Question

我有两个不同的JSON对象，一个

[{"pmid":"29092960","MemberID":"33"},{"pmid":"28652336","MemberID":"33"}]

和另一个类似

{
    "meta": {
        "refs": 0, 
        "pmids": "29092960,28652336"
    }, 
    "links": {
        "self": "https://icite.od.nih.gov/api/pubs?refs=0&pmids=23456789%2C27599104"
    }, 
    "data": [
        {
            "pmid": 29092960, 
            "doi": "10.1002/cncr.27976", 
            "authors": "ABC, XYZ", 
            "journal": "Cancer", 
            "year": 2013
        }, 
        {
            "pmid": 28652336, 
            "doi": "10.1371/journal.pbio.1002541", 
            "authors": "Bbbb", 
            "journal": "PLoS Biol.", 
            "year": 2016
        }
    ]
}

是否可以将这些组合？我创建的第一个对象是通过循环网格和添加项目。第二个来自我需要添加其他数据的api。这看起来与我发现的任何其他示例的结构都不一样。我想基于pmid值作为键进行链接

我希望输出为

       {"pmid": 29092960, 
        "doi": "10.1002/cncr.27976", 
        "authors": "ABC, XYZ", 
        "journal": "Cancer", 
        "year": 2013,
         "MemberID":"33"},{
         "pmid": 29092960, 
        "doi": "10.1002/cncr.27976", 
        "authors": "ABC, XYZ", 
        "journal": "Cancer", 
        "year": 2013,
       "MemberID": "33"}

Answer 1

合并两个JSON对象javaScript

let arr1 =[{"pmid":"29092960","MemberID":"33"},{"pmid":"28652336","MemberID":"33"}];


let arr2 = {
    "meta": {
        "refs": 0, 
        "pmids": "29092960,28652336"
    }, 
    "links": {
        "self": "https://icite.od.nih.gov/api/pubs?refs=0&pmids=23456789%2C27599104"
    }, 
    "data": [
        {
            "pmid": 29092960, 
            "doi": "10.1002/cncr.27976", 
            "authors": "ABC, XYZ", 
            "journal": "Cancer", 
            "year": 2013
        }, 
        {
            "pmid": 28652336, 
            "doi": "10.1371/journal.pbio.1002541", 
            "authors": "Bbbb", 
            "journal": "PLoS Biol.", 
            "year": 2016
        }
    ]
};

let arr3 = [];

arr1.forEach((itm, i) => {
  arr3.push(Object.assign({}, itm, arr2.data.find(k => k.pmid==itm.pmid)));
});
console.log(arr3);

Answer 2

您确定可以组合对象。

在循环中，这将帮助您完成工作，并且在处理对象时将其放在口袋中是一件好事。

我在下面提供了一个解决方案。

var obj1 = [{"pmid":"29092960","MemberID":"33"},{"pmid":"28652336","MemberID":"33"}];

var obj2 = {
    "meta": {
        "refs": 0, 
        "pmids": "29092960,28652336"
    }, 
    "links": {
        "self": "https://icite.od.nih.gov/api/pubs?refs=0&pmids=23456789%2C27599104"
    }, 
    "data": [
        {
            "pmid": 29092960, 
            "doi": "10.1002/cncr.27976", 
            "authors": "ABC, XYZ", 
            "journal": "Cancer", 
            "year": 2013
        }, 
        {
            "pmid": 28652336, 
            "doi": "10.1371/journal.pbio.1002541", 
            "authors": "Bbbb", 
            "journal": "PLoS Biol.", 
            "year": 2016
        }
    ]
};


for(var char in obj1){
	for(var innerArr in obj2.data){
       // obj2 pmid is a number, so convert it to a string for the compare
		if(obj1[char].pmid === obj2.data[char].pmid.toString()){
			obj2.data[char].MemberID = obj1[char].MemberID;
		}
	}
}

console.log(JSON.stringify(obj2.data));

Answer 3

结合使用 Array.map() 和 Object.Assign() ：

let arr = [
  {"pmid":"29092960","MemberID":"33"},
  {"pmid":"28652336","MemberID":"33"}
];

let json = {
  "meta": {
    "refs": 0, 
    "pmids": "29092960,28652336"
  }, 
  "links": {
    "self": "https://icite.od.nih.gov/api/pubs?refs=0&pmids=23456789%2C27599104"
  }, 
  "data": [
    {
      "pmid": 29092960, 
      "doi": "10.1002/cncr.27976", 
      "authors": "ABC, XYZ", 
      "journal": "Cancer", 
      "year": 2013
    }, 
    {
      "pmid": 28652336, 
      "doi": "10.1371/journal.pbio.1002541", 
      "authors": "Bbbb", 
      "journal": "PLoS Biol.", 
      "year": 2016
    }
  ]
};

let res = arr.map(({pmid, MemberID}) =>
{
    return Object.assign({pmid, MemberID}, json.data.find(o => o.pmid === +pmid));
});

console.log(res);

请注意， unary plus operator 用于将原始数组的pmid强制（或强制转换）为一个数字。从参考中您可以阅读：

一元加号是将某物转换为数字的最快且首选的方法

但是，与其他答案一样，您可以使用o.pmid == pmid代替o.pmid === +pmid。

结合两个JSON对象javaScript

3 个答案: