我的json数据如下:
[
{
"id": "i_1",
"name": "abc",
"address": [
{
"city": [
"city1",
"city2"
]
},
{
"city": [
"city1",
"city2"
]
}
]
},
{
"id": "i_2",
"name": "def",
"address": [
{
"city": []
},
{
"city": []
}
]
}
]
现在,我只希望city
数组不是null
的那些数据。因此,在上面的示例中,输出应为第一个元素,即ID为i_1
。
如何使用jmespath库过滤此json
?
答案 0 :(得分:1)
您可以这样做:
var arr = [
{
"id": "i_1",
"name": "abc",
"address": [
{
"city": [
"city1",
"city2"
]
},
{
"city": [
"city1",
"city2"
]
}
]
},
{
"id": "i_2",
"name": "def",
"address": [
{
"city": []
},
{
"city": []
}
]
}
];
console.log(jmespath.search(arr,"[?not_null(address[].city[])]"));
<script src="https://cdnjs.cloudflare.com/ajax/libs/jmespath/0.15.0/jmespath.js"></script>
答案 1 :(得分:0)
您可以使用filter
和every
在纯JavaScript中完成此操作
const items=[{"id":"i_1","name":"abc","address":[{"city":["city1","city2"]},{"city":["city1","city2"]}]},{"id":"i_2","name":"def","address":[{"city":[]},{"city":[]}]}]
const filtered = items.filter(i => i.address.every(a => a.city && a.city.length > 0))
console.log(filtered)
仅当address
中的每个对象具有非空的城市数组时,此函数才返回。
答案 2 :(得分:0)
您不需要使用 jmespath库,也可以使用filter
和`香草JS中的所有内容。效率更高。
let jsonTxt = '{"data":[{"id":"i_1","name":"abc","address":[{"city":["city1","city2"]},{"city":["city1","city2"]}]},{"id":"i_2","name":"def","address":[{"city":[]},{"city":[]}]}]}'
let jsonData = JSON.parse(jsonTxt);
let items = jsonData.data;
const result = items.filter(i => i.address.every(a => a.city && a.city.length))
console.log('id: ', result[0].id);
//using jmespath
console.log(jmespath.search({data: items}, "data[*].address[*].city"));
<script src="https://cdnjs.cloudflare.com/ajax/libs/jmespath/0.15.0/jmespath.js"></script>
答案 3 :(得分:0)
var arr = [
{
"id": "i_1",
"name": "abc",
"address": [
{
"city": [
"city1",
"city2"
]
},
{
"city": [
"city1",
"city2"
]
}
]
},
{
"id": "i_2",
"name": "def",
"address": [
{
"city": []
},
{
"city": []
}
]
}
];
console.log(jmespath.search({ c: arr}, "not_null(c[].address[].city[])"));
<script src="https://cdnjs.cloudflare.com/ajax/libs/jmespath/0.15.0/jmespath.js"></script>
我不知道你的结果如何。你能解释得更好吗?