Kou*_*nha 4 javascript node.js angularjs gulp
我有一个gulp任务,在index.html中注入几个依赖项,如下所示:
return gulp.src(config.serve.html)
.pipe($.inject(gulp.src(prependSource(config.inject.source.indexes)), {
read: false,
starttag: config.inject.indexes,
addRootSlash: false,
addPrefix: prefix,
ignorePath: ignore
}))
.pipe($.inject(gulp.src(managerList), {
read: false,
starttag: config.inject.managers,
addRootSlash: false,
addPrefix: prefix,
ignorePath: ignore
}))
.pipe($.inject(gulp.src(prependSource(config.inject.source.directives)), {
read: false,
starttag: config.inject.directives,
addRootSlash: false,
addPrefix: prefix,
ignorePath: ignore
}))
.pipe($.inject(gulp.src(prependSource(config.inject.source.routes)), {
read: false,
starttag: config.inject.routes,
addRootSlash: false,
addPrefix: prefix,
ignorePath: ignore
}))
.pipe($.inject(gulp.src(prependSource(config.inject.source.css)), {
read: false,
starttag: config.inject.css,
addRootSlash: false,
addPrefix: prefix,
ignorePath: ignore
}))
.pipe($.size())
.pipe(gulp.dest(dest));
};
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如您所见,所有管道都以某种方式重复(除了managerList
).所以我想要的是在一系列注入的帮助下将重复管道合并到一个单独的管道中.如何实现相同?
gulp.src返回一个流(就像每个.pipe一样).如果你考虑每个.pipe调用正在做什么,你可以简单地把它变成for循环.像这样的东西:
config = { /* your config object */ };
var stream = gulp.src(/* source glob */);
for (var i = 0; i < injects.length; i++) {
stream = stream.pipe($.inject(gulp.src(prependSource(config.inject.source[i])), config.inject.starttag[i]));
}
stream.pipe($.size())
.pipe(gulp.dest(dest));
return stream;
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