解析包含JSON元素的对象数组

Question

首先让我细分数据：

我有一个包含3个元素的数组...

每个元素都是一个以name和arrayOfJSON作为键的对象...

在arrayOfJSON内可以有任意数量的JSON字符串作为元素...

我需要捕获数组Alex@gmail和mess都出现arrayOfJSON的位置

结果应为：

position_of_mess = [0,2]

position_of_arrayOfJSON_for_position_of_mess_0 = [0]

position_of_arrayOfJSON_for_position_of_mess_2 = [1]

我目前正在尝试什么：

用于遍历混乱，用于遍历arrayOfJSON，以及用于Alex @ gmail的JSON.parse（）。 要花几分钟时间进行更新。

如果大家都认为不需要for循环就可以做到。

更新：快到了

mess = [{
    "name": "user1",
    "arrayOfJSON": `[{"email":"Alex@gmail","hobby":"coding"},{"email":"bob@gmail","hobby":"coocking"}]`
  },
  {
    "name": "user2",
    "arrayOfJSON": `[{"email":"Chris@gmail","hobby":"coding"},{"email":"bob@gmail","hobby":"coocking"}]`
  },
  {
    "name": "user3",
    "arrayOfJSON": `[{"email":"bob@gmail","hobby":"coocking"},{"email":"Alex@gmail","hobby":"coding"}]`
  }
]


console.log(mess)

for (i = 0; i < mess.length; i++) {
  console.log(JSON.parse(mess[i].arrayOfJSON))
  
  for (m = 0; m < (JSON.parse(mess[i].arrayOfJSON)).length; m++) {
    console.log("almost")
    console.log((JSON.parse(mess[i].arrayOfJSON))[m])
  } 
} 

Answer 1

1. 解析您的数据
   • 您希望能够访问内部对象“字符串”内的键
2. 遍历您的数据
   • 在访问键值对时，建立一个范围，以便以后返回

// Adapted from: https://gist.github.com/sphvn/dcdf9d683458f879f593
const traverse = function(o, fn, scope = []) {
  for (let i in o) {
    fn.apply(this, [i, o[i], scope]);
    if (o[i] !== null && typeof o[i] === "object") {
      traverse(o[i], fn, scope.concat(i));
    }
  }
}

const mess = [{
  "name": "user1",
  "arrayOfJSON": `[{"email":"Alex@gmail","hobby":"coding"},{"email":"bob@gmail","hobby":"coocking"}]`
}, {
  "name": "user2",
  "arrayOfJSON": `[{"email":"Chris@gmail","hobby":"coding"},{"email":"bob@gmail","hobby":"coocking"}]`
}, {
  "name": "user3",
  "arrayOfJSON": `[{"email":"bob@gmail","hobby":"coocking"},{"email":"Alex@gmail","hobby":"coding"}]`
}];

// Parse...
mess.forEach(item => {
  if (item.arrayOfJSON) {
    item.arrayOfJSON = JSON.parse(item.arrayOfJSON);
  }
});

traverse(mess, (key, value, scope) => {
  if (value === 'Alex@gmail') {
    console.log(
      `Position: mess[${scope.concat(key).map(k => isNaN(k) ? `'${k}'` : k).join('][')}]`
    );
  }
});

.as-console-wrapper {
  top: 0;
  max-height: 100% !important;
}

Answer 2

mess = [{
    "name": "user1",
    "arrayOfJSON": `[{"email":"Alex@gmail","hobby":"coding"},{"email":"bob@gmail","hobby":"coocking"}]`
  },
  {
    "name": "user2",
    "arrayOfJSON": `[{"email":"Chris@gmail","hobby":"coding"},{"email":"bob@gmail","hobby":"coocking"}]`
  },
  {
    "name": "user3",
    "arrayOfJSON": `[{"email":"bob@gmail","hobby":"coocking"},{"email":"Alex@gmail","hobby":"coding"}]`
  }
]


console.log(mess)
holdMessPosition = []

for (i = 0; i < mess.length; i++) {

  var pos = (JSON.parse(mess[i].arrayOfJSON)).map(function(e) {
      return e.email;
    })
    .indexOf("Alex@gmail");
  console.log("user position is " + pos);

  if (pos !== -1) {
    holdMessPosition.push(i)
  }

}

console.log(holdMessPosition)