unzip

问题 I'd like to unzip a compressed .mdb file in the www folder of my shiny app, query it for data, and then remove it. Unzip() works on my local machine, but when I deploy the app at shinyapps.io, it has issues unzipping the file. Because I'm not able to read.table() the resulting file (it's an .mdb) I don't think unz() will work. This code works when run on my local machine Server: require(shiny) shinyServer(function(input, output) { observeEvent(input$run,{ #Run Button dbName=unzip('www/test

问题 I tried to unzip large zipfiles from my Drive into my Colab and got this error: BadZipFile: zipfiles that span multiple disks are not supported How do I unzip large files from Drive into Colab? 回答1: Im doing that with two options: !unzip file_location -d file_destination #-d is for quite option. Or other option more elaborate: import zipfile from google.colab import drive drive.mount('/content/drive/') zip_ref = zipfile.ZipFile("/content/drive/My Drive/ML/DataSet.zip", 'r') zip_ref.extractall

问题 I'm working on hybrid mobile app using html5/js. It has a function download zip file then unzip them. The download function is not problem but I don't know how to unzip file (using javascript). Many people refer to zip.js but it seems only reading zip file (not unzip/extract to new folder) Very appreciate if someone could help me !!! 回答1: Have a look at zip.js documentation and demo page. Also notice the use of JavaScript filesystem API to read/write files and create temporary files. If the

问题 I have an Excel file (.xls extension) that is inside a zipped folder that I would like to read as a dataframe into R. I loaded the gdata library and set up my working directory to the folder that houses the zipped folder. When I type in the following syntax: data_frame1 <- read.xls( unz("./Data/Project1.zip","schools.xls")) I get the following error messages: Error in path.expand(xls) : invalid 'path' argument Error in file.exists(tfn) : invalid 'file' argument I'm guessing that I'm missing

问题 Does AWS apigateway change http body? How can I stop it from doing this? My application: (1) A front end "UI" that sends a "http request" using "POST method" that contains a "zip file" in "body" through "form-data". (2) AWS "apigateway" receives this request and forward it to "Lambda Proxy" (3) AWS "Lambda" implemented by python coding receives this request and decompresses this zip file to a temporary folder. The problem I'm facing: (1) and (2) works fine, but in (3) the pythong program at

问题 I have .gz zipped files within multiple folders that are all within a main folder called "usa". I was able to extract an individual file using the code below. import gzip import shutil source=r"C:\usauc300.dbf.gz" output=r"C:\usauc300.dbf" with gzip.open(source,"rb") as f_in, open(output,"wb") as f_out: shutil.copyfileobj(f_in, f_out) I have searched high and low but can't find an equivalent to the command line option gzip -dr..... which means "decompress recursive" and will go through each

问题 I'm having a node app which needs to get some zip file from client Postman and extract it to a folder in my fileSystem,Im using express I did the following which doesnt work, what am I missing here? I've created sample node app to simulate the issue. var express = require('express'); var upload = require('multer')({ dest: 'uploads/' }); var admZip = require('adm-zip'); var app = express(); app.post('/',upload.single('file'),function(req,res){ debugger; var zip = new admZip(req.file); zip

问题 Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 26 days ago . Improve this question Is there any way to perform diff operetion on two files in two zips without extracting them? If not - any other workaround to compare them without extracting? Thanks. 回答1: Combining the responses so far, the following bash function will compare the file listings

问题 I'm trying to decompress a zip file (with multiple files inside) using Azure Data Factory v2. The zip file is located in Azure File Storage. The ADF Copy task just copies the original zip file without decompressing it. Any suggestion on how to make this work? This is the current configuration: The zip file source was setup as a binary dataset with Compression Type = ZipDeflate. The target folder was also setup as a binary dataset but with Compression Type = None. A pipeline with a single Copy

问题 i used zipfile lib to extract file from zip and now after unzip the directory i found the permission of my file has been corrupted , import zipfile fh = open('sample.zip', 'rb') z = zipfile.ZipFile(fh) print z.namelist() for name in z.namelist(): z.extract(name, '/tmp/') fh.close() but when i use linux unzip tools this issue don't happen i try to use os.system('unzip sample.zip') but i still want to do this with zipfile 回答1: The related Python issues provide some insight as to why the issue