javascript - 等待 Promise.all 中的 array.map 迭代

Question

如果客户不存在，我有以下代码应该为客户添加项目。执行应该是并行的。

await Promise.all(
    customers.map(async (customer) => {
        return customer.items.map(async (item) => {
            return new Promise(async (resolve) => {
                const productExists = someArray.some(
                    (arrayValue) => arrayValue === item.id
                );
                if (!productExists) {
                    logger.info(
                    `customer item ${item.id} does not exist, creating...`
                    );
                    await createCustomerItem(item.id);
                    logger.info(`customer item ${item.id} created.`);

                    someArray.push(item.id);
                } else {
                    logger.info(`customer item ${item.id} already exists, skipping...`);
                }
                resolve(true);
            });
        });
    })

);

logger.info(`All items should now be present`);

问题是在以下情况下执行不等待createCustomerItem解决!productExists)

这是日志

customer item 32310 does not exist, creating...
customer item ao does not exist, creating...
customer item ute does not exist, creating...
All items should not be present
customer item ao created.
customer item ute created.
customer item 32310 created.

自然All items should not be present应该排在最后。

当所有项目都已经存在时，该过程看起来不错。

标签： javascriptnode.jsasynchronouspromise

Answer 1

你可以做这样的事情

const fruitsToGet = ['apple', 'grape', 'pear']

const mapLoop = async () => {
  console.log('Start')

  const promises = await fruitsToGet.map(async fruit => {
    const numFruit = new Promise((resolve, reject) => {
      setTimeout(() => resolve(fruit), 1000)
    });
    return numFruit
  })
  const numFruits = await Promise.all(promises)
  console.log(numFruits)

  console.log('End')
}

mapLoop();

结果

Start
["apple", "grape", "pear"]
End

源码 演示