Question

我试图通过运行任务将所有css文件连接成一个已发布的css文件：

gulp.task('task one', () => {  
    gulp.src([Dir + "**/*.css"])
        .pipe(concat(outFilename + ".css"))
        .pipe(gulp.dest(destDir));
});

之后，因为输出css文件中的字体路径不正确，所以我需要修改路径。

gulp.task('task two', () => {
    var urlAdjuster = require('gulp-css-url-adjuster');

    //change font path for released css files
    gulp.src([releasePath + "/*.css"])
        .pipe(urlAdjuster({
            replace: ['../../fonts', '../fonts'],
        }))
        .pipe(gulp.dest(releasePath + "/"));
);

为了实现这一目标，我需要逐个分别运行这两个任务。如何按顺序运行这两个任务并将它们放入一个任务中。

我试过了： series = require('gulp-series'); ----无法正常工作

gulp.task('task two', ['task one'], function () {
    // Gulp. src (...
});

-----不工作

---更新------------------------------------------- ------------- 我使用了返回，但它现在正在工作。实际代码是：

function procHTMLForRelease(stream, file) {
    gulp.src([Dir + "**/*.css"])
        .pipe(concat(outFilename + ".css"))
        .pipe(gulp.dest(destDir));

}

function processRelease() {


    // Grab all html files in the folder.
    // Loop over them, process them, then spit em out.
    gulp.src([paths.Source + "/*.html"])
        .pipe(foreach(procHTMLForRelease))
        .pipe(gulp.dest(paths.Release));

    // Process image files
    gulp.src([paths.Source + "images/*.*"])
        .pipe(gulp.dest(paths.Release + "images/"));

    // Copy font files
    gulp.src([paths.html + "fonts/*.*", paths.html + "fonts/*/*.*"])
        .pipe(gulp.dest(paths.release + "/fonts/"));


}  



gulp.task('task one', () => { return processRelease();
});

Answer 1

// Model file

<?php
defined('BASEPATH') OR exit('No direct script access allowed');
class User_model extends CI_Model
{
  public function insertuser($data)
  {
    return $this->db->insert('candidates', $data);
  }
  public function verifyemail($key)
  {
    
        $this->db->where('md5(email)', $key);
        return $this->db->update('candidates', $data);
  }
  public function check_user($email,$pass)
  {
    $sql = "SELECT  id , fname FROM candidates where email = ? and password = ?";
    $data = $this->db->query($sql, array($email,$pass));
        return ($data->result_array()) ;
  }

  
      // Function To Fetch Selected Record
 public function show_user_id($data){
      $this->db->select('*');
      $this->db->from('candidates');
      $this->db->where('email', $data);
      $q=$this->db->get('candidates');
      return $q->row_array();
      
      }
      // Update Query For Selected Student
 public function update_user_id1($id,$data){
      $this->db->where('email', $id);
      $this->db->update('candidates', $data);
      }
}

?>

这是正确的，它将gulp.task('task two', ['task one'], function () { // gulp.src( ... });设置为task one的依赖项。但是，要实现此目的，task two需要task one。

return

Answer 2

您可以像这样修改更新的代码，使其成为return。

gulp.task('taks one', function () {

    // Grab all html files in the folder.
    // Loop over them, process them, then spit em out.
    var one = gulp.src([paths.Source + "/*.html"])
        .pipe(foreach(procHTMLForRelease))
        .pipe(gulp.dest(paths.Release));

    // Process image files
    var two = gulp.src([paths.Source + "images/*.*"])
        .pipe(gulp.dest(paths.Release + "images/"));

    // Copy font files
    var three = gulp.src([paths.html + "fonts/*.*", paths.html + "fonts/*/*.*"])
        .pipe(gulp.dest(paths.release + "/fonts/"));

   // Return
   return [ one, two, three];

});

函数内部的任务将以异步方式运行，但您可以创建另一个函数来依赖所有已完成的任务。

gulp.task('task two', ['task one'], function () {
    // Gulp. src (...
});

Answer 3

对于任何人都有同样的问题并寻找解决方案：以下是我要解决的问题： https://www.npmjs.com/package/run-sequence

例如：

// This will run in this order:
// * build-clean
// * build-scripts and build-styles in parallel
// * build-html
// * Finally call the callback function
gulp.task('build', function(callback) {
  runSequence('build-clean',
              ['build-scripts', 'build-styles'],
              'build-html',
              callback);
});

如何按顺序运行gulp任务？

3 个答案: