Question

给出路径“documents / settings / user”

如何将其作为嵌套对象

结果

SELECT main.*
FROM
(
    SELECT
        CASE WHEN On_The_Table.Backpack_Number IS NOT NULL THEN On_The_Table.Backpack_Number ELSE Logs.Backpack_Number END AS Backpack_Number
        , On_The_Table.Pokemon_Number
        , On_The_Table.Name
        , On_The_Table.DateCaught
        , ROW_NUMBER() OVER() AS RANK

    FROM On_The_Table

  LEFT JOIN 
  (
      SELECT DISTINCT
            Backpack_Number,
            Pokemon_Number,
            DateCaught
        FROM What_We_Wrote_Down
        WHERE Backpack_Number IS NOT NULL AND Action = 'Caught'
  ) Logs
 ON On_The_Table.Pokemon_Number = Logs.Pokemon_Number AND On_The_Table.Date_Caught = Logs.Date_Caught


) main
WHERE RANK >= @searchParm1 AND RANK <= @searchParm2

我想不出如何引用每个先前的路径

{
   documents : {
         settings : {
               user : {}
          } 
    }
}

Answer 1

使用.reduce()实际上很容易做到这一点。

＆＃13;

var dir = {}

var paths = "documents/settings/user".split('/')

paths.reduce(function(dir, path) {
  return dir[path] = {}
}, dir)

console.log(dir)

＆＃13;

因为第一个参数始终是返回的最后一个值（或提供的第一个值），所以我们需要做的就是返回分配给当前路径的对象。并且因为赋值导致赋值，我们可以将其用作return语句本身的表达式。

如果需要，您可以防止由于相邻分隔符而导致的空路径名称。

＆＃13;

var dir = {}

var paths = "documents///settings/user".split('/')

paths.reduce(function(dir, path) {
  return path ? (dir[path] = {}) : dir
}, dir)

console.log(dir)

＆＃13;

Answer 2

Object通过引用传递。您可以使用此功能并执行此类操作。

Array.forEach

＆＃13;

var paths = "documents/settings/user".split('/')
var r = {};
var _tmp = r;
paths.forEach(function(el){
  _tmp[el] = {};
  _tmp = _tmp[el];
});

console.log(r)

＆＃13;

对于

＆＃13;

var paths = "documents/settings/user".split('/')
var r = {};
var _tmp = r;
for(var i=0; i<paths.length; i++){
  _tmp = (_tmp[paths[i]] = {});
};

console.log(r)

＆＃13;

Answer 3

另一种优雅的解决方案来构建具有价值的对象：

const buildObjWithValue = (path, value = '') => {
    const paths = path.split('.');
    return paths.reduceRight((acc, item, index) => ({
        [item]: index === paths.length - 1
            ? value
            : acc
    }), {});
}

例如 buildObjWithValue('very.deep.lake', 'Baikal') 给我们

{
  very: {
    deep: {
      lake: 'Bailkal'
    }
  }
}

根据文件路径字符串创建对象

3 个答案:

Array.forEach

对于