传递回调函数的问题，写错props.me is not a function，我只能展示部分代码

Question

有一个parent和一个child（组件）我想获取从child到parent的值，我把函数扔了自己进入组件，在 child 我写了 return 目标值来自 input

parent

const [m, setM] = useState("")
    const me = (m) => {
        setM(m)
        console.log(m)
    }

    const [dey, setD] = useState("")
    const de = (dey) => {
        setD(dey)
    }

    const [y, setY] = useState("")
    const ye = (y) => {
        setY(y)
    }

其companent connect in jsx

<CustomSelect
    className="flex-1"
    btnClass="inp"
    checkedOpt={dey}
    options={d}
    de={de}
/>
<CustomSelect
    className="flex-1 ms-2 ms-xxl-4"
    btnClass="inp"
    checkedOpt={m}
    options={month}
    me={me}
/>
<CustomSelect
    className="flex-1 ms-2 ms-xxl-4"
    btnClass="inp"
    checkedOpt={y}
    options={years}
    ye={ye}
/>

child

 const handleChange = (e) => {
        props.de(e.target.value);
        props.me(e.target.value);
        props.ye(e.target.value);
    }

jsx

<input
    type="radio"
    name="type"
    value={item}
    checked={(item === props.checkedOpt)}
    onChange={handleChange}
/>

Answer 1

由于 props.me 并未传递给每个组件，您只需检查它是否作为函数传递：

 const handleChange = (e) => {
        props.de?.(e.target.value);
        props.me?.(e.target.value);
        props.ye?.(e.target.value);
    }

传递回调函数的问题，写错props.me is not a function，我只能展示部分代码

The problem with passing callback functions, writes an error props.me is not a function, I can only show part of the code

javascript

reactjs

react-native