Javascript 按键值过滤 JSON
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Filter JSON by key value
提问by JordanBarber
I have the following JSON and I want to keep the array in the same format but only include objects with type "ar". How can I filter this JSON with JavaScript to only include the types I need?
我有以下 JSON，我想以相同的格式保留数组，但只包含类型为 的对象"ar"。如何使用 JavaScript 过滤此 JSON 以仅包含我需要的类型？
[{"time":"2016-07-26 09:02:27","type":"aa"},
{"time":"2016-04-21 20:35:07","type":"ae"},
{"time":"2016-08-20 03:31:57","type":"ar"},
{"time":"2017-01-19 22:58:06","type":"ae"},
{"time":"2016-08-28 10:19:27","type":"ae"},
{"time":"2016-12-06 10:36:22","type":"ar"},
{"time":"2016-07-09 12:14:03","type":"ar"},
{"time":"2016-10-25 05:05:37","type":"ae"},
{"time":"2016-06-05 07:57:18","type":"ae"},
{"time":"2016-10-08 22:03:03","type":"aa"},
{"time":"2016-08-13 21:27:37","type":"ae"},
{"time":"2016-04-09 07:36:16","type":"ar"},
{"time":"2016-12-30 17:20:08","type":"aa"},
{"time":"2016-03-11 17:31:46","type":"aa"},
{"time":"2016-05-04 14:08:25","type":"ar"},
{"time":"2016-11-29 05:21:02","type":"ar"},
{"time":"2016-03-08 05:46:01","type":"ar"},
]
回答by Mihai Alexandru-Ionut
You should use filtermethod.
你应该使用filter方法。
The filter()method creates a new arraywith all elements that pass the test implemented by the provided function.
该filter()方法array使用通过提供的函数实现的测试的所有元素创建一个新元素。
Providedfunction is a callbackwhich is applied to each element of the array.
提供的函数是一个callback应用于数组的每个元素。
var arr=[{"time":"2016-07-26 09:02:27","type":"aa"}, {"time":"2016-04-21 20:35:07","type":"ae"}, {"time":"2016-08-20 03:31:57","type":"ar"}, {"time":"2017-01-19 22:58:06","type":"ae"}, {"time":"2016-08-28 10:19:27","type":"ae"}, {"time":"2016-12-06 10:36:22","type":"ar"}, {"time":"2016-07-09 12:14:03","type":"ar"}, {"time":"2016-10-25 05:05:37","type":"ae"}, {"time":"2016-06-05 07:57:18","type":"ae"}, {"time":"2016-10-08 22:03:03","type":"aa"}, {"time":"2016-08-13 21:27:37","type":"ae"}, {"time":"2016-04-09 07:36:16","type":"ar"}, {"time":"2016-12-30 17:20:08","type":"aa"}, {"time":"2016-03-11 17:31:46","type":"aa"}, {"time":"2016-05-04 14:08:25","type":"ar"}, {"time":"2016-11-29 05:21:02","type":"ar"}, {"time":"2016-03-08 05:46:01","type":"ar"}, ];
console.log(arr.filter(function(item){
return item.type == "ar";
}));Also, you can use a shorterway with arrowfunctions:
此外，您可以使用shorter带有arrow函数的方法：
var filtered = arr.filter(a=>a.type=="ar");
回答by Zakaria Acharki
You could use filterfunction. Check browser supportfirst:
你可以使用filter函数。首先检查浏览器支持：
var result = a.filter(function(v){
return v.type==='ar';
})
Hope this helps.
希望这可以帮助。
var list = [
{"time":"2016-07-26 09:02:27","type":"aa"},
{"time":"2016-04-21 20:35:07","type":"ae"},
{"time":"2016-08-20 03:31:57","type":"ar"},
{"time":"2017-01-19 22:58:06","type":"ae"},
{"time":"2016-08-28 10:19:27","type":"ae"},
{"time":"2016-12-06 10:36:22","type":"ar"},
{"time":"2016-07-09 12:14:03","type":"ar"},
{"time":"2016-10-25 05:05:37","type":"ae"},
{"time":"2016-06-05 07:57:18","type":"ae"},
{"time":"2016-10-08 22:03:03","type":"aa"},
{"time":"2016-08-13 21:27:37","type":"ae"},
{"time":"2016-04-09 07:36:16","type":"ar"},
{"time":"2016-12-30 17:20:08","type":"aa"},
{"time":"2016-03-11 17:31:46","type":"aa"},
{"time":"2016-05-04 14:08:25","type":"ar"},
{"time":"2016-11-29 05:21:02","type":"ar"},
{"time":"2016-03-08 05:4601","type":"ar"},
];
var result = list.filter(function(obj, index){
return obj.type==='ar';
})
console.log(result);回答by Abhinav Galodha
Using filtermethod you can filter the array to return only those elements which match a particular condition data.filter((x)=>x.type === "ar");
使用filter方法可以过滤数组以仅返回匹配特定条件的那些元素data.filter((x)=>x.type === "ar");
The filtermethod creates a new array with all elements that will pass the condition x.type === "ar"
该filter方法创建一个新数组，其中包含将通过条件的所有元素 x.type === "ar"
var data =[{"time":"2016-07-26 09:02:27","type":"aa"},
{"time":"2016-04-21 20:35:07","type":"ae"},
{"time":"2016-08-20 03:31:57","type":"ar"},
{"time":"2017-01-19 22:58:06","type":"ae"},
{"time":"2016-08-28 10:19:27","type":"ae"},
{"time":"2016-12-06 10:36:22","type":"ar"},
{"time":"2016-07-09 12:14:03","type":"ar"},
{"time":"2016-10-25 05:05:37","type":"ae"},
{"time":"2016-06-05 07:57:18","type":"ae"},
{"time":"2016-10-08 22:03:03","type":"aa"},
{"time":"2016-08-13 21:27:37","type":"ae"},
{"time":"2016-04-09 07:36:16","type":"ar"},
{"time":"2016-12-30 17:20:08","type":"aa"},
{"time":"2016-03-11 17:31:46","type":"aa"},
{"time":"2016-05-04 14:08:25","type":"ar"},
{"time":"2016-11-29 05:21:02","type":"ar"},
{"time":"2016-03-08 05:46:01","type":"ar"},
];
var result = data.filter((x)=>x.type === "ar");
console.log(result);回答by user2085143
Not as clean as the answer above, but will still do the job
不像上面的答案那么干净，但仍然可以完成这项工作
var list = [{"time":"2016-07-26 09:02:27","type":"aa"},
{"time":"2016-04-21 20:35:07","type":"ae"},
{"time":"2016-08-20 03:31:57","type":"ar"},
{"time":"2017-01-19 22:58:06","type":"ae"},
{"time":"2016-08-28 10:19:27","type":"ae"},
{"time":"2016-12-06 10:36:22","type":"ar"},
{"time":"2016-07-09 12:14:03","type":"ar"},
{"time":"2016-10-25 05:05:37","type":"ae"},
{"time":"2016-06-05 07:57:18","type":"ae"},
{"time":"2016-10-08 22:03:03","type":"aa"},
{"time":"2016-08-13 21:27:37","type":"ae"},
{"time":"2016-04-09 07:36:16","type":"ar"},
{"time":"2016-12-30 17:20:08","type":"aa"},
{"time":"2016-03-11 17:31:46","type":"aa"},
{"time":"2016-05-04 14:08:25","type":"ar"},
{"time":"2016-11-29 05:21:02","type":"ar"},
{"time":"2016-03-08 05:46:01","type":"ar"},
]
var newList = [];
for (var i = 0; i < list.length; i++) {
if (list[i].type === 'ar') {
newList.push(list[i]);
}
}
console.log(newList);
