从当前文件夹压缩应用程序
Zip application from the current folder
我有 node.js 个应用程序,我需要使用以下命令压缩所有当前文件夹
并在根
上获取 zip
为此,我想使用 archiver npm 包,但我不明白以下内容:
- where I put the current folder (since I want to zip all the application )
- where should I put the name of the zip (the zip that should be created when execute the command)
我的应用程序具有以下结构
MyApp
Node_modules
server.js
app.js
package.json
arc.js
在 arc.js
中,我已经放置了所有 zip 逻辑,所以我想我需要提供 zipPath(在我的例子中是“./”)
和 myZip 之类的 zip 名称......
我尝试了以下但没有成功,知道吗?
var fs = require('fs');
var archiver = require('archiver');
// create a file to stream archive data to.
var output = fs.createWriteStream(__dirname + '/.');
var archive = archiver('zip', {
zlib: { level: 9 } // Sets the compression level.
});
// listen for all archive data to be written
output.on('close', function() {
console.log(archive.pointer() + ' total bytes');
console.log('archiver has been finalized and the output file descriptor has closed.');
});
archive.on('warning', function(err) {
if (err.code === 'ENOENT') {
// log warning
} else {
// throw error
throw err;
}
});
// good practice to catch this error explicitly
archive.on('error', function(err) {
throw err;
});
// pipe archive data to the file
archive.pipe(output);
当我打开命令行时需要它
folder->myApp->
运行 zip arc 并将在当前路径下创建压缩文件(在本例中为根目录....)
在github of node-archiver there is an example folder
where I put the current folder (since I want to zip all the
application )
示例:
var file1 = __dirname + '/fixtures/file1.txt';
var file2 = __dirname + '/fixtures/file2.txt';
archive
.append(fs.createReadStream(file1), { name: 'file1.txt' })
.append(fs.createReadStream(file2), { name: 'file2.txt' })
.finalize();
具体关于您的案例和目录,您可以使用 .directory() method of archiver
where should I put the name of the zip (the zip that should be created
when execute the command)
示例:
var output = fs.createWriteStream(__dirname + '/example-output.zip');
您可以使用 glob
方法,但一定要排除 *.zip
文件。否则 zip 文件本身将成为存档的一部分。
这是一个例子:
// require modules
var fs = require('fs');
var archiver = require('archiver');
// create a file to stream archive data to.
var output = fs.createWriteStream(__dirname + '/example.zip');
var archive = archiver('zip', {
zlib: { level: 9 } // Sets the compression level.
});
// listen for all archive data to be written
output.on('close', function () {
console.log(archive.pointer() + ' total bytes');
console.log('archiver has been finalized and the output file descriptor has closed.');
});
// good practice to catch warnings (ie stat failures and other non-blocking errors)
archive.on('warning', function (err) {
if (err.code === 'ENOENT') {
// log warning
} else {
// throw error
throw err;
}
});
// good practice to catch this error explicitly
archive.on('error', function (err) {
throw err;
});
// pipe archive data to the file
archive.pipe(output);
archive.glob('**/*', { ignore: ['*.zip'] });
archive.finalize();
我有 node.js 个应用程序,我需要使用以下命令压缩所有当前文件夹 并在根
上获取 zip为此,我想使用 archiver npm 包,但我不明白以下内容:
- where I put the current folder (since I want to zip all the application )
- where should I put the name of the zip (the zip that should be created when execute the command)
我的应用程序具有以下结构
MyApp
Node_modules
server.js
app.js
package.json
arc.js
在 arc.js
中,我已经放置了所有 zip 逻辑,所以我想我需要提供 zipPath(在我的例子中是“./”)
和 myZip 之类的 zip 名称......
我尝试了以下但没有成功,知道吗?
var fs = require('fs');
var archiver = require('archiver');
// create a file to stream archive data to.
var output = fs.createWriteStream(__dirname + '/.');
var archive = archiver('zip', {
zlib: { level: 9 } // Sets the compression level.
});
// listen for all archive data to be written
output.on('close', function() {
console.log(archive.pointer() + ' total bytes');
console.log('archiver has been finalized and the output file descriptor has closed.');
});
archive.on('warning', function(err) {
if (err.code === 'ENOENT') {
// log warning
} else {
// throw error
throw err;
}
});
// good practice to catch this error explicitly
archive.on('error', function(err) {
throw err;
});
// pipe archive data to the file
archive.pipe(output);
当我打开命令行时需要它
folder->myApp->
运行 zip arc 并将在当前路径下创建压缩文件(在本例中为根目录....)
在github of node-archiver there is an example folder
where I put the current folder (since I want to zip all the application )
示例:
var file1 = __dirname + '/fixtures/file1.txt';
var file2 = __dirname + '/fixtures/file2.txt';
archive
.append(fs.createReadStream(file1), { name: 'file1.txt' })
.append(fs.createReadStream(file2), { name: 'file2.txt' })
.finalize();
具体关于您的案例和目录,您可以使用 .directory() method of archiver
where should I put the name of the zip (the zip that should be created when execute the command)
示例:
var output = fs.createWriteStream(__dirname + '/example-output.zip');
您可以使用 glob
方法,但一定要排除 *.zip
文件。否则 zip 文件本身将成为存档的一部分。
这是一个例子:
// require modules
var fs = require('fs');
var archiver = require('archiver');
// create a file to stream archive data to.
var output = fs.createWriteStream(__dirname + '/example.zip');
var archive = archiver('zip', {
zlib: { level: 9 } // Sets the compression level.
});
// listen for all archive data to be written
output.on('close', function () {
console.log(archive.pointer() + ' total bytes');
console.log('archiver has been finalized and the output file descriptor has closed.');
});
// good practice to catch warnings (ie stat failures and other non-blocking errors)
archive.on('warning', function (err) {
if (err.code === 'ENOENT') {
// log warning
} else {
// throw error
throw err;
}
});
// good practice to catch this error explicitly
archive.on('error', function (err) {
throw err;
});
// pipe archive data to the file
archive.pipe(output);
archive.glob('**/*', { ignore: ['*.zip'] });
archive.finalize();