【发布时间】:2021-07-24 18:33:57
【问题描述】:
在我的index.js 内部,我有一个这样结束的函数:
return res.send(response).then(saveProductID(uid, product.id));
在 Firebase 控制台中显示:
TypeError: res.send(...).then 不是函数
完整代码示例:
exports.createConnectAccount = functions.https.onRequest(async (req, res) => {
var data = req.body
console.log(data, "<-- clark jable")
var uid = data.userID
console.log(uid, "<-- this is the uid")
var email = data.email
var response = {}
strip.accounts.create({
type: 'express',
country: 'US',
requested_capabilities: [
'transfers',
],
business_type: 'individual',
},
(err, account) => {
if (err) {
console.log("Couldn't create stripe account: " + err)
return res.send(err)
}
// createStripe_customers(uid, customer, intent)
console.log("ACCOUNT: " + account.id)
response.body = {
success: account.id
}
//createStripe_customers()
return res.send(response).then(createStripe_Accounts(uid, account));
}
);
});
function createStripe_Accounts(uid, account) {
console.log(uid, " did the createStripe_Accounts Run? ", account.id)
const userRef = admin.database().ref('Stripe_Accounts').child(uid) //.child(uid)
return userRef.set({
account_id: account.id,
});
}
.then() 以前(并将继续)用于许多其他功能。 那么为什么createConnectAccount 会出现这个错误?
【问题讨论】:
标签: javascript ios firebase google-cloud-functions