Question

我正在努力将xml转换为json对象，然后从转换后的对象中提取节点。我正在使用busyboy从服务器上传的文件中读取。之后我使用inspect将xml转换为json，然后打印json对象。最终输出似乎是

{ declaration: { attributes: { version: '1.0', encoding: 'utf-8' } },
  root: 
   { name: 'order',
     attributes: 
      { orderid: '123456',
        xmlns: 'http://www.someRandomNameSpace.com' },
     children: 
      [ { name: 'orderperson',
          attributes: {},
          children: [],
          content: 'str1234' },
        { name: 'shipto',
          attributes: {},
          children: 
           [ { name: 'name',
               attributes: {},
               children: [],
               content: 'Adnan Ali' },

我想从这个对象中读取'name'='Adnan Ali'，如何在nodejs中完成？我的意思是我怎样才能找到名称='name'和content ='Adnan Ali'的对象。

打印命令是 console.log(inspect(order, {colors: true, depth: Infinity}));

Answer 1

您应该能够使用此路径content: 'Adnan Ali' data.root.children[1].children[0]与const data = { declaration: { attributes: { version: '1.0', encoding: 'utf-8' } }, root: { name: 'order', attributes: { orderid: '123456', xmlns: 'http://www.someRandomNameSpace.com' }, children: [{ name: 'orderperson', attributes: {}, children: [], content: 'str1234' }, { name: 'shipto', attributes: {}, children: [{ name: 'name', attributes: {}, children: [], content: 'Adnan Ali' }] }] } }; console.log(data.root.children[1].children[0])对象：

data

说明：

root是一个包含root对象的对象。 children是一个包含root.children数组的对象。 1（索引children）中的第二个元素是一个对象，它包含另一个0数组，其中包含您在第一个索引（chgrp www-data file.php chmod 770 file.php）处查找的对象。

Answer 2

由于您正在使用NodeJS，或许尝试使用JSONPath是一个好主意。然后你可以做这样的事情：

var jp = require("JSONPath");
var tobj = { "declaration": { "attributes": { "version": '1.0', "encoding": 'utf-8' } },
  "root": 
   { "name": 'order',
     "attributes": 
      { "orderid": '123456',
        "xmlns": 'http://www.someRandomNameSpace.com' },
     "children": 
      [ { "name": 'orderperson',
          "attributes": {},
          "children": [],
          "content": 'str1234' },
        { "name": 'shipto',
          "attributes": {},
          "children": 
           [ { "name": 'name',
               "attributes": {},
               "children": [],
               "content": 'Adnan Ali' 
               }
           ]
        }
    ]
}};
var result = jp.eval(tobj, "$..children[?(@.name === 'name' && @.content === 'Adnan Ali')]");
console.log(result);

示例输出：

[ { name: 'name',
    attributes: {},
    children: [],
    content: 'Adnan Ali' } ]

（别忘了安装JSONPath; - ））

来源：
https://www.npmjs.com/package/JSONPath
http://goessner.net/articles/JsonPath/

Answer 3

您需要在对象数组中搜索您感兴趣的对象。有多种方法可以做到这一点，包括Array.prototype.find（不确定它是否在所有Node.js版本中都可用）和{{ 3}}

使用Array.prototype.filter解决方案看起来像这样（未经测试）：

function findObject(array, key, value) {
    var filtered = array.filter(obj => (obj[key] === value));
    if (filtered.length !== 1) throw new Error('Found ' + filtered.length + ' objects with `' + key + '`=`' + value + '`, expected to find 1.');
    return filtered[0];
}

var shipto = findObject(input.root.children, 'name', 'shipto');
var name = findObject(shipto.children, 'name', 'name').content;

console.log(name);

Answer 4

请考虑使用object-scan。将头包裹起来，它会非常强大。

const objectScan = require('object-scan');

const find = (input) => objectScan(['**'], {
  abort: true,
  rtn: 'value',
  filterFn: ({ value }) => value.content === 'Adnan Ali' && value.name === 'name'
})(input);

const tobj = {"declaration":{"attributes":{"version":"1.0","encoding":"utf-8"}},"root":{"name":"order","attributes":{"orderid":"123456","xmlns":"http://www.someRandomNameSpace.com"},"children":[{"name":"orderperson","attributes":{},"children":[],"content":"str1234"},{"name":"shipto","attributes":{},"children":[{"name":"name","attributes":{},"children":[],"content":"Adnan Ali"}]}]}};

console.log(find(tobj));
// => { name: 'name', attributes: {}, children: [], content: 'Adnan Ali' }

从json对象中提取子对象

4 个答案: