尽管在下一行中明确定义了方法,但我遇到了以下错误.
undefined method `before_filter' for AuthorsController:Class
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我正在学习本教程.
代码段如下:
authors_controller.rb
class AuthorsController < ApplicationController
before_action :set_author, only: [:show, :edit, :update, :destroy]
before_filter :zero_authors_or_authenticated, only: [:new, :create]
def zero_authors_or_authenticated
# If either one of them is true this filter won’t do anything, allowing the requested user registration form to be rendered
unless Author.count == 0 || current_user # checking if there are either zero registered users OR if there is a user already logged in
redirect_to root_path # if neither …Run Code Online (Sandbox Code Playgroud) 我一直在尝试为 Flutter 编写平台代码以启动后台服务。在这里,我使用了一个没有完成实际工作的最小示例来表明应用程序根本无法运行。实际的颤振代码根本没有被修改。
主活动.java
public class MainActivity extends FlutterActivity {
Intent i = new Intent(this, MainService.class);
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
GeneratedPluginRegistrant.registerWith(this);
}
}
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主服务.java
public class MainService extends IntentService {
public MainService() {
super("MainService");
}
@Override
protected void onHandleIntent(Intent Intent) {
}
}
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AndroidManifest.xml
<service android:enabled="true"
android:name=".MainService"></service>
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buildVersion > 27 并且 Manifest 文件相应地service添加了标签。
编译并运行 withflutter run -v将显示以下消息:
..
[ +121 ms] Starting: Intent { act=android.intent.action.RUN flg=0x20000000 cmp=com.example.hello/.MainActivity (has extras)}
[ +1 …Run Code Online (Sandbox Code Playgroud) 此方法将接收Long并返回LongStream传递给方法的任何数字的素数.
factors.java
public LongStream factors(long x){
LongStream factorStream = LongStream.range(1, x+1).filter(n -> x%n == 0);
return factorStream;
}
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利用上述方法找到共同的因素首先是OK.
primeFactors.java
public LongStream primeFactors(long x){
LongStream primeFactorStream = factors(x).filter(n -> factors(n).count() == 0);
//doesn't work as factors.java returns a LongStream, which might include non-prime factors, which will not equate to zero.
return primeFactorStream;
}
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我理解这应该通过使用带谓词的简单isPrime()方法来轻松规避,但有没有办法为素数因子做同样的事情,但只能用一个方法?
我正在尝试在我的flutter应用程序中使用Google的Firebase身份验证.但是,在处理用户登录时,我似乎找不到"代码"来区分我要处理的异常类型.
Loginhandler.dart:
class LoginHandler {
String signup(_username, _password) async {
bool _valid = false;
final prefs = await SharedPreferences.getInstance();
FirebaseAuth.instance.createUserWithEmailAndPassword(
email: _username, password: _password)
.catchError((e) {
print(e);
print(e.message);
print(e.code);
print(e.details);
});
...
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错误输出:
W/BiChannelGoogleApi(26738): [FirebaseAuth: ] getGoogleApiForMethod() returned Gms: com.google.firebase.auth.api.internal.zzal@4e727e7
W/IInputConnectionWrapper(26738): getCursorCapsMode on inactive InputConnection
I/flutter (26738): PlatformException(exception, The email address is already in use by another account., null)
I/flutter (26738): The email address is already in use by another account.
I/flutter (26738): exception
I/flutter (26738): null
I/flutter (26738): PlatformException(exception, …Run Code Online (Sandbox Code Playgroud) 我有一个 Flutter 应用程序,它由一个带有滑块和 tabview 的脚手架组成。
tabview 包含一个列表,它显示在每个选项卡上,如下图所示。
List<Widget> widgetList = <Widget>[
Post(),
Feed(),
Location(),
HomePage(),
Feed(),
];
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现在我想在移动滑块时刷新屏幕上的当前选项卡。但是,由于这些类是私有的,即_HomePageState,我不知道如何访问该refreshList()方法,如下面的代码片段所示。
主页.dart:
class HomePage extends StatefulWidget {
@override
_HomePageState createState() => new _HomePageState();
}
class _HomePageState extends State<HomePage> {
var list;
var random;
var refreshKey = GlobalKey<RefreshIndicatorState>();
@override
void initState() {
super.initState();
random = Random();
refreshList();
}
Future<Null> refreshList() async {
refreshKey.currentState?.show(atTop: false);
await Future.delayed(Duration(seconds: 2));
setState(() {
list = List.generate(random.nextInt(10), (i) => "Item $i");
});
return null;
} …Run Code Online (Sandbox Code Playgroud) onChange当用户在输入字段中键入内容时,我试图消除事件的反跳。
我正在引用这些线程:
我有以下代码片段,我尝试复制上面线程中提供的解决方案:
const handler = useCallback(debounce(setSearchQuery(value), 500), []);
useEffect(() => {
document.addEventListener('keydown', handleDocumentKeyDown);
handler(value);
return () => document.removeEventListener('keydown', handleDocumentKeyDown);
}, [isOpen, handleDocumentKeyDown, handler, value]);
// ...
const handleChange = (event) => {
setValue(event.target.value);
};
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错误:
未捕获的类型错误:处理程序不是函数
当用户在输入字段中键入内容时,setSerachQuery()如何消除反跳?500ms
我正在按照本指南尝试设置 mongoDB 数据库。
mongoClient.listDatabaseNames().forEach(System.out::println);
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getDatabaseNames() 已弃用并替换。
然而,这一行给出了以下错误:
error: reference to forEach is ambiguous
mongoClient.listDatabaseNames().forEach(System.out::println);
^
both method forEach(Consumer<? super T>) in Iterable and method forEach(Block<? super TResult>) in MongoIterable match
where T,TResult are type-variables:
T extends Object declared in interface Iterable
TResult extends Object declared in interface MongoIterable
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文档指出 listDatabaseNames() 返回 a ListDatabasesIterable,为什么我不能遍历这个列表?
我有一个 React 应用程序,它读取几个与 API 相关的环境变量。
但是,使用映像和configmap在 Kubernetes 中创建 pod不起作用 - 应用程序运行但未设置环境变量。
pod.yaml
...
spec:
containers:
- command:
- sleep
- "3600"
envFrom:
- configMapRef:
name: configmap
image: xxxxx
imagePullPolicy: IfNotPresent
...
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配置图
apiVersion: v1
data:
API_HOST: xxxxxxx
SOME_ID: abcdef
NODE_ENV: development
PROVIDER: GCP
kind: ConfigMap
metadata:
creationTimestamp: xxxx
name: configmap
namespace: xxxx
resourceVersion: xxxx
selfLink: xxxx
uid: xxxx
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反应片段
if(!process.env.SOME_ID) {
console.log('ID')
}
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我的麻烦在于将环境变量传递给 React 应用程序。我确信 Pod 中的环境变量设置正确,但客户端 React 应用程序似乎没有这些变量(即不console.log打印任何内容)。 …
我希望通过 Azure AD使用OAuth2进行身份验证。
服务器.js
const express = require("express");
const app = express();
const port = process.env.PORT || 5000;
var bodyParser = require("body-parser");
app.use(bodyParser.urlencoded({ extended: false }));
app.use(bodyParser.json());
app.get("/authorize", async (req, res) => {
const credentials = {
client: {
id: "xxx",
secret: "xxx"
},
auth: {
tokenHost:
"xxx"
}
};
const oauth2 = require("simple-oauth2").create(credentials);
const tokenConfig = {
scope: "<scope>"
};
const httpOptions = {};
try {
const result = await oauth2.clientCredentials.getToken(
tokenConfig,
httpOptions
);
const accessToken = …Run Code Online (Sandbox Code Playgroud) 我有两个存储库fe和craft,分别是应用程序库和组件库。
两个存储库都可以自行编译。但是,当我尝试通过组件库使用组件时,如果组件本身props.theme在styled-components.
Main.tsx来自fe
import React from "react";
import { Button, HeadlineOne } from "handcraft";
const Page = (props: any) => {
console.log(props.theme);
return (
<>
<Button onClick={() => {}}>Sample</Button> // This causes TypeError: Cannot read property 'primary' of undefined.
<HeadlineOne>Sample</HeadlineOne> // This is fine.
</>
);
};
export default Page.
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来自Button.tsxcraft(不行,访问 props)。
import React, { ReactNode } from "react";
import { Button as BaseButton } from "@material-ui/core"; …Run Code Online (Sandbox Code Playgroud) dart ×3
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android ×1
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