Get the current file name in gulp src

Question

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

User · Answer

Here is another simple way    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              gulp task  do   function      return gulp src    examples   html       pipe logFile es       pipe gulp dest    build

User · Answer

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

User · Answer

I found this plugin to be doing what I was expecting  gulp-using  Simple usage example  Search all files in project with  jsx extension  gulp task  reactify   function            gulp src           jsx                  pipe using                              Output    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

User · Answer

If you want to use  OverZealous  answer  https   stackoverflow com a 21806974 1019307  in Typescript  you need to import instead of require   import   as debug from  gulp-debug             return gulp src    examples   html            pipe debug  title   example src               pipe gulp dest    build        I also added a title

User · Answer

You can use the gulp-filenames module to get the array of paths  You can even group them by namespaces    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

User · Answer

For my case gulp-ignore was perfect  As option you may pass a function there   function condition file        do whatever with file path     return boolean true if needed to exclude file      And the task would look like this   var gulpIgnore   require  gulp-ignore     gulp task  task   function       gulp src         js        pipe gulpIgnore exclude condition        pipe gulp dest    dist

[gulp] Get the current file name in gulp.src()

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