我有以下代码:
const Test = obj && obj.length > 0 && obj.map((myguids2, key) => {
const url2 = myguids2.predicatList.map(mypredicats => {
const url = mypredicats.p
var arr = url.split('/')[3]
return (
<div>{arr}</div>
)
})
const content = myguids2.guidList.map((myqueriedguids, key) => {
return (
<tr>
<td>{myqueriedguids.guid.includes('http://linkedbuildingdata.net/ifc/resources20201208_005325/') ? <MyPopUp myproperties={guidList} myguidnames={myqueriedguids.guid}>{myqueriedguids.guid}</MyPopUp> : myqueriedguids.guid}</td>
</tr>
)
})
return content
})
我想把url2的输出和来自内容的输出组合成一个两列表,url2在左表列,内容在右表中
有没有办法以这种方式构建函数?
只返回一个对象
const Test = obj && obj.length > 0 && obj.map((myguids2, key) => {
const url2 = myguids2.predicatList.map(mypredicats => {
const url = mypredicats.p
var arr = url.split('/')[3]
return (
<div>{arr}</div>
)
})
const content = myguids2.guidList.map((myqueriedguids, key) => {
return (
<tr>
<td>{myqueriedguids.guid.includes('http://linkedbuildingdata.net/ifc/resources20201208_005325/') ? <MyPopUp myproperties={guidList} myguidnames={myqueriedguids.guid}>{myqueriedguids.guid}</MyPopUp> : myqueriedguids.guid}</td>
</tr>
)
})
return { content, url2 }
})
然后您可以通过访问属性
Test.content
或Test.url2
因此,目前最好的方法是将Const测试拆分为两个变量,一个包含url2,另一个包含内容。使用返回语句,我设法放入{url2}:{Test}以获得两列输出