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：

obj1 = JSON.parse(jsondata);

现在我只想要包含网站的名称和网站数据等于"yahoo"。

Answer 1

这就是你应该怎么做的：(对于谷歌搜索)

$([
  {"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：(萨勒曼)

$.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

不需要jQuery，除非您针对旧的浏览器，并且不想使用shims。

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

以下代码适用于我：

​

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)

​