JSON从子对象获取父对象

Question

如果brand_id =='983',我怎样才能获得折扣价值.

示例JSON:

{
     "prods": [
               {
            "info": {
                  "rate": 100
                    },
            "grocery": [
                     {
                      "brand": "A",
                      "brand_id": "983"
                     },
                     {
                      "brand": "B",
                      "brand_id": "253"
                     }
                     ],
             "discount": "20"
         }
     ]
}

我到现在为止尝试的是

$.prods[*].grocery[?(@.brand_id=='983')]

这将返回匹配对象的列表/数组.但我无法回到树上.对此有何帮助？

Answer 1

事实上，JSONPath 在这方面并不是很擅长，所以我用我自己的小库解决了这类问题；所以，这是你的例子的小提琴：

https://jsfiddle.net/YSharpLanguage/j9oetwnn/3

在哪里：

var products = {
     "prods": [
        {
            "info": {
                  "rate": 85
                    },
            "grocery": [
                     {
                      "brand": "C",
                      "brand_id": "984"
                     },
                     {
                      "brand": "D",
                      "brand_id": "254"
                     }
                     ],
             "discount": "15"
        },
        {
            "info": {
                  "rate": 100
                    },
            "grocery": [
                     {
                      "brand": "A",
                      "brand_id": "983"
                     },
                     {
                      "brand": "B",
                      "brand_id": "253"
                     }
                     ],
             "discount": "20"
         }
     ]
};

function GroceryItem(obj) {
  return (typeof obj.brand === "string") && (typeof obj.brand_id === "string");
}

    // last parameter set to "true", to grab all the "GroceryItem" instances
    // at any depth:
var itemsAndDiscounts = [ products ].nodeset(GroceryItem, true).
    map(
      function(node) {
        var item = node.value, // node.value: the current "GroceryItem" (aka "$.prods[*].grocery[*]")

            discount = node.parent. // node.parent: the array of "GroceryItem" (aka "$.prods[*].grocery")
                       parent. // node.parent.parent: the product (aka "$.prods[*]")
                       discount; // node.parent.parent.discount: the product discount

        // finally, project into an easy-to-filter form:
        return { id: item.brand_id, discount: discount };
      }
    ),
    discountOfItem983;

discountOfItem983 = itemsAndDiscounts.
  filter
  (
    function(mapped) {
      return mapped.id === "983";
    }
  )
  [0].discount;

console.log("All items and discounts: " + JSON.stringify(itemsAndDiscounts, null, 2));

console.log("Discount of #983: " + discountOfItem983);

给出：

All items and discounts: [
  {
    "id": "984",
    "discount": "15"
  },
  {
    "id": "254",
    "discount": "15"
  },
  {
    "id": "983",
    "discount": "20"
  },
  {
    "id": "253",
    "discount": "20"
  }
]
Discount of #983: 20

以下是其他示例/用例：

JSON 到某些标记

JSON 转换，重新审视（XSLT 类似）

（位于： https: //jsfiddle.net/YSharpLanguage/kj9pk8oz/10）

JSON 到 JSON

超轻量级 JSON 到 JSON 转换

（位于： https: //jsfiddle.net/YSharpLanguage/ppfmmu15/10）

JavaScript 等价于...

XSLT 3.0 REC 第 14.4 节示例：根据公共值对节点进行分组

（位于： http: //jsfiddle.net/YSharpLanguage/8bqcd0ey/1）

比照。https://www.w3.org/TR/xslt-30/#grouping-examples

JavaScript 等价于...

JSONiq 用例第 1.1.2 节。JSON 的分组查询

（位于： https: //jsfiddle.net/YSharpLanguage/hvo24hmk/3）

比照。http://jsoniq.org/docs/JSONiq-usecases/html-single/index.html#jsongrouping

'希望这可以帮助，