Javascript JSX中.map循环中的If-else语句

Javascript JSX中.map循环中的If-else语句,javascript,html,reactjs,jsx,Javascript,Html,Reactjs,Jsx,Instagram克隆: 我想检查我是否跟随我的跟随者然后显示unfollow,或者如果不显示follow var ifollowandhfollowme; myfollowers.map(follower=>{ 返回( {follower.userName} {IfollowAndHeFollowMe= myfollowers.filter((follower)=>follower.userName==follower.userName) } //这不管用 {ifollowandheflowm

Instagram克隆:

我想检查我是否跟随我的跟随者然后显示
unfollow
,或者如果不显示
follow

var ifollowandhfollowme;
myfollowers.map(follower=>{
返回(
{follower.userName}
{IfollowAndHeFollowMe=
myfollowers.filter((follower)=>follower.userName==follower.userName)
}
//这不管用
{ifollowandheflowme.length>0?返回(unfollow):返回(follow)}
//这也不起作用
{返回ifollowandheflowme.length>0?取消follow:follow}
)
})

//https://instagram-app-clone.netlify.app/ ---仅针对电话--

内部的JSX代码{}应作为语句编写

const followers=['John','Hanna'];
函数RenderMap(){
返回(
{followers.map(follower=>(
{跟随者}
{followers.length>0(
取消关注
) : (
跟随
)}
))}
);
}

//现在没有错误,但它的所有显示按钮都带有以下文本

{Data[2][0].followers.map(follower => {
                idFollowers++;

                return (

                    <div className="follower" id={idFollowers}  >
                        <img src={follower.profilna} className="profilnaFollower" alt="" />

                        <div className="userInfoFollower">
                            <h3>{follower.userName}</h3>
                            <p>{follower.name}</p>
                        </div>

                        {dbFollow = Object.entries(Data[2][0].following).filter(following => following.userName == follower.userName)}
                        {dbFollow.length > 0 ? (<button>unfoldasdaslow</button>) : (<button>fodsadsllow</button>)}
                    </div>
                )
            })}
        </div>
{Data[2][0]。followers.map(follower=>{
idFollowers++;
返回(
{follower.userName}
{follower.name}

{dbFollow=Object.entries(数据[2][0]。following.filter(following=>following.userName==follower.userName)} {dbFollow.length>0?(展开asdaslow):(fodsadslow)} ) })}
不返回??是的,Javascript三元运算符就是这样工作的。未捕获错误:对象作为React子对象无效(找到:具有键{userName,name,profImg,posts}的对象)。如果你想呈现一组子对象,请改用数组。{IfollowAndHeFollowMe=MyFollowers.filter(follower=>follower.userName==follower.userName)}这很糟糕,你能帮个忙吗?@Eugene…….看起来不错,有什么错误吗?它的dbFollower不能是object,因为Follower1是object。必须是数组或字符串,它将使用.filter()设置follow1,但我希望dbFollow是follow1.userName非objectobject对象作为React子对象无效(找到:具有键{userName,name,profilna,posts}的对象)。如果你想呈现一组孩子,用数组代替。你是最好的兄弟,谢谢!!!!!!!!!!
{Data[2][0].followers.map(follower => {
                idFollowers++;

                return (

                    <div className="follower" id={idFollowers}  >
                        <img src={follower.profilna} className="profilnaFollower" alt="" />

                        <div className="userInfoFollower">
                            <h3>{follower.userName}</h3>
                            <p>{follower.name}</p>
                        </div>

                        {dbFollow = Object.entries(Data[2][0].following).filter(following => following.userName == follower.userName)}
                        {dbFollow.length > 0 ? (<button>unfoldasdaslow</button>) : (<button>fodsadsllow</button>)}
                    </div>
                )
            })}
        </div>