NodeJS - Promise Return 未正确接收值

Question

问题：我应该从 promises 中得到正确的对象。但是我最后得到的是空对象。 SQL 我正在执行的查询运行完美。唯一的问题是正确的 return 值。

调试：当我尝试在解析之前控制对象值时；正在说“[Promise {}]”。

我得到的输出：

{
    "status": true,
    "data": [
        {},
        {}
    ],
    "errors": null,
    "msg": "Hahaha"
}

代码

export const executeAllQueries = (allQueries) => {
    return allQueries.map ( async productQuery => {
        try {
            let res = await executeProductQuery (productQuery);
            console.log(res); 
            return res;
        } catch (error) {
            console.log(error);
            return error;
        }
    })
}

const executeProductQuery = async (productQuery) => {

    let parentResponse = await new Promise ((resolve, reject) => {
        mysql.query( productQuery.parent.sql , productQuery.parent.data, (parentError, parentResult) => {
            if (parentError)
                reject ( { status: false, parent_sku: productQuery.parent.data[1], error: parentError.message } )
            else {
                resolve( { status: true,  parent_sku: productQuery.parent.data[1], parent_id: parentResult.insertId , error: null } )
            }
        })
    }); 

    if (parentResponse.status === true)
    {
        let variantPromises = productQuery.varients.map (async varientQuery => {
            try {
                return await executeVariantQuery(varientQuery, parentResponse.parent_id)
            } catch (error) {
                console.log(error);
                return error;
            }
        }); 

        return { parent: parentResponse, varient: variantPromises }
    }
}

const executeVariantQuery = async ( varientQuery, parent_id ) => {

    let varientResponse = await new Promise ((resolve, reject) => {
        varientQuery.varient.data[11] = parent_id; 
        mysql.query(varientQuery.varient.sql, varientQuery.varient.data, (varientError, varientResult) => {
            if (varientError)
                reject ( { status: false, error: varientError.message } )
            else {
                resolve( { status: true, varient_id: varientResult.insertId , error: null } )
            }
        })
    })

    if (varientResponse.status === true) {
        let attributePromises = varientQuery.attributes.map(async attribQuery => {
            try {
                return await executeAttributeQuery(attribQuery);
            } catch (error) {
                console.log(error);
                return error; 
            }
        })
        return { varient: varientResponse, attributes: attributePromises }
    }
}

const executeAttributeQuery = async (attribQuery) => {
    let attribResponse = await new Promise((resolve, reject) => {
        mysql.query(attribQuery.sql, attribQuery.data, (attribError, attribResult) => {
            if (attribError)
                reject ( { status: false, error: attribError.message } )
            else {
                resolve( { status: true, varient_id: attribResult.insertId , error: null } )
            }
        })
    })
    return attribResponse; 
}

执行 top 方法 executeAllQueries() 如下：

let completeVariants = isolateVariantProducts(fileRows); // All products are isolated in separate groups. Each group has a parent and it's variants.
let allQueries = generateCompleteQueriesForVariant(completeVariants);
let responses = executeAllQueries(allQueries);

res.status(200).json({status: true, data: responses, errors: null, msg: 'Hahaha'})

可运行的 NodeJS 代码段：

let allQueries = [{parent: 1, varients: [1,2,3,4]}, {parent: 2, varient: [5,6,7,8]}]

const executeAllQueries = (allQueries) => {
    return allQueries.map ( async productQuery => {
        try {
            return await executeProductQuery (productQuery);
        } catch (error) {
            return error;
        }
    })
}

const executeProductQuery = async (productQuery) => {

    let parentResponse = await new Promise ((resolve, reject) => {
        if (true)
            resolve({ status: true, parent: productQuery.parent, msg: "Resolved"})
        else
            reject({status: false, parent: productQuery.parent, msg: "Rejected"})
    }); 

    if (parentResponse.status === true)
    {
        let variantPromises = productQuery.varients.map (async varientQuery => {
            try {
                return await executeVariantQuery(varientQuery)
            } catch (error) {
                return error;
            }
        }); 

        return { parent: parentResponse, varient: variantPromises }
    }
}

const executeVariantQuery = async ( varientQuery ) => {

    let varientResponse = await new Promise ((resolve, reject) => {
        if (true)
            resolve ( { status: true, msg: "Resolved" } )
        else {
            reject ( { status: false, msg: "Rejected" } )
        }
    });
    return varientResponse; 
}

console.log(executeAllQueries(allQueries));

Answer 1

executeAllQueries returns 承诺数组。您应该 Promise.all() 才能得到结果。

// Instead of
let responses = executeAllQueries(allQueries);

// do

let responses = await Promise.all(executeAllQueries(allQueries));

NodeJS - Promise Return 未正确接收值

NodeJS - Promise Return Values are not received properly

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