriString) We can get a file name with the extension;If you do not need an extension, you can also use BaseName ($uriString, $extString) to filter the extension and simply return the