Gulp合并来自不同文件夹的json文件，同时保持文件夹结构 [英] Gulp merge json files from different folders while keeping folder structure

问题描述

输入

  /lang/module1/file1.json 
 /lang/module1/file2.json 
 /lang/module2/file1.json 
 / lang /module2/subModule1/file1.json 
 
 /lang_additional/module1/file1.json 
 /lang_additional/module1/file2.json 
 /lang_additional/module2/file1.json 
 /lang_additional/module2/subModule1/file1.json 
  

预期产量

  /dist/lang/module1/file1.json（均合并file1.json）
 /dist/lang/module1/file2.json file2.json的合并）
 /dist/lang/module2/file1.json（两个file1.json的合并）
 /dist/lang/module2/subModule1/file1.json（两个file1.json的合并）
  

是否可以用gulp来实现这一点？可能有多个文件夹和子文件夹。我希望指定哪个父文件夹在合并冲突解决方案中具有优先权。 这适用于您的示例输入和我相信每个文件夹中的任何数量的子文件夹。如果在另一个目录中有一个匹配的文件，它将被添加到一个流数组中，然后这个数组将被用于启动必要数量的gulp流，以保留文件夹结构，将其发送到'gulp-merge-json'和'dist'输出。

与gulp mergeJSON一起使用

  var gulp = require '吞'）; 
 var fs = require（'fs'）; 
 var path = require（'path'）; 
 var merge = require（'gulp-merge-json'）; 
 
 const files = []; 
 const parentFolders = []; 
 let streams = []; 
 const baseNames = []; 
 
 //注意'gulp-merge-json'的作用方式是与合并
中的前一个文件'wins'具有相同键的最后一个文件//因此无论哪个目录是在文件夹数组中最后列出的将具有PRECEDENCE 
 //'gulp-merge-json'也将采用'编辑'功能选项
 
 // [在你说的评论中你想要保留在'./dist'
 //但是'lang_additional'的优先顺序] 
 
 //保留的'lang'文件夹结构顺便说一下，它可以处理两个以上的目录，或者它可以工作与
 //不同级别的文件夹，如：
 // const folders = ['lang'，'lang_additional / module1 / subModule2']; 
 //只将这些文件合并到最深
 //级别的相同目录结构中。所以在上面的例子中，只能在subModule2和下面。 
 
 const folders = ['lang'，'lang_additional']; 
 $ b gulp.task（'mergeJSON'，function（）{
 
 getFiles（文件夹）; 
 makeStreams（）; 
 
 / / streams [1] = lang'\\module1\file2.json，lang_additional\module1\file2.json 
 
 //启动多个流，不是真正的流，直到吞噬。 src创建一个
 streams.forEach（函数（流）{
 
 //从一个流文件中获取fileName，它们都以相同的文件结束
 let fileName = path.basename（stream [stream.length  -  1]）; 
 
 //获取其中一个流的目录，流中的所有文件都具有相同的目录
 let dirName = path.dirname（stream [stream.length  -  1]）; 
 //剥离第一个目录，保留所有子目录
 dirName = dirName.substr（dirName.indexOf（path.sep））; 
 
 return gulp.src（stream）
 .pipe（merge（{fileName：fileName}））
 
 //因为问题需要dist / lang /子文件夹hiera rchy 
.pipe（gulp.dest（path.join（'./ dist'，'lang'，dirName）））; 
}）; 
}）; 
 
 // getFiles是递归的，如果fs.readdirSync检索到的文件是一个目录
 
函数getFiles（文件夹）{
 
 let possibleDirectory; 
 
 folders.forEach（function（folder，index）{
 
 //从每个目录读取文件列表
 let tempFiles = fs.readdirSync（'./ '文件夹）; 
 
 tempFiles.forEach（function（fileOrDirectory）{
 
 possibleDirectory = path.join（folder，fileOrDirectory）; 
 if（fs.lstatSync （possibleDirectory）.isDirectory（））{
 getFiles（[possibleDirectory]）; 
} 
 else {
 
 //文件[]将包含所有在（文件或目录））; 
 
 if $ b //如果该文件名元素尚不存在于baseName数组中
 //仅包含真实文件的基本名称数组
aseNames.push（fileOrDirectory）; 
} 
} 
}）; 
}）; 
 
 $ b $函数makeStreams（）{
 
 //对于每个文件，找到并保存其父目录而不使用文件夹[]root目录
 files.forEach（function（file）{
 
 let thisParentFolders = path.dirname（file）.substr（file.indexOf（path.sep））; 
 
 if（ parentFolders.indexOf（thisParentFolders）=== -1）{
 
 //如果该父文件夹不存在于baseName数组
 parentFolders.push（thisParentFolders）; 
} 
}）; 
 
 //现在通过所有独特的目录循环查找与每个parentFolder与baseName附加
 parentFolders.forEach（函数（文件夹）{
 
 let foldersFile = folder.substr（folder.indexOf（path.sep））; 
 
 baseNames.forEach（function（baseName）{
 
 streams.push（files.filter（function（file ）{
 return file.endsWith（path.join（foldersFile，baseName））; 
}））; 
}）; 
}）; 
 
 //在这里：删除所有只有一个文件的streams（数组子元素），length == 1 
 //但是现在这个过滤器是必须的，因为一个未定义的条目。 
 streams = streams.filter（stream => stream.length> = 1）; （（stream，index）=> console.log（streams [+ index +] =+ stream））; 
 
 streams.forEach 
} 
  

I would like to merge multiple .json files from different folders while keeping the general folder structure.

Input

/lang/module1/file1.json
/lang/module1/file2.json
/lang/module2/file1.json
/lang/module2/subModule1/file1.json

/lang_additional/module1/file1.json
/lang_additional/module1/file2.json
/lang_additional/module2/file1.json
/lang_additional/module2/subModule1/file1.json

Expected Output

/dist/lang/module1/file1.json                (both file1.json's merged)
/dist/lang/module1/file2.json                (both file2.json's merged)
/dist/lang/module2/file1.json                (both file1.json's merged)
/dist/lang/module2/subModule1/file1.json     (both file1.json's merged)

Is it possible to achieve this with gulp? There may be multiple folders and subfolders. I wish to designate which parent folder has precedence in the merge conflict resolution.

解决方案

This works for your example input and I believe any number of subfolders in each folder. If there is a matching file in another directory it is added to a streams array that will then be used to start the necessary number of gulp streams to send to 'gulp-merge-json' and your 'dist' output with the folder structure retained.

Use with "gulp mergeJSON"

var gulp = require('gulp');
var fs = require('fs');
var path = require('path');
var merge = require('gulp-merge-json');

const files = [];
const parentFolders = [];
let   streams = [];
const baseNames = [];

//    note that the way 'gulp-merge-json' works is the last file having the same key as an earlier file 'wins' on the merge
//    so whichever directory is listed last in the folders array will have PRECEDENCE
//   'gulp-merge-json' will also take an 'edit' function option

// [  in the comments you said you wanted the 'lang' folder structure preserved in './dist'
//     but with 'lang_additional' precedence  ]

//  By the way, this works with more than two directories or it can work with
// folders of different levels such as:
//         const folders = ['lang', 'lang_additional/module1/subModule2'];
// only merging those files with the same directory structure starting at the deepest
// levels only. So in the above example, only within subModule2 and below.

const folders = ['lang', 'lang_additional'];

gulp.task('mergeJSON', function () {

  getFiles(folders);
  makeStreams();

    // streams[1] =   lang\module1\file2.json,   lang_additional\module1\file2.json

  // spin up multiple "streams", not really a stream yet until gulp.src creates one
  streams.forEach(function (stream) {

    //  get the fileName from one of the stream files, they all end with the same file
    let fileName = path.basename(stream[stream.length - 1]);

    //  get the directories of one of the streams, again all files within a stream have the same directories
    let dirName = path.dirname(stream[stream.length - 1]);
    // strip off the first directory, leaving all subdirectories
    dirName = dirName.substr(dirName.indexOf(path.sep));

    return gulp.src(stream)
      .pipe(merge({ fileName: fileName }))

      // since the question wanted a dist/lang/subfolders hierarchy
      .pipe(gulp.dest(path.join('./dist', 'lang', dirName)));
  });
});

// getFiles is recursive, if a "file" retrieved by fs.readdirSync is a directory

function getFiles(folders) {

  let possibleDirectory;

  folders.forEach(function (folder, index) {

    // read the file list from each directory
    let tempFiles = fs.readdirSync('./' + folder);

    tempFiles.forEach(function (fileOrDirectory) {

      possibleDirectory = path.join(folder, fileOrDirectory);
      if (fs.lstatSync(possibleDirectory).isDirectory())  {
        getFiles([possibleDirectory]);
      }
      else {

        // files[] will include all files found under the folders array
        files.push(path.join(folder, fileOrDirectory));

        if (baseNames.indexOf(fileOrDirectory) === -1) {

          // if that file name element doesn't already exist in baseName array
          // an array of just the basenames of true files
          baseNames.push(fileOrDirectory);
        }
      }
    });
  });
}

function makeStreams() {

  // for each file, find and save its parent directories without the folders[] "root" directories
  files.forEach(function (file) {

    let thisParentFolders = path.dirname(file).substr(file.indexOf(path.sep));

    if (parentFolders.indexOf(thisParentFolders) === -1) {

      // if that parentfolder  doesn't already exist in baseName array
      parentFolders.push(thisParentFolders);
    }
  });

  // now loop through all unique directories looking for those files with each parentFolder with baseName attached
  parentFolders.forEach(function (folder) {

    let foldersFile = folder.substr(folder.indexOf(path.sep));

    baseNames.forEach(function (baseName) {

      streams.push(files.filter(function (file) {
         return file.endsWith(path.join(foldersFile, baseName));
      }));
    });
  });

  //    Here: remove any "streams" (array sub-elements) that have only one file in them, length == 1
  //    But for now this filter is necessary due to a undefined entry in .
  streams = streams.filter( stream => stream.length >= 1);

  streams.forEach( (stream, index) => console.log("streams[" + index + "] = " + stream));
}

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