You can use the `suffix` or `extname` parameters of [gulp-rename][1] like: 

    gulp.task(&#39;compressjs&#39;, function () {
      gulp.src(&#39;app/**/*.js&#39;)
        .pipe(uglify())
        .pipe(rename({ suffix: &#39;.min&#39; }))
        .pipe(gulp.dest(&#39;dist&#39;))
    })

  [1]: https://github.com/hparra/gulp-rename

I want to parameterise the following computation using `dplyr` that finds which values of `Sepal.Length` are associated with more than one value of `Sepal.Width`:

    library(dplyr)

    iris %&gt;%
        group_by(Sepal.Length) %&gt;%
        summarise(n.uniq=n_distinct(Sepal.Width)) %&gt;%
        filter(n.uniq &gt; 1)

Normally I would write something like this:

    not.uniq.per.group &lt;- function(data, group.var, uniq.var) {
        iris %&gt;%
            group_by(group.var) %&gt;%
            summarise(n.uniq=n_distinct(uniq.var)) %&gt;%
            filter(n.uniq &gt; 1)
    }

However, this approach throws errors because `dplyr` uses [non-standard evaluation](http://adv-r.had.co.nz/Computing-on-the-language.html). How should this function be written?

Pass arguments to dplyr functions

Those are my first steps in IOS applications development and I&#39;m facing some problem that I can&#39;t figure out.

    error: Embedded binary is not signed with the same certificate as the parent app. Verify the embedded binary target&#39;s code sign settings match the parent app&#39;s.

		Embedded Binary Signing Certificate:	Not Code Signed
		Parent App Signing Certificate:		    iPhone Developer: Emil Adz (9QNEF95395)

I can&#39;t understand, what is a **Embedded Binary Signing Certificate**?

I went over the questions with the same error here, but none of them were related to the problem presented here (**Not Code Signed**).

I tried to revoke my certificate from the Apple Developer Member Center and request another certificate, but the issue persists.

Does some know how it could be fixed?


iOS error &quot;Embedded binary is not signed with the same certificate as the parent app&quot;

The js file being compressed is output with the same filename.

    gulp.task(&#39;compressjs&#39;, function () {
      gulp.src(&#39;app/**/*.js&#39;)
        .pipe(uglify())
        .pipe(gulp.dest(&#39;dist&#39;));
    });

How can I have it output the file with `min.js` at the back?

Gulp uglify output min.js

The js file being compressed is output with the same filename.
<pre><code class="hljs language-javascript">gulp.task('compressjs', function () {
 gulp.src('app/**/*.js')
 .pipe(uglify())
 .pipe(gulp.dest('dist'));
});
</code></pre>
How can I have it output the file with <code>min.js</code> at the back?

I&#39;ve been trying for a day to write two pipe functions, one that compiles less files and another one that concats these files. I want to learn how to write transform streams/pipes for more complex plugins.

So I want to know how to read data from another pipe, and how to alter that data and send it to the next pipe. This is what I have so far:

     gulp.src(sources)
       .pipe(through.obj(function (chunk, enc, cb) {

         var t = this;
         // console.log(&quot;chunk&quot;, chunk.path);
         fs.readFile(chunk.path, enc, function (err,data) {
           if (err) { cb(err); }
                            
           less.render(data, {
             filename : chunk.path,
             sourceMap : {
               sourceMapRootpath : true
             }
           })
           .then(function (outputCss) {
              // console.log(&quot;less result&quot;,outputCss);
              t.push(chunk);// or this.push(outputCss) same result
              cb();
           });

         });

       }))
       .pipe(through.obj(function (chunk, enc, cb) {
         console.log(&quot;chunk&quot;, chunk.path); // not event getting called.
         cb();
       }))

I can&#39;t get the `outputCSS` for each file in the second pipe. How can I send it?

How can I write a simple gulp pipe function?

I want to log to stdout (the config environment) when a [gulp][1] task is running or has run.

Something like this:

    gulp.task(&#39;scripts&#39;, function () {
      var enviroment = argv.env || &#39;development&#39;;
      var config = gulp.src(&#39;config/&#39; + enviroment + &#39;.json&#39;)
          .pipe(ngConstant({name: &#39;app.config&#39;}));
      var scripts = gulp.src(&#39;js/*&#39;);
    
      return es.merge(config, scripts)
        .pipe(concat(&#39;app.js&#39;))
        .pipe(gulp.dest(&#39;app/dist&#39;))
        .on(&#39;success&#39;, function() { 
          console.log(&#39;Configured environment: &#39; + environment);
        });
    });

I am not sure what event I should be responding to or where to find a list of these. Any pointers? Many thanks.


  [1]: http://gulpjs.com

console.log to stdout on gulp events

I would like to use google web starter kit. I installed node.js v0.12.0, `node-sass` &amp; `gulp`.

And then ran:

    $ sudo npm install

When I typed `gulp serve` then got this error:

    Using gulpfile ~/web-starter-kit/gulpfile.js Starting &#39;styles&#39;... &#39;styles&#39; errored after 93 ms 
    Error: `libsass` bindings not found. Try reinstalling `node-sass`? at getBinding

I reinstalled node and gulp but this doesn&#39;t help.

What should I do next?

Try reinstalling `node-sass` on node 0.12?

Normally we can run gulp task from console via something like `gulp mytask`. Is there anyway that I can pass in parameter to gulp task? If possible, please show example how it can be done.

Pass Parameter to Gulp Task

I&#39;m extremely new to Gulp.  I&#39;m basically trying to watch for a modified JavaScript file, and then make a new copy of it with a new name.  (eventually there&#39;ll be some processing on it, but Rome wasn&#39;t built in a day).

My (naive) attempt is this:

    gulp.task(&#39;default&#39;, function() {
    
        return gulp.watch(&#39;../**/**.js&#39;, function(obj){
            gulp.src(obj.path)
                .pipe(gulp.dest(&#39;foobar.js&#39;));
        });
    
    });

This takes the modified file and successfully copies it into a folder now called foobar.js.  Is there anything simple I can replace `gulp.dest(&#39;foobar.js&#39;)` with that will simply copy and rename the src file in place?

---

EDIT

By copy in place, I mean I want to take the modified file, and make a copy of it right **where it currently is** with a new name.  The equivalent of clicking the file (in windows) and hitting control-c control-v, then renaming the resulting file.

Gulp - copy and rename a file

i&#39;m pretty new to gulp , and I follow tutorials in http://leveluptuts.com/tutorials/learning-gulp , I get this error:

    events.js:141
      throw er; // Unhandled &#39;error&#39; event
      ^
    Error
    at new JS_Parse_Error (/home/kid/node_modules/gulp-uglify/node_modules/uglify-js/lib/parse.js:196:18)
    at js_error (/home/kid/node_modules/gulp-uglify/node_modules/uglify-js/lib/parse.js:204:11)
    at croak (/home/kid/node_modules/gulp-uglify/node_modules/uglify-js/lib/parse.js:680:9)
    at token_error (/home/kid/node_modules/gulp-uglify/node_modules/uglify-js/lib/parse.js:688:9)
    at unexpected (/home/kid/node_modules/gulp-uglify/node_modules/uglify-js/lib/parse.js:694:9)
    at expr_atom (/home/kid/node_modules/gulp-uglify/node_modules/uglify-js/lib/parse.js:1201:9)
    at maybe_unary (/home/kid/node_modules/gulp-uglify/node_modules/uglify-js/lib/parse.js:1363:19)
    at expr_ops (/home/kid/node_modules/gulp-uglify/node_modules/uglify-js/lib/parse.js:1398:24)
    at maybe_conditional (/home/kid/node_modules/gulp-uglify/node_modules/uglify-js/lib/parse.js:1403:20)
    at maybe_assign (/home/kid/node_modules/gulp-uglify/node_modules/uglify-js/lib/parse.js:1427:20)


here is my code :
   
     var gulp = require(&#39;gulp&#39;)
	    uglify = require(&#39;gulp-uglify&#39;);

    gulp.task(&#39;default&#39;, function() {
	    // body...
	    gulp.src(&#39;js/*.js&#39;)
	    .pipe(uglify())
	    .pipe(gulp.dest(&#39;minjs&#39;));

    });

and the tree of directory, just simple 


    -gulp
         -js/
         -gulpfile.js


thanks a lot


Gulp-uglify : throw er; // Unhandled &#39;error&#39; event

I am trying to minify my script files for which i am using gulp task runner 
And I am trying gulp-uglify plugin

Code:


 

     gulp.task(&#39;concat&#39;, function() {
        return gulp.src(&#39;app/**/*.js&#39;)
            // .pipe(concat(&#39;script.js&#39;))
            .pipe(uglify())
            .pipe(gulp.dest(&#39;./dist/&#39;))
    });


but i am getting error as

[![enter image description here][1]][1]


  [1]: https://i.stack.imgur.com/pypfL.png

when i try to run gulp task as gulp concat
Any help would be appreciated

GulpUglifyError:Unable to minify JavaScript

I am trying to use gulp in order to minify a folder containing JS files. However, one of the files has the above error, preventing it from being minified.

I managed to catch and print the error, which I&#39;ve partially printed here:

    JS_Parse_Error {
     message: &#39;SyntaxError: Unexpected token: punc ())&#39;,
     filename: &#39;ex.js&#39;,
     line: 189,
     col: 25,
     pos: 6482,
     stack: Error\n    at new JS_Parse_Error (eval at &lt;anonymous&gt; ... ) 
     plugin: &#39;gulp-uglify&#39;,
     fileName: &#39;.../js/ex.js&#39;,
     showStack: false
    }

The file in question contains the following, shortened:

    function() {
      ...
      $.confirm({
		buttons: {
			confirm: function() {
				$.post(&#39;/ajax-handler&#39;, {
						...
					})
					.done( function(response) {
						var data = filterResponse(response);
						if (data[&#39;status&#39;] == &#39;success&#39;) {
							sleep(1000).then(() =&gt; {
						*	    ...
							});
							sleep(5000).then(() =&gt; {
								...  
							});
							
						} else {
							console.log(&#39;Oops!&#39;);
						}
					})
					.fail( function(err, status, response) {
						...
				});
			},
			cancel: function() {}
		}
	 });
      ...
    }

I added the &quot;*&quot; above in order to indicate the exact position listed by JS_Parse_Error.

Uglify SyntaxError: Unexpected token: punc ())

When I run gulp I get the following error:
    
    [12:54:14] { [GulpUglifyError: unable to minify JavaScript]
    cause:
    { [SyntaxError: Unexpected token: operator (&gt;)]
     message: &#39;Unexpected token: operator (&gt;)&#39;,
     filename: &#39;bundle.js&#39;,
     line: 3284,
     col: 46,
     pos: 126739 },
    plugin: &#39;gulp-uglify&#39;,
    fileName: &#39;C:\\servers\\vagrant\\workspace\\awesome\\web\\tool\\bundle.js&#39;,
    showStack: false }

The offending line contains an arrow function:
    
    let zeroCount = numberArray.filter(v =&gt; v === 0).length

I know I can replace it with the following to remedy the minification error by abandoning ES6 syntax:
        
    let zeroCount = numberArray.filter(function(v) {return v === 0;}).length

How can I minify code containing ES6 features via gulp?



Content Type	Original Author	Original Content on Stackoverflow
Question	user1995781	View Question on Stackoverflow
Solution 1 - Gulp	Preview	View Answer on Stackoverflow

Gulp uglify output min.js

Gulp Problem Overview

Gulp Solutions

Solution 1 - Gulp

iOS error "Embedded binary is not signed with the same certificate as the parent app"

Pass arguments to dplyr functions

Attributions