如何在JavaScript或jQuery中过滤JSON数据？

Question

如何使用Javascript或jQuery过滤JSON数据？

这是我的JSON数据:

[{"name":"Lenovo Thinkpad 41A4298","website":"google"},
{"name":"Lenovo Thinkpad 41A2222","website":"google"},
{"name":"Lenovo Thinkpad 41Awww33","website":"yahoo"},
{"name":"Lenovo Thinkpad 41A424448","website":"google"},
{"name":"Lenovo Thinkpad 41A429rr8","website":"ebay"},
{"name":"Lenovo Thinkpad 41A429ff8","website":"ebay"},
{"name":"Lenovo Thinkpad 41A429ss8","website":"rediff"},
{"name":"Lenovo Thinkpad 41A429sg8","website":"yahoo"}]

JavaScript的:

obj1 = JSON.parse(jsondata);

现在我只想要包含网站的名称和网站数据等于"雅虎"

Answer 1

这是你应该这样做的:(谷歌查找)

$([
  {"name":"Lenovo Thinkpad 41A4298","website":"google222"},
  {"name":"Lenovo Thinkpad 41A2222","website":"google"}
  ])
    .filter(function (i,n){
        return n.website==='google';
    });

更好的解决方案:(萨尔曼)

$.grep( [{"name":"Lenovo Thinkpad 41A4298","website":"google"},{"name":"Lenovo Thinkpad 41A2222","website":"google"}], function( n, i ) {
  return n.website==='google';
});

http://jsbin.com/yakubixi/4/edit

Answer 2

除非您针对旧浏览器并且不想使用填充程序,否则不需要jQuery.

var yahooOnly = JSON.parse(jsondata).filter(function (entry) {
    return entry.website === 'yahoo';
});

在ES2015中:

const yahooOnly = JSON.parse(jsondata).filter(({website}) => website === 'yahoo');

Answer 3

试试这种方式,让您甚至可以通过其他键进行过滤

数据:

var my_data = [{"name":"Lenovo Thinkpad 41A4298","website":"google"},
{"name":"Lenovo Thinkpad 41A2222","website":"google"},
{"name":"Lenovo Thinkpad 41Awww33","website":"yahoo"},
{"name":"Lenovo Thinkpad 41A424448","website":"google"},
{"name":"Lenovo Thinkpad 41A429rr8","website":"ebay"},
{"name":"Lenovo Thinkpad 41A429ff8","website":"ebay"},
{"name":"Lenovo Thinkpad 41A429ss8","website":"rediff"},
{"name":"Lenovo Thinkpad 41A429sg8","website":"yahoo"}];

用法:

//We do that to ensure to get a correct JSON
var my_json = JSON.stringify(my_data)
//We can use {'name': 'Lenovo Thinkpad 41A429ff8'} as criteria too
var filtered_json = find_in_object(JSON.parse(my_json), {website: 'yahoo'});

过滤功能

function find_in_object(my_object, my_criteria){

  return my_object.filter(function(obj) {
    return Object.keys(my_criteria).every(function(c) {
      return obj[c] == my_criteria[c];
    });
  });

}

Answer 4

以下代码对我有用：

var data = [{"name":"Lenovo Thinkpad 41A4298","website":"google"},
{"name":"Lenovo Thinkpad 41A2222","website":"google"},
{"name":"Lenovo Thinkpad 41Awww33","website":"yahoo"},
{"name":"Lenovo Thinkpad 41A424448","website":"google"},
{"name":"Lenovo Thinkpad 41A429rr8","website":"ebay"},
{"name":"Lenovo Thinkpad 41A429ff8","website":"ebay"},
{"name":"Lenovo Thinkpad 41A429ss8","website":"rediff"},
{"name":"Lenovo Thinkpad 41A429sg8","website":"yahoo"}]

var data_filter = data.filter( element => element.website =="yahoo")
console.log(data_filter)