我正在使用云端防火墙进行电话验证.在firestore db中,我正在存储用户电话号码和uid.这是我尝试过手机号码登录的代码:
@IBAction func signUp(_ sender: Any) {
// dismiss keyboard
view.endEditing(true)
if sendOTP == false {
let mobileNumber = "+91" + phoneNumberTextField.text!
self.Userdefaults.set(mobileNumber, forKey: "mobileNumber")
print("mobileNumber::::\(mobileNumber)")
sendOTPCode()
sendOTP = true
} else {
let codestring = OTPCodeTextField.text
if codestring?.count == 6 {
loginusingOTP(OTPtext: codestring!)
} else {
print("Enter 6 digit code")
}
}
func sendOTPCode() {
let mymobilenumber = Userdefaults.string(forKey: "mobileNumber")
PhoneAuthProvider.provider().verifyPhoneNumber(mymobilenumber!) { (verificationID, error) in
self.Userdefaults.set(verificationID, forKey: "authVerificationID")
if error != nil
{
print ("insde SendCode, there is …
Run Code Online (Sandbox Code Playgroud) 我目前devServer{proxy:{...}}
在vue.config.js中使用它来配置api调用的代理,以避免应用程序中的CORS问题。当我npm run serve
在本地主机上运行时,它工作正常。
现在,我需要将应用程序部署到主机,因此我可以运行npm run build
,更改Ajax调用的url,并且该URL无法运行...因此,我真正需要的是配置代理以进行部署(构建),而不是为devServer进行配置。
正确的方法是什么?
我已经尝试过:
server{proxy:{...}}
和build{proxy:{...}}
,但是在运行时都不允许使用它们npm run build
。
谢谢!
在我的应用程序中,我使用电话身份验证登录并存储用户登录的用户电话号码和数据库中的uid.当使用现有用户电话号码登录时,它会创建具有相同uid和电话号码的新文档ID,您可以查看图像1和3.
这是我用于登录的代码,如果用户存在则意味着它应该检查并导航到主控制器,否则签署用户详细控制器.
@IBAction func loginbtn(_ sender: UIButton) {
let defaults = UserDefaults.standard
let credential: PhoneAuthCredential = PhoneAuthProvider.provider().credential(withVerificationID: defaults.string(forKey: "authVID")!,
verificationCode: otpText.text!)
Auth.auth().signIn(with: credential)
{
(user, error) in
if error != nil
{
print("error: \(String(describing: error?.localizedDescription))")
}
else if user != nil
{
print("Phone number: \(String(describing: user?.phoneNumber))")
let userInfo = user?.providerData[0]
print("Provider ID: \(String(describing: userInfo?.providerID))")
let currentUser = Auth.auth().currentUser?.uid
var ref: DocumentReference? = nil
ref = self.db.collection("Users").addDocument(data: [
"phoneNumber" : user?.phoneNumber as Any,
"UID" : user?.uid as Any
]) { err …
Run Code Online (Sandbox Code Playgroud) ios firebase google-cloud-datastore swift firebase-authentication
在过去的两天里,我一直试图将360度相机,单鱼眼图像转换为节点js中的equirectangular viewer.在stackoverflow中,在伪代码中询问和回答相同的问题.我一直在尝试将伪代码转换为节点js并清除了一些错误.现在项目运行没有错误,但输出图像为空白.
从那个伪,我不知道polar_w,polar_h和geo_w,geo_h,geo和极值,所以,它给出了静态值来显示输出.这是我将伪代码转换为节点js的链接. 如何将球面坐标转换为equirectangular投影坐标?.
这是我尝试将球形图像转换为equirectangular查看器的代码:
exports.sphereImage=(request, response)=>{
var Jimp = require('jimp');
// Photo resolution
var img_w_px = 1280;
var img_h_px = 720;
var polar_w = 1280;
var polar_h = 720;
var geo_w = 1280;
var geo_h = 720;
var img_h_deg = 70;
var img_w_deg = 30;
// Camera field-of-view angles
var img_ha_deg = 70;
var img_va_deg = 40;
// Camera rotation angles
var hcam_deg = 230;
var vcam_deg = 60;
// Camera rotation angles in radians
var …
Run Code Online (Sandbox Code Playgroud) 我有不同的故事板叫做Main,tabBar,home,map等.在tabBar的故事栏中,我使用了SWRevealViewController视图并作为初始视图启动.在主要的Storyboard中,只有两个用于登录和注册控制器.
我的问题是当用户注销并返回时它自动进入主屏幕而不是登录屏幕[此问题是由于SWRevealViewController是初始视图控制器].
我试过这里的代码:在符号视图控制器中
在viewDidAppear中检查firebase中是否有可用的用户
override func viewDidAppear(_ animated: Bool) {
super.viewDidAppear(animated)
if API.User.CURRENT_USER != nil {
// segue to the Tab Bar Controller
self.performSegue(withIdentifier: "signInToTabBar", sender: nil)
}
}
Run Code Online (Sandbox Code Playgroud)
登录动作:
@IBAction func SignInButton(_ sender: Any) {
view.endEditing(true)
guard
let email = emailTextField.text, !email.isEmpty,
let password = passwordTextField.text, !password.isEmpty
else {
self.showErrorAlert(message: "Username or email or passowrd should not be empty")
return
}
// show the progress to the user
ProgressHUD.show("Starting sign-in...", interaction: false)
// use the signIn class …
Run Code Online (Sandbox Code Playgroud) xcode ios swrevealviewcontroller swift firebase-authentication
现在我能够SCNPlane
在检测到卡片时显示不同的内容.显示后SCNPlanes
,用户触摸任何平面以显示新的SCNPlane
.但是现在触摸工作正常但新SCNPlane
的没有显示.
这是我尝试过的代码:
var cake_1_PlaneNode : SCNNode? = nil
func renderer(_ renderer: SCNSceneRenderer, didAdd node: SCNNode, for anchor: ARAnchor) {
guard let imageAnchor = anchor as? ARImageAnchor else { return }
if let imageName = imageAnchor.referenceImage.name {
print(imageName)
if imageName == "menu" {
// Check To See The Detected Size Of Our menu Card (Should By 5cm*3cm)
let menuCardWidth = imageAnchor.referenceImage.physicalSize.width
let menuCardHeight = imageAnchor.referenceImage.physicalSize.height
print(
"""
We Have Detected menu Card With …
Run Code Online (Sandbox Code Playgroud) 我一直试图将车身与车轮一起移动.我创建了增强现实项目,将汽车模型放置在水平面上,汽车由四个按钮控制,即加速,转向,倒车和制动.汽车左,右是控制器通过转向加速和反向.
实际上现在我能够将3D汽车模型放在水平面上,但不知道如何像车身那样向前和向后旋转车轮.
这是我一直在尝试放置3d对象的代码:
override func touchesBegan(_ touches: Set<UITouch>, with event: UIEvent?) {
if let touch = touches.first {
// gives us the location of where we touched on the 2D screen.
let touchLocation = touch.location(in: sceneView)
// hitTest is performed to get the 3D coordinates corresponding to the 2D coordinates that we got from touching the screen.
// That 3d coordinate will only be considered when it is on the existing plane that we detected.
let results = sceneView.hitTest(touchLocation, types: …
Run Code Online (Sandbox Code Playgroud) 在这里,我使用alamofire向后端发送参数值,如username,user toke,post id等.如果状态成功,则通知将从后端发送.在postnotification函数内部我已经尝试使用alamofire和datatask方法发布方法代码, 但它不起作用.在控制台我得到请求超时或没有.
这是我的代码:
func postNotification(postItem: String, post: Post) {
// declare parameter as a dictionary which contains string as key and value combination. considering inputs are valid
print("Get token from post:::",post.token)
print(postItem)
let token = UserDefaults.standard.string(forKey: "token")
//create the url with URL
var parameters = [String:Any]()
parameters["count"] = post.likeCount!
parameters["likedby"] = currentName
parameters["postId"] = postItem
parameters["token"] = post.token!
let Url = String(format: "http://highavenue.co:9000/likesnotification")
guard let serviceUrl = URL(string: Url) else { …
Run Code Online (Sandbox Code Playgroud) 我在vuejs中使用了google登录弹出方法。在实时谷歌登录中,弹出窗口闪烁并且从未登录。但在本地一切正常。
这是 firebase google 登录和注册方法代码:
const database = firebase.initializeApp(config);
const firestore = database.firestore();
var provider = new firebase.auth.GoogleAuthProvider();
database.signIn = async (email, password) => {
try {
await firebase.auth().signInWithPopup(provider).then((result) => {
// This gives you a Google Access Token. You can use it to access Google API.
var token = result.credential.accessToken;
// The signed-in user info.
var user = result.user;
console.log("user",user.displayName);
console.log('result google',result.user);
// ...
store.commit("setCurrentUser", result.user);
}).catch(function(error) {
// Handle Errors here.
var errorCode = error.code;
var errorMessage = …
Run Code Online (Sandbox Code Playgroud) javascript firebase vue.js firebase-authentication google-signin
在我的项目中,我有三个 tabBar 项目主页、通知和个人资料。侧面菜单控制器有主页、预订、个人资料和注销。侧栏菜单控制器是通过使用 SWRevealViewController cocopods 完成的。
当我将侧栏菜单导航到主页选项卡栏索引时,正确选择并正确导航。从预订导航时,它可以正确导航,但再次导航家庭应用程序崩溃。控制台输出显示无法将“UINavigationController”(0x10ef79420)类型的值转换为“UITabBarController”(0x10ef79970)。
由于预订控制器是自定义视图控制器,其余是标签栏控制器。当导航到预订屏幕视图控制器标签栏时,应该隐藏,用户再次点击菜单按钮并导航到主页或任何其他控制器。
由于预订控制器没有标签栏索引而崩溃。那么如何在不崩溃的情况下导航到带有选定索引项的自定义控制器和标签栏控制器。
这是我的截图:
我的故事板截图:
这是我尝试过的代码:
func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) {
// tableView.deselectRow(at: indexPath, animated: true)
let row = indexPath.row
if row == 0{
let tabBarController = revealViewController().frontViewController as! UITabBarController
let storyboard = UIStoryboard(name: "Home", bundle: nil)
let obj = storyboard.instantiateViewController(withIdentifier: "HomeViewController") as! HomeViewController
let navController = UINavigationController.init(rootViewController: obj)
tabBarController.selectedIndex = (indexPath as NSIndexPath).row
tabBarController.tabBar.isHidden = false
self.revealViewController().pushFrontViewController(tabBarController, animated: true)
} else if row == 1{
let tabBarController = revealViewController().frontViewController as! …
Run Code Online (Sandbox Code Playgroud) uitabbarcontroller uitableview ios swrevealviewcontroller swift
swift ×7
ios ×6
firebase ×3
arkit ×2
javascript ×2
node.js ×2
vue.js ×2
xcode ×2
alamofire ×1
cors ×1
cross-domain ×1
post ×1
projection ×1
proxy ×1
scenekit ×1
scnnode ×1
three.js ×1
uitableview ×1