在 Supabase 重新分配查询变量时出现打字稿错误

Question

我正在使用react和supabase。我编写了以下代码。我的工作是更新或创建。

export const createEditCabin = async (newCabin: createCabinType, id?: number) => {
    let imagePath = "";
    let imageName = "";
    if (typeof newCabin.image === "object") {
        imageName = `${Math.floor(Math.random()) * 10}-${newCabin.image.name}`.replaceAll("/", "");
        imagePath = `${supabaseUrl}/storage/v1/object/public/cabin-images/${imageName}`;
    } else {
        imagePath = newCabin.image
    }

    let query =supabase.from("Cabins");


    // Create 
    if (!id) {
        query = supabase.from("Cabins").insert([
            { ...newCabin, image: imagePath }
        ]);
    }
    // Edit
    if (id) {
        query = supabase.from("Cabins")
            .update({ ...newCabin, image: imagePath })
            .eq('id', id);
    }

    const { data, error } = await query.select().single();
    return data;

调用

createEditCabin

函数时，查询出现以下打字稿错误：

类型“PostgrestFilterBuilder”缺少类型“PostgrestQueryBuilder”中的以下属性：插入、更新插入、更新、删除(2739)

如何解决？

Answer 1

这是因为您正在重用 query 变量来选择和改变数据。您可以尝试分离逻辑，例如：

export const createEditCabin = async (newCabin: createCabinType, id?: number) => {
    let imagePath = "";
    let imageName = "";
    if (typeof newCabin.image === "object") {
        imageName = `${Math.floor(Math.random() * 10)}-${newCabin.image.name}`.replaceAll("/", "");
        imagePath = `${supabaseUrl}/storage/v1/object/public/cabin-images/${imageName}`;
    } else {
        imagePath = newCabin.image;
    }

    if (!id) {
        const { data, error } = await supabase
            .from("Cabins")
            .insert([{ ...newCabin, image: imagePath }])
            .select()
            .single();

        if (error) throw error;
        return data;
    }

    if (id) {
        const { data, error } = await supabase
            .from("Cabins")
            .update({ ...newCabin, image: imagePath })
            .eq("id", id)
            .select()
            .single();

        if (error) throw error;
        return data;
    }

    return null;
};