We can create a zip new file and add files using Go Language.
But, how to add a new file with existing zip file using GoLang?
If we can use Create function, how to get the zip.writer reference?
Bit confused.
After more analysis, i found that, it is not possible to add any files with the existing zip file.
But, I was able to add files with tar file by following the hack given in this URL.
you can:
copy old zip items into a new zip file;
add new files into the new zip file;
zipReader, err := zip.OpenReader(zipPath)
targetFile, err := os.Create(targetFilePath)
targetZipWriter := zip.NewWriter(targetFile)
for _, zipItem := range zipReader.File {
zipItemReader, err := zipItem.Open()
header, err := zip.FileInfoHeader(zipItem.FileInfo())
header.Name = zipItem.Name
targetItem, err := targetZipWriter.CreateHeader(header)
_, err = io.Copy(targetItem, zipItemReader)
}
addNewFiles(targetZipWriter) // IMPLEMENT YOUR LOGIC
Although I have not attempted this yet with a zip file that already exists and then writing to it, I believe you should be able to add files to it.
This is code I have written to create a conglomerate zip file containing multiple files in order to expedite uploading the data to another location. I hope it helps!
type fileData struct {
Filename string
Body []byte
}
func main() {
outputFilename := "path/to/file.zip"
// whatever you want as filenames and bodies
fileDatas := createFileDatas()
// create zip file
conglomerateZip, err := os.Create(outputFilename)
if err != nil {
return err
}
defer conglomerateZip.Close()
zipWriter := zip.NewWriter(conglomerateZip)
defer zipWriter.Close()
// populate zip file with multiple files
err = populateZipfile(zipWriter, fileDatas)
if err != nil {
return err
}
}
func populateZipfile(w *zip.Writer, fileDatas []*fileData) error {
for _, fd := range fileDatas {
f, err := w.Create(fd.Filename)
if err != nil {
return err
}
_, err = f.Write([]byte(fd.Body))
if err != nil {
return err
}
err = w.Flush()
if err != nil {
return err
}
}
return nil
}
This is a bit old and already has an answer, but if performance isn't a key concern for you (making the zip file isn't on a hot path for example) you can do this with the archive/zip library by creating a new writer and copying the existing files into it then adding your new content. Something like this:
zw := // new zip writer from buffer or temp file
newFileName := // file name to add
reader, _ := zip.NewReader(bytes.NewReader(existingFile), int64(len(existingFile)))
for _, file := range reader.File {
if file.Name == newFileName {
continue // don't copy the old file over to avoid duplicates
}
fw, _ := zw.Create(file.Name)
fr, _ := file.Open()
io.Copy(fw, fr)
fr.Close()
}
Then you would return the new writer and append files as needed. If you aren't sure which files might overlap you can turn that if check into a function with a list of file names you will eventually add. You can also use this logic to remove a file from an existing archive.
Now in 2021, there is still no support for appending files to an existing archive.
But at least it is now possible to add already-compressed files, i.e. we don't anymore have to decompress & re-compress files when duplicating them from old archive to new one.
(NOTE: this only applies to Go 1.17+)
So, based on examples by #wongoo and #Michael, here is how I would implement appending files now with the minimum performance overhead (you'll want to add error handling though):
zr, err := zip.OpenReader(zipPath)
defer zr.Close()
zwf, err := os.Create(targetFilePath)
defer zwf.Close()
zw := zip.NewWriter(zwf)
defer zwf.Close() // or not... since it will try to wrote central directory
for _, zipItem := range zrw.File {
if isOneOfNamesWeWillAdd(zipItem.Name) {
continue // avoid duplicate files!
}
zipItemReader, err := zipItem.OpenRaw()
header := zipItem.FileHeader // clone header data
targetItem, err := targetZipWriter.CreateRaw(&header) // use cloned data
_, err = io.Copy(targetItem, zipItemReader)
}
addNewFiles(zw) // IMPLEMENT YOUR LOGIC
Related
I have a 7z archive of a number of .txt files. I am trying to list all the files in the archive and upload them to an s3 bucket. But I'm having trouble with extracting .7z archives on Go. To do this, I found a package github.com/gen2brain/go-unarr (imported as extractor) and this is what I have so far
content, err := ioutil.ReadFile("sample_archive.7z")
if err != nil {
fmt.Printf("err: %+v", err)
}
a, err := extractor.NewArchiveFromMemory(content)
if err != nil {
fmt.Printf("err: %+v", err)
}
lst, _ := a.List()
fmt.Printf("lst: %+v", last)
This prints a list of all the files in the archive. But this has two issues.
It reads files from local using ioutil and the input of NewArchiveFromMemory must be of type []byte. But I can't read from local and will have to use a file from memory of type os.file. So I will either have to find a different method or convert the os.file to []byte. There's another method NewArchiveFromReader(r io.Reader). But this is returning an error saying Bad File Descriptor.
file, err := os.OpenFile(
path,
os.O_WRONLY|os.O_TRUNC|os.O_CREATE,
0666,
)
a, err := extractor.NewArchiveFromReader(file)
if err != nil {
fmt.Printf("ERROR: %+v", err)
}
lst, _ := a.List()
fmt.Printf("files: %+v\n", lst)
I am able to get the list of the files in the archive. And using Extract(destinaltion_path string), I can also extract it to a local directory. But I want the extracted files also in os.file format ( ie. a list of os.file since there will be multiple files ).
How can I change my current code to achieve both the above targets? Is there any other library to do this?
os.File implements the io.Reader interface (because it has a Read([]byte) (int, error) method defined), so you can use NewArchiveFromReader(file) without any conversions needed. You can read up on Go interfaces for more background on why that works.
If you're okay with extracting to a local directory, you can do that and then read the files back in (warning, may contain typos):
func extractAndOpenAll(*extractor.Archive) ([]*os.File, error) {
err := a.Extract("/tmp/path") // consider using ioutil.TempDir()
if err != nil {
return nil, err
}
filestats, err := ioutil.ReadDir("/tmp/path")
if err != nil {
return nil, err
}
# warning: all these file handles must be closed by the caller,
# which is why even the error case here returns the list of files.
# if you forget, your process might leak file handles.
files := make([]*os.File, 0)
for _, fs := range(filestats) {
file, err := os.Open(fs.Name())
if err != nil {
return files, err
}
files = append(files, file)
}
return files, nil
}
It is possible to use the archived files without writing back to disk (https://github.com/gen2brain/go-unarr#read-all-entries-from-archive), but whether or not you should do that instead depends on what your next step is.
I'm trying to use os.open(fileDir) to read a file, then upload that file to an s3 bucket. Here's what I have so far.
func addFileToS3(s *session.Session, fileDir string) error {
file, err := os.Open(fileDir)
if err != nil {
return err
}
defer file.Close()
// Get file size and read the file content into a buffer
fileInfo, _ := file.Stat()
var size int64 = fileInfo.Size()
buffer := make([]byte, size)
file.Read(buffer)
// code to upload to s3
return nil
My directory structure is like
|--project-root
|--test
|--functional
|--assets
|-- good
|--fileINeed
But my code is running inside
|--project-root
|--services
|--service
|--test
|--myGoCode
How do a I pass in the correct fileDir? I need a solution that works locally and when the code gets deployed. I looked at the package path/filepath but I wasn't sure whether to get the absolute path first, then go down the hierarchy or something else.
You can add the following small function to get the expected file path.
var (
_, file, _, _ = runtime.Caller(0)
baseDir = filepath.Dir(file)
projectDir = filepath.Join(baseDir, "../../../")
)
func getFileINeedDirectory() string {
fileINeedDir := path.Join(projectDir, "test/functional/assets/good/fileINeed")
return fileINeedDir // project-dir/test/functional/assets/good/fileINeed
}
I've found lots of examples on how to extract all files from .zip, but I can't figure out how to extract a single file without iterating over all files in the .zip file.
Is it possible in Go to extract a single file from a .zip archive without iterating over all files in the .zip file?
For example, if a zip file contained:
folder1/file1.txt
folder1/file2.txt
folder1/file3.txt
folder2/file1.txt
How would I extract only folder2/file1.txt?
zip.Reader provides you the content of the archive, the files as a slice (of zip.File). There is no helper method to get a file by name, you have to iterate over the files with a loop. You don't need to open / extract the files, but to find a file by name, you have to use a loop.
For example:
r, err := zip.OpenReader("testdata/readme.zip")
if err != nil {
log.Fatal(err)
}
defer r.Close()
for _, f := range r.File {
if f.Name != "folder2/file1.txt" {
continue
}
// Found it, print its content to terminal:
rc, err := f.Open()
if err != nil {
log.Fatal(err)
}
_, err = io.Copy(os.Stdout, rc)
if err != nil {
log.Fatal(err)
}
rc.Close()
fmt.Println()
break
}
func SimpleUploader(r *http.Request, w http.ResponseWriter) {
// temp folder path
chunkDirPath := "./creatives/.uploads/" + userUUID
// create folder
err = os.MkdirAll(chunkDirPath, 02750)
// Get file handle from multipart request
var file io.Reader
mr, err := r.MultipartReader()
var fileName string
// Read multipart body until the "file" part
for {
part, err := mr.NextPart()
if err == io.EOF {
break
}
if part.FormName() == "file" {
file = part
fileName = part.FileName()
fmt.Println(fileName)
break
}
}
// Create files
tempFile := chunkDirPath + "/" + fileName
dst, err := os.Create(tempFile)
defer dst.Close()
buf := make([]byte, 1024*1024)
file.Read(buf)
// write/save buffer to disk
ioutil.WriteFile(tempFile, buf, os.ModeAppend)
if http.DetectContentType(buf) != "video/mp4" {
response, _ := json.Marshal(&Response{"File upload cancelled"})
settings.WriteResponse(w, http.StatusInternalServerError, response)
return
}
// joinedFile := io.MultiReader(bytes.NewReader(buf), file)
_, err = io.Copy(dst, file)
if err != nil {
settings.LogError(err, methodName, "Error copying file")
}
response, _ := json.Marshal(&Response{"File uploaded successfully"})
settings.WriteResponse(w, http.StatusInternalServerError, response)
}
I am uploading a Video file.
Before uploading the entire file I want to do some checks so I save the first 1mb to a file :
buf := make([]byte, 1024*1024)
file.Read(buf)
// write/save buffer to disk
ioutil.WriteFile(tempFile, buf, os.ModeAppend)
Then if the checks pass I want to upload the rest of the file dst is the same file used to save the 1st 1 mb so basically i am trying to append to the file :
_, err = io.Copy(dst, file)
The uploaded file size is correct but the file is corrupted(can't play the video).
What else have I tried? : Joining both the readers and saving to a new file. But with this approach the file size increases by 1 mb and is corrupted.
joinedFile := io.MultiReader(bytes.NewReader(buf), file)
_, err = io.Copy(newDst, joinedFile)
Kindly help.
You've basically opened the file twice by doing os.Create and ioutil.WriteFile
the issue being is that os.Create's return value (dst) is like a pointer to the beginning of that file. WriteFile doesn't move where dst points to.
You are basically doing WriteFile, then io.Copy on top of the first set of bytes WriteFile wrote.
Try doing WriteFile first (with Create flag), and then os.OpenFile (instead of os.Create) that same file with Append flag to append the remaining bytes to the end.
Also, it's extremely risky to allow a client to give you the filename as it could be ../../.bashrc (for example), to which you'd overwrite your shell init with whatever the user decided to upload.
It would be much safer if you computed a filename yourself, and if you need to remember the user's selected filename, store that in your database or even a metadata.json type file that you load later.
I am trying to take input from the keyboard and then store it in a text file but I am a bit confused on how to actually do it.
My current code is as follow at the moment:
// reads the file txt.txt
bs, err := ioutil.ReadFile("text.txt")
if err != nil {
panic(err)
}
// Prints out content
textInFile := string(bs)
fmt.Println(textInFile)
// Standard input from keyboard
var userInput string
fmt.Scanln(&userInput)
//Now I want to write input back to file text.txt
//func WriteFile(filename string, data []byte, perm os.FileMode) error
inputData := make([]byte, len(userInput))
err := ioutil.WriteFile("text.txt", inputData, )
There are so many functions in the "os" and "io" packages. I am very confused about which one I actually should use for this purpose.
I am also confused about what the third argument in the WriteFile function should be. In the documentation is says of type " perm os.FileMode" but since I am new to programming and Go I am a bit clueless.
Does anybody have any tips on how to proced?
Thanks in advance,
Marie
// reads the file txt.txt
bs, err := ioutil.ReadFile("text.txt")
if err != nil { //may want logic to create the file if it doesn't exist
panic(err)
}
var userInput []string
var err error = nil
var n int
//read in multiple lines from user input
//until user enters the EOF char
for ln := ""; err == nil; n, err = fmt.Scanln(ln) {
if n > 0 { //we actually read something into the string
userInput = append(userInput, ln)
} //if we didn't read anything, err is probably set
}
//open the file to append to it
//0666 corresponds to unix perms rw-rw-rw-,
//which means anyone can read or write it
out, err := os.OpenFile("text.txt", os.O_APPEND, 0666)
defer out.Close() //we'll close this file as we leave scope, no matter what
if err != nil { //assuming the file didn't somehow break
//write each of the user input lines followed by a newline
for _, outLn := range userInput {
io.WriteString(out, outLn+"\n")
}
}
I've made sure this compiles and runs on play.golang.org, but I'm not at my dev machine, so I can't verify that it's interacting with Stdin and the file entirely correctly. This should get you started though.
For example,
package main
import (
"fmt"
"io/ioutil"
"os"
)
func main() {
fname := "text.txt"
// print text file
textin, err := ioutil.ReadFile(fname)
if err == nil {
fmt.Println(string(textin))
}
// append text to file
f, err := os.OpenFile(fname, os.O_CREATE|os.O_APPEND|os.O_WRONLY, 0666)
if err != nil {
panic(err)
}
var textout string
fmt.Scanln(&textout)
_, err = f.Write([]byte(textout))
if err != nil {
panic(err)
}
f.Close()
// print text file
textin, err = ioutil.ReadFile(fname)
if err != nil {
panic(err)
}
fmt.Println(string(textin))
}
If you simply want to append the user's input to a text file, you could just read the
input as you've already done and use ioutil.WriteFile, as you've tried to do.
So you already got the right idea.
To make your way go, the simplified solution would be this:
// Read old text
current, err := ioutil.ReadFile("text.txt")
// Standard input from keyboard
var userInput string
fmt.Scanln(&userInput)
// Append the new input to the old using builtin `append`
newContent := append(current, []byte(userInput)...)
// Now write the input back to file text.txt
err = ioutil.WriteFile("text.txt", newContent, 0666)
The last parameter of WriteFile is a flag which specifies the various options for
files. The higher bits are options like file type (os.ModeDir, for example) and the lower
bits represent the permissions in form of UNIX permissions (0666, in octal format, stands for user rw, group rw, others rw). See the documentation for more details.
Now that your code works, we can improve it. For example by keeping the file open
instead of opening it twice:
// Open the file for read and write (O_RDRW), append to it if it has
// content, create it if it does not exit, use 0666 for permissions
// on creation.
file, err := os.OpenFile("text.txt", os.O_RDWR|os.O_APPEND|os.O_CREATE, 0666)
// Close the file when the surrounding function exists
defer file.Close()
// Read old content
current, err := ioutil.ReadAll(file)
// Do something with that old content, for example, print it
fmt.Println(string(current))
// Standard input from keyboard
var userInput string
fmt.Scanln(&userInput)
// Now write the input back to file text.txt
_, err = file.WriteString(userInput)
The magic here is, that you use the flag os.O_APPEND while opening the file,
which makes file.WriteString() append. Note that you need to close the file after
opening it, which we do after the function exists using the defer keyword.