Angular:未捕获的类型错误:无法读取未定义的属性“调用”

问题描述

我正在处理一个 angular 9 项目,在该项目中,我通过以下步骤使用 AWS Amplify 实现了自行设计的用户身份验证:https://gerard-sans.medium.com/build-your-first-full-stack-serverless-app-with-angular-and-aws-amplify-d2e4716de9bd。此外,我在后端集成了 graphQL。并且整个项目在本地运行良好,并在部署时引发错误。它在部署时显示以下错误

enter image description here

我正在使用以下角度依赖项:

"dependencies": {
    "@angular/animations": "~9.0.7","@angular/cdk": "^10.2.7","@angular/common": "~9.0.7","@angular/compiler": "~9.0.7","@angular/core": "~9.0.7","@angular/forms": "~9.0.7","@angular/localize": "^9.1.12","@angular/material": "^10.2.7","@angular/platform-browser": "~9.0.7","@angular/platform-browser-dynamic": "~9.0.7","@angular/router": "~9.0.7","@apollo/client": "^3.0.0","@fortawesome/angular-fontawesome": "^0.7.0","@fortawesome/fontawesome-svg-core": "^1.2.31","@fortawesome/free-solid-svg-icons": "^5.15.0","@nestjs/graphql": "^7.9.4","@ng-bootstrap/ng-bootstrap": "^5.1.4","@swimlane/ngx-datatable": "^16.0.2","angular-calendar": "^0.28.22","angularx-flatpickr": "^6.5.1","apollo-angular": "^2.1.0","apollo-server-express": "^2.19.1","aws-amplify": "^3.3.13","aws-amplify-angular": "^5.0.42","bcryptjs": "^2.4.3","bootstrap": "^4.5.2","date-fns": "^2.16.1","flatpickr": "^4.6.6","graphql": "^15.4.0","graphql-tools": "^7.0.2","jsonwebtoken": "^8.5.1","ngx-cookie-service": "^11.0.2","rxjs": "~6.5.4","tslib": "^1.10.0","zone.js": "~0.10.2"
  },"devDependencies": {
    "@angular-devkit/build-angular": "~0.900.7","@angular/cli": "~9.0.7","@angular/compiler-cli": "~9.0.7","@angular/language-service": "~9.0.7","@types/jasmine": "~3.5.0","@types/jasminewd2": "~2.0.3","@types/node": "^12.11.1","codelyzer": "^5.1.2","graphql-cli": "^4.1.0","jasmine-core": "~3.5.0","jasmine-spec-reporter": "~4.2.1","karma": "~4.3.0","karma-chrome-launcher": "~3.1.0","karma-coverage-istanbul-reporter": "~2.1.0","karma-jasmine": "~2.0.1","karma-jasmine-html-reporter": "^1.4.2","protractor": "~5.4.3","ts-node": "~8.3.0","tslint": "~5.18.0","typescript": "~3.7.5"
  },

解决方法

将我的 Angular 版本从 9 更新到 11 对我有用。

相关问答

Selenium Web驱动程序和Java。元素在(x,y)点处不可单击。其...
Python-如何使用点“。” 访问字典成员?
Java 字符串是不可变的。到底是什么意思?
Java中的“ final”关键字如何工作?(我仍然可以修改对象。...
“loop:”在Java代码中。这是什么,为什么要编译?
java.lang.ClassNotFoundException:sun.jdbc.odbc.JdbcOdbc...