Question

我们说我有以下两个数据集：

{ "fruits": {
    "apple": {
        "isFruit": "yes"
        },
    "orange": {
        "isFruit": "yes"
       },
    "tomato": {
        "isFruit": "yes"
       }
    }
}

和

{ "food": {
        "apple": {
            "color": "red",
            "isFruit": "yes"
            },
        "orange": {
            "color": "orange",
            "isFruit": "yes"
            },
        "potato": {
            "color": "brown"
            }
        "tomato": {
            "color": "red",
            "isFruit": "yes"
            }
        }
    }

有没有办法将这两个组合起来，使它们最终看起来像：

{{1}}

我认为必须有一些方法来解析这两组并将它们组合起来，但我还没有弄清楚如何。

Answer 1

var foodsData = { "food": {
    "apple": {
        "color": "red"
        },
    "orange": {
        "color": "orange"
        },
    "potato": {
        "color": "brown"
        },
    "tomato": {
        "color": "red"
        }
    }
};

var fruitsData = { "fruits": {
    "apple": {
        "isFruit": "yes"
        },
    "orange": {
        "isFruit": "yes"
       },
    "tomato": {
        "isFruit": "yes"
       }
    }
};


for (f in foodsData.food) {
  if (fruitsData.fruits.hasOwnProperty(f)) {
    foodsData.food[f].isFruit = fruitsData.fruits[f].isFruit
  } 
} 
console.log(foodsData);

Answer 2

   var foodsData = {
  "food": {
    "apple": {
      "color": "red"
    },
    "orange": {
      "color": "orange"
    },
    "potato": {
      "color": "brown"
    },
    "tomato": {
      "color": "red"
    }
  }
};

var fruitsData = {
  "fruits": {
    "apple": {
      "isFruit": "yes",
      "isRoot": 'no',
      "seasonal": 'Yes',
    },
    "orange": {
      "isFruit": "yes",
      "isRoot": 'no'
    },
    "tomato": {
      "isFruit": "yes",
      "isRoot": 'No',
      "seasonal": 'No',
    }
  }
};


for (fruit in foodsData.food) {
  if (fruitsData.fruits.hasOwnProperty(fruit)) {
    var allProperties = Object.keys(fruitsData.fruits[fruit]);
    for (var i = 0; i < allProperties.length; i++) {
      var property = allProperties[i];
      foodsData.food[fruit][property] = fruitsData.fruits[fruit][property];
    }
  }
}
console.log(foodsData);

选中此fiddle

即使你在fruitData中为fruit添加了一些属性，这也会有效。

Answer 3

尝试JavaScript ternary operator ：

var firstJSON = {
	"food": {
		"apple": {
			"color": "red"
		},
		"orange": {
			"color": "orange"
		},
		"potato": {
			"color": "brown"
		},
		"tomato": {
			"color": "red"
		}
	}
};

var secondJSON = {
	"fruits": {
		"apple": {
			"isFruit": "yes"
		},
		"orange": {
			"isFruit": "yes"
		},
		"tomato": {
			"isFruit": "yes"
		}
	}
};

for(var i in firstJSON.food) {
  secondJSON.fruits.hasOwnProperty(i) ? firstJSON.food[i].isFruit = secondJSON.fruits[i].isFruit : '';
}

console.log(firstJSON);

Answer 4

尝试以下解决方案，它将以通用方式工作所有JSON对象。而且我也处理了物体的动态长度。

我希望这会对你有所帮助。

var obj1 = { "food": {
            "apple": {
                "color": "red"
                },
            "orange": {
                "color": "orange"
                },
            "potato": {
                "color": "brown"
                },
            "tomato": {
                "color": "red"
                }
            }
        }
var obj2 = { "fruits": {
                "apple": {
                    "isFruit": "yes"
                    },
                "orange": {
                    "isFruit": "yes"
                   },
                "tomato": {
                    "isFruit": "yes"
                   }
                }
            }

function result() {
    var resObj = [];

    if(objSize(obj1[Object.keys(obj1)]) >= objSize(obj2[Object.keys(obj2)])) {
        var indexObj1 = 0;
        var indexObj2 = 0;
        for(var i in obj1[Object.keys(obj1)]) {
            var flag= false;
            indexObj1++;
            indexObj2 = 0;
            for(var j in obj2[Object.keys(obj2)]) {
                indexObj2++;
                if(indexObj2 == indexObj1){
                    if(i == j) {
                        var item = [];
                        var item1 = {};

                        item.push(obj1[Object.keys(obj1)][i]);
                        item.push(obj2[Object.keys(obj2)][j]);
                        item1[i] = item;            

                        resObj.push(item1);
                        flag = true;
                        break;
                    } else {
                        var item = [];
                        var item1 = {};

                        item.push(obj1[Object.keys(obj1)][i]);
                        item1[i] = item;

                        resObj.push(item1);

                        item = [];
                        item1 = {};
                        item.push(obj2[Object.keys(obj2)][j]);
                        item1[j] = item;            

                        resObj.push(item1);
                        flag = true;
                        break;
                    }
                }
            }
            if(!flag) {
                var item1 = {};

                item.push(obj1[Object.keys(obj1)][i]);
                item1[i] = item;            

                resObj.push(item1);
                flag = true;
            }

        }
    } else {
        for(var i in obj2[Object.keys(obj2)]) {
            var flag= false;
            for(var j in obj1[Object.keys(obj1)]) {
                if(i == j) {
                    var item = [];
                    var item1 = {};

                    item.push(obj1[Object.keys(obj1)][i]);
                    item.push(obj2[Object.keys(obj2)][j]);
                    item1[i] = item;            

                    resObj.push(item1);
                    flag = true;
                    break;
                } 
            }
            if(!flag) {
                var item1 = {};

                item.push(obj1[Object.keys(obj1)][i]);
                item1[i] = item;            

                resObj.push(item1);
                flag = true;
            }

        }
    }

    return resObj;
}

function objSize(obj) {
    var size = 0, key;
    for (key in obj) {
        if (obj.hasOwnProperty(key)) size++;
    }
    return size;
}

 console.log(result());

请在下面的小提琴中找到工作演示。

Working Demo

组合JSON数据集

4 个答案: