我不能只顯示："Mona", "Paul","Bastian"，我可以顯示：
"Mona", "Paul","Bastian","Average"但我不想顯示“平均”。并在我的第二個集團中顯示“平均” radars [{...}, { name: 'Average',dataKey: 'Average'}。請問有什么問題嗎？？
const { names, locations } = graphs ?? {}
function RadarNames() {
const data = Object.entries(graphs.names[0]).map(([key, value]) => ({
"name": key,
"age": value,
}))
return creacteChart({
data: data,
polarAngleAxisOptions: { dataKey: 'name' },
radars: [
{
name: 'name',
dataKey: 'age',
},
{
name: 'Average',
dataKey: 'Average',
},
],
})
}
我試著做：
const data = Object.entries(graphs.names[0]!=='Average'? graphs.names[0].map(([key, value]) => ({"name": key,"age": value})):'')
但它不作業...
我的 api 中的 json 是：
{
"names": [
{
"Mona": 63,
"Paul": 29,
"Bastian": 31,
"Average": 75
}
],
"locations": [
{
"location": "Dublin",
"country":"Irland"
}
]
}
uj5u.com熱心網友回復：
使用filter：
const data = Object.entries(graphs.names[0])
.filter(([key]) => key != "Average")
.map(([name, age]) => ({name, age}));
另一種方法是擴展物件以提取平均值（并忽略它）并將其余部分放入新物件中：
const {Average, ...rest} = graphs.names[0];
const data = Object.entries(rest)
.map(([name, age]) => ({name, age}));
uj5u.com熱心網友回復：
const data = Object.entries(graphs.names[0]).filter(([key]) => key != "Average").map(([key, value]) => ({
"name": key,
"age": value,
}))
// then add
const dataAv = Object.entries(graphs.names[0]).filter(([key]) => key == "Average").map(([key, value]) => ({
"Average": key,
"average": value,
}))
return ({
data: data, dataAv,
polarAngleAxisOptions: { dataKey: 'name' },
radars: [
{
name: 'name',
dataKey: 'age',
},
{
name: 'Average',
dataKey: 'average',
},
],
})
轉載請註明出處，本文鏈接：https://www.uj5u.com/caozuo/497525.html
標籤：javascript html 反应 重新图表
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