rename

问题 trying to rename internal file within a zip file without having to extract and then re-zip programatically. example. test.zip contains test.txt, i want to change it so that test.zip will contain newtest.txt(test.txt renamed to newtest.txt, contents remain the same) came across this link that works but unfortunately it expects test.txt to exist on the system. In the example the srcfile should exist on the server. Blockquote Rename file in zip with zip4j Then icame across zipnote on Linux that

问题 I am trying to name my series "Points" but it is not showing up as Points. Points = pd.Series([1,2,3]) print(Points.name) output: None I even tried renaming it but it still shows "None" Points.rename("Points") print(Points.name) output: None What am i doing wrong? 回答1: Points is your variable assigned the your series. Points = Points.rename("Points") Not "s" in the above example. print(Points) 0 1 1 2 2 3 Name: Points, dtype: int64 and print(Points.name) 'Points' 来源： https://stackoverflow.com

问题 Let's say I am editing blah.txt with Emacs and I decide to open dired to rename the file blah.txt. When I press C-x d RET (or C-x C-f RET ), a dired buffer will show up to display the content of the directory containing blah.txt , but the cursor will not be on blah.txt . So I need to search my file first ( C-s blah.txt ) to place my cursor on it and then I can rename it ( R ). How do I automate or remove the step C-s blah.txt ? 回答1: dired-jump is exactly what you want. (autoload 'dired-jump

问题 const a = { b: { c: 'Hi!' } }; const { b: { c } } = a; Is it possible rename b in this case? I want get c and also rename b . 回答1: You could destructure with a renaming and take the same property for destructuring. const a = { b: { c: 'Hi!' } }; const { b: formerB, b: { c } } = a; console.log(formerB) console.log(c); 回答2: You can destructure the same property multiple times, onto different targets: const { b: {c}, b: d } = a; This assigns a.b.c to c and a.b to d . 来源： https://stackoverflow

问题 const a = { b: { c: 'Hi!' } }; const { b: { c } } = a; Is it possible rename b in this case? I want get c and also rename b . 回答1: You could destructure with a renaming and take the same property for destructuring. const a = { b: { c: 'Hi!' } }; const { b: formerB, b: { c } } = a; console.log(formerB) console.log(c); 回答2: You can destructure the same property multiple times, onto different targets: const { b: {c}, b: d } = a; This assigns a.b.c to c and a.b to d . 来源： https://stackoverflow

问题 const a = { b: { c: 'Hi!' } }; const { b: { c } } = a; Is it possible rename b in this case? I want get c and also rename b . 回答1: You could destructure with a renaming and take the same property for destructuring. const a = { b: { c: 'Hi!' } }; const { b: formerB, b: { c } } = a; console.log(formerB) console.log(c); 回答2: You can destructure the same property multiple times, onto different targets: const { b: {c}, b: d } = a; This assigns a.b.c to c and a.b to d . 来源： https://stackoverflow

问题 I an new to bash and could do with some help. I am looking to create a command that will instantly duplicate the last created file with a new new. So far I come up with the following command; find /home/ian/Desktop/TEST/ -type f -mmin -1 -exec echo cp {} /home/ian/Desktop/TEST/ \; But for the life or me I cannot find a way to rename the file and I have looked around and tried various methods but none seemed to work. I would like to have control over the renaming so that I can add '_backup' to

问题 I have 1000 images that I want to rename them from 1 to 1000. I found this solution (the most voted answer): dirData = dir('*.png'); %# Get the selected file data fileNames = {dirData.name}; %# Create a cell array of file names for iFile = 1:numel(fileNames) %# Loop over the file names newName = sprintf('%04d.png',iFile); %# Make the new name movefile(fileNames{iFile},newName); %# Rename the file end But it falls short when the number of digits from the original file name changes.

问题 I have a file structure like this: 00000010 000000001.file1 000000001.file2 00000020 00000003.file1 00000003.file2 00000003.file3 ... So there are folders with 8-digit names containing one ore more files with name starting with 8-digit numbers. But theses filenames are – let's say – out of sync. So Now I try to rename them recursively in bash to archive: 00000010 000000010.file1 000000010.file2 00000020 00000020.file1 00000020.file2 00000020.file3 ... My script does look like: #! /bin/bash

问题 I have a file structure like this: 00000010 000000001.file1 000000001.file2 00000020 00000003.file1 00000003.file2 00000003.file3 ... So there are folders with 8-digit names containing one ore more files with name starting with 8-digit numbers. But theses filenames are – let's say – out of sync. So Now I try to rename them recursively in bash to archive: 00000010 000000010.file1 000000010.file2 00000020 00000020.file1 00000020.file2 00000020.file3 ... My script does look like: #! /bin/bash