JS Fiddle:
https://jsfiddle.net/eywraw8t/529272/
我有两个数组:
(1)价格表
[
{
"n_bubble": "2",
"size_1": "40.00",
"size_2": "72.00",
"size_3": "112.00"
},
{
"n_bubble": "5",
"size_1": "65.00",
"size_2": "98.00",
"size_3": "144.00"
},
{
"n_bubble": "10",
"size_1": "90.00",
"size_2": "138.00",
"size_3": "183.00"
}
...
]
〔2〕部分
[
{
"id": 1,
"bubble_size": "1",
"n_bubble": "0",
"price": "0",
},
{
"id": 2,
"bubble_size": "2",
"n_bubble": "7",
"price": "0",
},
{
"id": 3,
"bubble_size": "1",
"n_bubble": "0",
"prezzo": "0",
},
{
"id": 4,
"bubble_size": "1",
"n_bubble": "0",
"prezzo": "0",
}
]
我正在循环浏览“部件”,基于连接到
bubble_size
我必须选择
n_bubble
从
price list
数组并选择匹配大小:
<div v-for="(part, index) in parts" :key="part.id">
<label>Bubble size</label><br>
<label>1
<input type="radio" v-model="part.bubble_size" :checked="part.bubble_size == 1" value="1" :name="'bubble_size['+index+']'" /></label>
<label>2 <input type="radio" v-model="part.bubble_size" :checked="part.bubble_size == 2" value="2" :name="'bubble_size['+index+']'" /></label>
<label>3 <input type="radio" v-model="part.bubble_size" :checked="part.bubble_size == 3" value="3" :name="'bubble_size['+index+']'" /></label><br>
<label>Bubble number <input type="number" min="0" v-model.number="part.n_bubble" /></label><br>
<label for="">Price <input type="number" min="0" v-model.number="price_list[0]['size_'+part.bubble_size]" /></label>
<br><br>
</div>
我正在努力解决这个问题:
<input v-model.number="price_list[0]['size_'+part.bubble_size]" type="number" />
我暂时放了一个
0
索引,因为我不知道如何根据
pricelist.n_bubble
. 我需要选择最近的
less than
基于
parts.n_bubble
预期结果:
如果我选择收音机
bubble_size => 2
和
n_bubble => 7
,
price
应该设置
98
因为它应该接收第二个数组
价格表
,因为
7
小于
n_bubble 10
:
{
"n_bubble": "5",
"size_1": "65.00",
"size_2": "98.00",
"size_3": "144.00"
},