glob匹配，排除所有JS文件 [英] Glob matching, exclude all JS files

问题描述

我是 gulp.js 的新用户.我想将所有非 JavaScript 文件移动到构建目录.我现在得到的是这样的:

I'm a new user to gulp.js. I'd like to move all of my non-javascript files to a build directory. What I've got right now is this:

//Test copy
gulp.task('test-copy', function() {
    gulp.src(['myProject/src/**/*.!(js|map|src)'])
        .pipe(gulp.dest('myProject/build'));
});


//Results for various files
myProject/css/style.css //Copied - GOOD
myProject/html/index.html //Copied - GOOD
myProject/js/foo.js //Not Copied - GOOD
myProject/js/bar.min.js //Copied - BAD!
myProject/js/jquery-2.0.3.min.js //Copied - BAD!
myProject/js/jquery-2.0.3.min.map //Copied - BAD!

如您所见，它只匹配文件路径字符串中的第一个点之后，而不是最后一个点，如我所愿.如何修改 glob 搜索字符串以使其行为符合我的意愿?

As you can see, it only matches after the first dot in the file path string, not the last one, as I'd like. How can I modify the glob search string to behave as I'd like?

推荐答案

试试这个 glob 模式:

Try this glob pattern:

myProject/src/**/!(*.js|*.map|*.src)

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