用于排除多个文件的 node.js glob 模式
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/23809897/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
node.js glob pattern for excluding multiple files
提问by ke_wa
I'm using the npm module node-glob.
我正在使用 npm 模块node-glob。
This snippet returns recursively all files in the current working directory.
此代码段递归返回当前工作目录中的所有文件。
var glob = require('glob');
glob('**/*', function(err, files) {
console.log(files);
});
sample output:
示例输出：
[ 'index.html', 'js', 'js/app.js', 'js/lib.js' ]
I want to exclude index.htmland js/lib.js. I tried to exclude these files with negative pattern '!'but without luck. Is there a way to achieve this only by using a pattern?
我想排除index.html和js/lib.js。我试图用否定模式“！”排除这些文件 但没有运气。有没有办法仅通过使用模式来实现这一目标？
采纳答案by Koji
Or without an external dependency:
或者没有外部依赖：
/**
Walk directory,
list tree without regex excludes
*/
var fs = require('fs');
var path = require('path');
var walk = function (dir, regExcludes, done) {
var results = [];
fs.readdir(dir, function (err, list) {
if (err) return done(err);
var pending = list.length;
if (!pending) return done(null, results);
list.forEach(function (file) {
file = path.join(dir, file);
var excluded = false;
var len = regExcludes.length;
var i = 0;
for (; i < len; i++) {
if (file.match(regExcludes[i])) {
excluded = true;
}
}
// Add if not in regExcludes
if(excluded === false) {
results.push(file);
// Check if its a folder
fs.stat(file, function (err, stat) {
if (stat && stat.isDirectory()) {
// If it is, walk again
walk(file, regExcludes, function (err, res) {
results = results.concat(res);
if (!--pending) { done(null, results); }
});
} else {
if (!--pending) { done(null, results); }
}
});
} else {
if (!--pending) { done(null, results); }
}
});
});
};
var regExcludes = [/index\.html/, /js\/lib\.js/, /node_modules/];
walk('.', regExcludes, function(err, results) {
if (err) {
throw err;
}
console.log(results);
});
回答by Sergei Panfilov
I suppose it's not actual anymore but i got stuck with the same question and found an answer.
我想它不再是真实的，但我遇到了同样的问题并找到了答案。
This can be done using only npm globmodule.
We need to use optionsas a second parameter to globfunction
这可以仅使用npm glob模块来完成。我们需要使用选项作为第二个参数来glob运行
glob('pattern', {options}, cb)
There is an options.ignorepattern for your needs.
有一种options.ignore模式可以满足您的需求。
var glob = require('glob');
glob("**/*",{"ignore":['index.html', 'js', 'js/app.js', 'js/lib.js']}, function (err, files) {
console.log(files);
})
回答by Sindre Sorhus
回答by stefanbuck
You can use node-globulefor that:
您可以使用node-globule：
var globule = require('globule');
var result = globule.find(['**/*', '!index.html', '!js/lib.js']);
console.log(result);
回答by Intervalia
Here is what I wrote for my project:
这是我为我的项目写的：
var glob = require('glob');
var minimatch = require("minimatch");
function globArray(patterns, options) {
var i, list = [];
if (!Array.isArray(patterns)) {
patterns = [patterns];
}
patterns.forEach(function (pattern) {
if (pattern[0] === "!") {
i = list.length-1;
while( i > -1) {
if (!minimatch(list[i], pattern)) {
list.splice(i,1);
}
i--;
}
}
else {
var newList = glob.sync(pattern, options);
newList.forEach(function(item){
if (list.indexOf(item)===-1) {
list.push(item);
}
});
}
});
return list;
}
And call it like this (Using an array):
并像这样调用它（使用数组）：
var paths = globArray(["**/*.css","**/*.js","!**/one.js"], {cwd: srcPath});
or this (Using a single string):
或者这个（使用单个字符串）：
var paths = globArray("**/*.js", {cwd: srcPath});
回答by Ivan Ferrer
A samples example with gulp:
gulp 的示例示例：
gulp.task('task_scripts', function(done){
glob("./assets/**/*.js", function (er, files) {
gulp.src(files)
.pipe(gulp.dest('./public/js/'))
.on('end', done);
});
});
