blocks|key|154117|text|我发现这个插件正在做我所期望的事情：gulp-using|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|154118|简单使用示例:使用.jsx扩展名搜索项目中的所有文件|154119|gulp.task('reactify',+function(){
++++++++gulp.src(['../**/*.jsx'])+
++++++++++++.pipe(using({}));
++++++++....
++++});|code-block|syntax|javascript|154120|输出：|154121|[gulp]+Using+gulpfile+/app/build/gulpfile.js
[gulp]+Starting+'reactify'...
[gulp]+Finished+'reactify'+after+2.92+ms
[gulp]+Using+file+/app/staging/web/content/view/logon.jsx
[gulp]+Using+file+/app/staging/web/content/view/components/rauth.jsx|154122|entityMap|0|LINK|mutability|MUTABLE|url|https://www.npmjs.org/package/gulp-using^0|I|A|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|11|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|C|$]]|$1|M|3|N|5|H|7|13|8|@]|9|@]|C|$I|J]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

I found this plugin to be doing what I was expecting: <a href="https://www.npmjs.org/package/gulp-using" rel="noreferrer">gulp-using</a>

Simple usage example: Search all files in project with .jsx extension

<pre><code>gulp.task('reactify', function(){
 gulp.src(['../**/*.jsx']) 
 .pipe(using({}));
 ....
 });
</code></pre>

Output:

<pre><code>[gulp] Using gulpfile /app/build/gulpfile.js
[gulp] Starting 'reactify'...
[gulp] Finished 'reactify' after 2.92 ms
[gulp] Using file /app/staging/web/content/view/logon.jsx
[gulp] Using file /app/staging/web/content/view/components/rauth.jsx
</code></pre>

blocks|key|4660633|text|这是另一种简单的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4660634|var+es,+log,+logFile;

es+=+require('event-stream');

log+=+require('gulp-util').log;

logFile+=+function(es)+{
++return+es.map(function(file,+cb)+{
++++log(file.path);
++++return+cb(null,+file);
++});
};

gulp.task("do",+function()+{
+return+gulp.src('./examples/*.html')
+++.pipe(logFile(es))
+++.pipe(gulp.dest('./build'));
});|code-block|syntax|javascript|4660635|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is another simple way.
<pre><code>var es, log, logFile;

es = require('event-stream');

log = require('gulp-util').log;

logFile = function(es) {
 return es.map(function(file, cb) {
 log(file.path);
 return cb(null, file);
 });
};

gulp.task(&quot;do&quot;, function() {
 return gulp.src('./examples/*.html')
 .pipe(logFile(es))
 .pipe(gulp.dest('./build'));
});
</code></pre>

blocks|key|4660653|text|您可以使用gulp-filenames模块来获取路径数组。您甚至可以按名称空间对它们进行分组：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4660654|var+filenames+=+require("gulp-filenames");

gulp.src("./src/*.coffee")
++++.pipe(filenames("coffeescript"))
++++.pipe(gulp.dest("./dist"));

gulp.src("./src/*.js")
++.pipe(filenames("javascript"))
++.pipe(gulp.dest("./dist"));

filenames.get("coffeescript")+//+["a.coffee","b.coffee"]++
++++++++++++++++++++++++++++++//+Do+Something+With+it+|code-block|syntax|javascript|4660655|entityMap|0|LINK|mutability|MUTABLE|url|https://www.npmjs.com/package/gulp-filenames^0|5|E|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

You can use the <a href="https://www.npmjs.com/package/gulp-filenames" rel="noreferrer">gulp-filenames</a> module to get the array of paths.
You can even group them by namespaces: 

<pre><code>var filenames = require("gulp-filenames");

gulp.src("./src/*.coffee")
 .pipe(filenames("coffeescript"))
 .pipe(gulp.dest("./dist"));

gulp.src("./src/*.js")
 .pipe(filenames("javascript"))
 .pipe(gulp.dest("./dist"));

filenames.get("coffeescript") // ["a.coffee","b.coffee"] 
 // Do Something With it 
</code></pre>

blocks|key|154162|text|在我的例子中，gulp-ignore是完美的。作为选项，您可以在其中传递一个函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|154163|function+condition(file)+{
+//+do+whatever+with+file.path
+//+return+boolean+true+if+needed+to+exclude+file+
}|code-block|syntax|javascript|154164|该任务将如下所示：|154165|var+gulpIgnore+=+require('gulp-ignore');

gulp.task('task',+function()+{
++gulp.src('./**/*.js')
++++.pipe(gulpIgnore.exclude(condition))
++++.pipe(gulp.dest('./dist/'));
});|154166|entityMap|0|LINK|mutability|MUTABLE|url|https://www.npmjs.com/package/gulp-ignore^0|7|B|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

For my case <a href="https://www.npmjs.com/package/gulp-ignore" rel="nofollow">gulp-ignore</a> was perfect.
As option you may pass a function there:

<pre><code>function condition(file) {
 // do whatever with file.path
 // return boolean true if needed to exclude file 
}
</code></pre>

And the task would look like this:

<pre><code>var gulpIgnore = require('gulp-ignore');

gulp.task('task', function() {
 gulp.src('./**/*.js')
 .pipe(gulpIgnore.exclude(condition))
 .pipe(gulp.dest('./dist/'));
});
</code></pre>

blocks|key|154171|text|如果你想在Typescript中使用@OverZealous‘answer+(https://stackoverflow.com/a/21806974/1019307)，你需要使用import而不是require|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|154172|import+*+as+debug+from+'gulp-debug';

...

++++return+gulp.src('./examples/*.html')
++++++++.pipe(debug({title:+'example+src:'}))
++++++++.pipe(gulp.dest('./build'));|code-block|syntax|javascript|154173|(我还添加了一个title)。|154174|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/21806974/1019307^0|2I|6|2R|7|13|18|0|0|0|8|5|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

If you want to use @OverZealous' answer (<a href="https://stackoverflow.com/a/21806974/1019307">https://stackoverflow.com/a/21806974/1019307</a>) in Typescript, you need to <code>import</code> instead of <code>require</code>:

<pre><code>import * as debug from 'gulp-debug';

...

 return gulp.src('./examples/*.html')
 .pipe(debug({title: 'example src:'}))
 .pipe(gulp.dest('./build'));
</code></pre>

(I also added a <code>title</code>).

In my gulp.js file I'm streaming all HTML files from the <code>examples</code> folder into the <code>build</code> folder. 

To create the gulp task is not difficult:

<pre><code>var gulp = require('gulp');

gulp.task('examples', function() {
 return gulp.src('./examples/*.html')
 .pipe(gulp.dest('./build'));
});
</code></pre>

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.

Get the current file name in gulp.src()

在我的gulp.js文件中，我将examples文件夹中的所有文件流式传输到build文件夹中。创建吞咽任务并不困难：var gulp = require('gulp');gulp.task('examples', function() {    return gulp.src('./examples/*.html')...

问获取gulp.src()中的当前文件名
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