node.js Glob / minimatch：如何 gulp.src() 所有内容，然后排除文件夹但保留一个文件
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Glob / minimatch: how to gulp.src() everything, then exclude folder but keep one file in it
提问by Borek Bernard
I have a project like this:
我有一个这样的项目：
root
|-incl1
|-incl2
|- ...
|-excl1
|-excl2
|- .gitignore <-- keep this one
|- (other files) <-- exclude them
I need to write gulp.src()that will include all folders except excl1and excl2butkeep the .gitignorefile.
我需要写gulp.src()将包括所有文件夹excl1，excl2但保留.gitignore文件。
This is my code that doesn't work:
这是我的代码不起作用：
gulp.src([
baseDir + '/**',
'!' + baseDir + '/{excl1, excl1/**}'
'!' + baseDir + '/excl2/{**, !.gitignore}' // <-- doesn't work
], {dot: true})
回答by Heikki
This seems to work:
这似乎有效：
gulp.src([
baseDir + '/**', // Include all
'!' + baseDir + '/excl1{,/**}', // Exclude excl1 dir
'!' + baseDir + '/excl2/**/!(.gitignore)', // Exclude excl2 dir, except .gitignore
], { dot: true });
Excluding single file from glob match was tricky because there's no similar examples in minimatch docs.
从 glob 匹配中排除单个文件很棘手，因为 minimatch 文档中没有类似的例子。
https://github.com/isaacs/minimatch
https://github.com/isaacs/minimatch
"If the pattern starts with a !character, then it is negated".
“如果模式以!字符开头，则它被否定”。
