获取gulp.src()中的当前文件名 [英] Get the current file name in gulp.src()
问题描述
在我的gulp.js文件中,我将所有HTML文件从示例
文件夹流式传输到 build
文件夹中。
创建一个gulp任务并不困难:
var gulp = require('gulp');
gulp.task('examples',function(){
return gulp.src('./ examples / *。html')
.pipe(gulp.dest './build'));
});
但我无法弄清楚如何检索任务中找到(和处理过的)文件名,或者我无法找到正确的插件。
我不确定您想如何使用文件名,但其中一个应该有所帮助:
-
如果您只想查看名称,可以使用
gulp-debug
,其中列出了乙烯基文件的详细信息。插入这个你想要列表的任何地方,如下所示:
var gulp = require('gulp'),
debug = require('gulp-debug');
$ b gulp.task('examples',function(){
return gulp.src('./ examples / *。html')
.pipe(debug())
.pipe(gulp.dest('./ build'));
});
-
另一个选项是
gulp-文件大小
,它会输出文件及其大小。 如果您想要更多控制,可以使用如
gulp-tap
,它可让您提供您的自己的功能,并查看管道中的文件。 In my gulp.js file I'm streaming all HTML files from the examples
folder into the build
folder.
To create the gulp task is not difficult:
var gulp = require('gulp');
gulp.task('examples', function() {
return gulp.src('./examples/*.html')
.pipe(gulp.dest('./build'));
});
But I can't figure out how retrieve the file names found (and processed) in the task, or I can't find the right plugin.
I'm not sure how you want to use the file names, but one of these should help:
If you just want to see the names, you can use something like
gulp-debug
, which lists the details of the vinyl file. Insert this anywhere you want a list, like so:var gulp = require('gulp'), debug = require('gulp-debug'); gulp.task('examples', function() { return gulp.src('./examples/*.html') .pipe(debug()) .pipe(gulp.dest('./build')); });
Another option is
gulp-filelog
, which I haven't used, but sounds similar (it might be a bit cleaner).Another options is
gulp-filesize
, which outputs both the file and it's size.If you want more control, you can use something like
gulp-tap
, which lets you provide your own function and look at the files in the pipe.
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