打字稿：type＆＃x27; Promise＆lt; void＆gt;＆＃x27;不能分配给类型＆＃x27; void |破坏者＆＃x27;

发布于 2025-01-28 12:56:55 字数 880 浏览 4 评论 0原文

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类型'Promise'不能分配给类型'void | destructor'。

对于CheckUserLoggedIn（），请在使用效果中调用。

我可以通过执行const checkuserloggedin来摆脱它：但是

，这也许不是最理想的解决方法...

如果我做const checkuserloggedin：Promise：Promise 然后，我得到不同的错误：

此表达不可呼应。输入“承诺”没有电话签名。

这不是我想要的...我希望它可以呼叫...我正在翻译/转换我的JavaScript文件为Typescript .....

useEffect(() => checkUserLoggedIn(), [])

    // Check if user is logged in
    const checkUserLoggedIn = async () => {
        console.log('checkUserLoggedIn')

        const res = await fetch(`${NEXT_URL}/api/user`)
        const data = await res.json()

        if (res.ok) {
            setUser(data.user)

            console.log('data.user', data.user)
            router.push('/account/dashboard')
        } else {
            setUser(null)
        }
    }

原文

The editor is showing

Type 'Promise' is not assignable to type 'void | Destructor'.

for the checkUserLoggedIn() call in useEffect.

I can get rid of it by doing const checkUserLoggedIn:any

However maybe that's not the most ideal way of solving this...

If I do const checkUserLoggedIn:Promise
I then get a different error:

This expression is not callable. Type 'Promise' has no call
signatures.

which is not what I want...I want it to be callable...I'm translating/converting my javascript file to typescript.....

useEffect(() => checkUserLoggedIn(), [])

    // Check if user is logged in
    const checkUserLoggedIn = async () => {
        console.log('checkUserLoggedIn')

        const res = await fetch(`${NEXT_URL}/api/user`)
        const data = await res.json()

        if (res.ok) {
            setUser(data.user)

            console.log('data.user', data.user)
            router.push('/account/dashboard')
        } else {
            setUser(null)
        }
    }

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ヤ经典坏疍 2025-02-04 12:56:55

箭头功能以您描述的方式写入的方式，隐式从函数返回值。例如，此功能（）=＆gt; 4返回值4。相反，当您将其包裹在块中时，必须明确定义返回值：（）=＆gt; {4; }此功能不会返回任何内容。

在您的情况下，您正在传递一个匿名箭头功能，该功能隐含地返回异步函数的结果，这是一个承诺。

要解决此问题，请将您传递给usefect的函数的内容包装在一个块中：

useEffect(() => {
  checkUserLoggedIn();
}, []);

Arrow functions written in the way you've described implicitly return the value from the function. For example, this function () => 4 returns the value 4. Conversely, when you wrap this in a block, you must explicitly define the return value: () => { 4; } this function does not return anything.

In your case, you are passing an anonymous arrow function that implicitly returns the result of an asynchronous function, which is a promise.

To fix this, wrap the content of the function you pass to useEffect in a block:

useEffect(() => {
  checkUserLoggedIn();
}, []);

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不念旧人 2025-02-04 12:56:55

    useEffect(() => {
        checkUserLoggedIn()
    }, [])

    // Check if user is logged in
    const checkUserLoggedIn:() => Promise<void> = async () => {
        console.log('checkUserLoggedIn')

        const res = await fetch(`${NEXT_URL}/api/user`)
        const data = await res.json()

        if (res.ok) {
            setUser(data.user)

            console.log('data.user', data.user)
            router.push('/account/dashboard')
        } else {
            setUser(null)
        }
    }

    useEffect(() => {
        checkUserLoggedIn()
    }, [])

    // Check if user is logged in
    const checkUserLoggedIn:() => Promise<void> = async () => {
        console.log('checkUserLoggedIn')

        const res = await fetch(`${NEXT_URL}/api/user`)
        const data = await res.json()

        if (res.ok) {
            setUser(data.user)

            console.log('data.user', data.user)
            router.push('/account/dashboard')
        } else {
            setUser(null)
        }
    }

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