Gulp - Zip the contents of subfolders & name these archives after them
0
I would like to use Gulp & gulp-zip to:
- Zip each subfolder of
./lessons/folder - Name each archive after the original folder name. So archived folder
./lessons/1-hermit-crab/should be named1-hermit-crab.zip. - Move all these archives into the
./lessons/folder.
I have started with this, but got stuck on not being able to retrieve the name of the ziped directory, so that I can use it for the archive name. So I keep the archives in the subfolder now.
Thanks for any help.
gulp.task('zip-lessons', function() {
// Get an array of subdirectories under ./lessons/
var subDirectories = glob.sync('./lessons/*/');
// For each directory…
subDirectories.forEach(function (subDirectory) {
return gulp.src(subDirectory + '**')
.pipe(zip('lesson.zip'))
.pipe(gulp.dest(subDirectory));
});
});
gulp gulp-zip
asked Jan 18 at 2:56
Petr ChutnyPetr Chutny
186
add a comment |
0
I would like to use Gulp & gulp-zip to:
- Zip each subfolder of
./lessons/folder - Name each archive after the original folder name. So archived folder
./lessons/1-hermit-crab/should be named1-hermit-crab.zip. - Move all these archives into the
./lessons/folder.
I have started with this, but got stuck on not being able to retrieve the name of the ziped directory, so that I can use it for the archive name. So I keep the archives in the subfolder now.
Thanks for any help.
gulp.task('zip-lessons', function() {
// Get an array of subdirectories under ./lessons/
var subDirectories = glob.sync('./lessons/*/');
// For each directory…
subDirectories.forEach(function (subDirectory) {
return gulp.src(subDirectory + '**')
.pipe(zip('lesson.zip'))
.pipe(gulp.dest(subDirectory));
});
});
gulp gulp-zip
asked Jan 18 at 2:56
Petr ChutnyPetr Chutny
186
add a comment |
0
0
0
I would like to use Gulp & gulp-zip to:
- Zip each subfolder of
./lessons/folder - Name each archive after the original folder name. So archived folder
./lessons/1-hermit-crab/should be named1-hermit-crab.zip. - Move all these archives into the
./lessons/folder.
I have started with this, but got stuck on not being able to retrieve the name of the ziped directory, so that I can use it for the archive name. So I keep the archives in the subfolder now.
Thanks for any help.
gulp.task('zip-lessons', function() {
// Get an array of subdirectories under ./lessons/
var subDirectories = glob.sync('./lessons/*/');
// For each directory…
subDirectories.forEach(function (subDirectory) {
return gulp.src(subDirectory + '**')
.pipe(zip('lesson.zip'))
.pipe(gulp.dest(subDirectory));
});
});
gulp gulp-zip
asked Jan 18 at 2:56
Petr ChutnyPetr Chutny
186
I would like to use Gulp & gulp-zip to:
- Zip each subfolder of
./lessons/folder - Name each archive after the original folder name. So archived folder
./lessons/1-hermit-crab/should be named1-hermit-crab.zip. - Move all these archives into the
./lessons/folder.
I have started with this, but got stuck on not being able to retrieve the name of the ziped directory, so that I can use it for the archive name. So I keep the archives in the subfolder now.
Thanks for any help.
gulp.task('zip-lessons', function() {
// Get an array of subdirectories under ./lessons/
var subDirectories = glob.sync('./lessons/*/');
// For each directory…
subDirectories.forEach(function (subDirectory) {
return gulp.src(subDirectory + '**')
.pipe(zip('lesson.zip'))
.pipe(gulp.dest(subDirectory));
});
});
gulp gulp-zip
gulp gulp-zip
asked Jan 18 at 2:56
Petr ChutnyPetr Chutny
186
asked Jan 18 at 2:56
Petr ChutnyPetr Chutny
186
asked Jan 18 at 2:56
Petr ChutnyPetr Chutny
186
asked Jan 18 at 2:56
Petr ChutnyPetr Chutny
186
asked Jan 18 at 2:56
Petr ChutnyPetr Chutny
186
186
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
2
You can use node's path module to get the name of the subdirectory like so:
const path = require('path');
const dirName = path.parse('./folderA/folderB').base // -> 'folderB'
And pass the dirName to zip():
const { task, src, dest } = require('gulp');
const path = require('path');
const zip = require('gulp-zip');
const glob = require('glob');
const subDirs = glob.sync('./lessons/*');
task('zipLessions', (done) => {
subDirs.forEach(subDir => {
const dirName = path.parse(subDir).base;
src(subDir + '/*')
.pipe(zip(`${dirName}.zip`))
.pipe(dest('./lessons'))
})
done()
})
edited Jan 18 at 16:00
Mark
11.8k33351
answered Jan 18 at 6:05
Derek NguyenDerek Nguyen
736111
add a comment |
-1
My colleague has offered me this solution that works, so now I am using it.
const gulp = require('gulp');
const zip = require('gulp-zip');
var fs = require('fs');
var path = require('path');
// Functions
function getFolders(dir) {
return fs.readdirSync(dir)
.filter(function(file) {
return fs.statSync(path.join(dir, file)).isDirectory();
});
}
// Tasks
// Zip each folder in /lessons and place the archive in the /lessons root
gulp.task('zip-lessons', () => {
let folders = getFolders('lessons/');
console.log(folders)
var tasks = folders.map(function(folder) {
gulp.src(`lessons/${folder}/**`)
.pipe(zip(`${folder}.zip`))
.pipe(gulp.dest('lessons'))
})
});
answered Jan 20 at 3:07
Petr ChutnyPetr Chutny
186
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
You can use node's path module to get the name of the subdirectory like so:
const path = require('path');
const dirName = path.parse('./folderA/folderB').base // -> 'folderB'
And pass the dirName to zip():
const { task, src, dest } = require('gulp');
const path = require('path');
const zip = require('gulp-zip');
const glob = require('glob');
const subDirs = glob.sync('./lessons/*');
task('zipLessions', (done) => {
subDirs.forEach(subDir => {
const dirName = path.parse(subDir).base;
src(subDir + '/*')
.pipe(zip(`${dirName}.zip`))
.pipe(dest('./lessons'))
})
done()
})
edited Jan 18 at 16:00
Mark
11.8k33351
answered Jan 18 at 6:05
Derek NguyenDerek Nguyen
736111
add a comment |
2
You can use node's path module to get the name of the subdirectory like so:
const path = require('path');
const dirName = path.parse('./folderA/folderB').base // -> 'folderB'
And pass the dirName to zip():
const { task, src, dest } = require('gulp');
const path = require('path');
const zip = require('gulp-zip');
const glob = require('glob');
const subDirs = glob.sync('./lessons/*');
task('zipLessions', (done) => {
subDirs.forEach(subDir => {
const dirName = path.parse(subDir).base;
src(subDir + '/*')
.pipe(zip(`${dirName}.zip`))
.pipe(dest('./lessons'))
})
done()
})
edited Jan 18 at 16:00
Mark
11.8k33351
answered Jan 18 at 6:05
Derek NguyenDerek Nguyen
736111
add a comment |
2
2
2
You can use node's path module to get the name of the subdirectory like so:
const path = require('path');
const dirName = path.parse('./folderA/folderB').base // -> 'folderB'
And pass the dirName to zip():
const { task, src, dest } = require('gulp');
const path = require('path');
const zip = require('gulp-zip');
const glob = require('glob');
const subDirs = glob.sync('./lessons/*');
task('zipLessions', (done) => {
subDirs.forEach(subDir => {
const dirName = path.parse(subDir).base;
src(subDir + '/*')
.pipe(zip(`${dirName}.zip`))
.pipe(dest('./lessons'))
})
done()
})
edited Jan 18 at 16:00
Mark
11.8k33351
answered Jan 18 at 6:05
Derek NguyenDerek Nguyen
736111
You can use node's path module to get the name of the subdirectory like so:
const path = require('path');
const dirName = path.parse('./folderA/folderB').base // -> 'folderB'
And pass the dirName to zip():
const { task, src, dest } = require('gulp');
const path = require('path');
const zip = require('gulp-zip');
const glob = require('glob');
const subDirs = glob.sync('./lessons/*');
task('zipLessions', (done) => {
subDirs.forEach(subDir => {
const dirName = path.parse(subDir).base;
src(subDir + '/*')
.pipe(zip(`${dirName}.zip`))
.pipe(dest('./lessons'))
})
done()
})
edited Jan 18 at 16:00
Mark
11.8k33351
answered Jan 18 at 6:05
Derek NguyenDerek Nguyen
736111
edited Jan 18 at 16:00
Mark
11.8k33351
edited Jan 18 at 16:00
Mark
11.8k33351
edited Jan 18 at 16:00
Mark
11.8k33351
11.8k33351
answered Jan 18 at 6:05
Derek NguyenDerek Nguyen
736111
answered Jan 18 at 6:05
Derek NguyenDerek Nguyen
736111
answered Jan 18 at 6:05
Derek NguyenDerek Nguyen
736111
736111
add a comment |
add a comment |
-1
My colleague has offered me this solution that works, so now I am using it.
const gulp = require('gulp');
const zip = require('gulp-zip');
var fs = require('fs');
var path = require('path');
// Functions
function getFolders(dir) {
return fs.readdirSync(dir)
.filter(function(file) {
return fs.statSync(path.join(dir, file)).isDirectory();
});
}
// Tasks
// Zip each folder in /lessons and place the archive in the /lessons root
gulp.task('zip-lessons', () => {
let folders = getFolders('lessons/');
console.log(folders)
var tasks = folders.map(function(folder) {
gulp.src(`lessons/${folder}/**`)
.pipe(zip(`${folder}.zip`))
.pipe(gulp.dest('lessons'))
})
});
answered Jan 20 at 3:07
Petr ChutnyPetr Chutny
186
add a comment |
-1
My colleague has offered me this solution that works, so now I am using it.
const gulp = require('gulp');
const zip = require('gulp-zip');
var fs = require('fs');
var path = require('path');
// Functions
function getFolders(dir) {
return fs.readdirSync(dir)
.filter(function(file) {
return fs.statSync(path.join(dir, file)).isDirectory();
});
}
// Tasks
// Zip each folder in /lessons and place the archive in the /lessons root
gulp.task('zip-lessons', () => {
let folders = getFolders('lessons/');
console.log(folders)
var tasks = folders.map(function(folder) {
gulp.src(`lessons/${folder}/**`)
.pipe(zip(`${folder}.zip`))
.pipe(gulp.dest('lessons'))
})
});
answered Jan 20 at 3:07
Petr ChutnyPetr Chutny
186
add a comment |
-1
-1
-1
My colleague has offered me this solution that works, so now I am using it.
const gulp = require('gulp');
const zip = require('gulp-zip');
var fs = require('fs');
var path = require('path');
// Functions
function getFolders(dir) {
return fs.readdirSync(dir)
.filter(function(file) {
return fs.statSync(path.join(dir, file)).isDirectory();
});
}
// Tasks
// Zip each folder in /lessons and place the archive in the /lessons root
gulp.task('zip-lessons', () => {
let folders = getFolders('lessons/');
console.log(folders)
var tasks = folders.map(function(folder) {
gulp.src(`lessons/${folder}/**`)
.pipe(zip(`${folder}.zip`))
.pipe(gulp.dest('lessons'))
})
});
answered Jan 20 at 3:07
Petr ChutnyPetr Chutny
186
My colleague has offered me this solution that works, so now I am using it.
const gulp = require('gulp');
const zip = require('gulp-zip');
var fs = require('fs');
var path = require('path');
// Functions
function getFolders(dir) {
return fs.readdirSync(dir)
.filter(function(file) {
return fs.statSync(path.join(dir, file)).isDirectory();
});
}
// Tasks
// Zip each folder in /lessons and place the archive in the /lessons root
gulp.task('zip-lessons', () => {
let folders = getFolders('lessons/');
console.log(folders)
var tasks = folders.map(function(folder) {
gulp.src(`lessons/${folder}/**`)
.pipe(zip(`${folder}.zip`))
.pipe(gulp.dest('lessons'))
})
});
answered Jan 20 at 3:07
Petr ChutnyPetr Chutny
186
answered Jan 20 at 3:07
Petr ChutnyPetr Chutny
186
answered Jan 20 at 3:07
Petr ChutnyPetr Chutny
186
answered Jan 20 at 3:07
Petr ChutnyPetr Chutny
186
186
add a comment |
add a comment |
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