我想在HTML页面中使用组件的静态变量.如何将组件的静态变量与角度2中的HTML元素绑定?
import { Component, OnInit } from '@angular/core';
import { Observable } from 'rxjs/Rx';
@Component({
moduleId: module.id,
selector: 'url',
templateUrl: 'url.component.html',
styleUrls: ['url.component.css']
})
export class UrlComponent {
static urlArray;
constructor() {
UrlComponent.urlArray=" Inside Contructor"
}
}
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<div>
url works!
{{urlArray}}
</div >
Run Code Online (Sandbox Code Playgroud) 没有任何 systemd 命令在 WSL(Ubuntu Bash 18.04)中工作。当我跑时sudo systemctl is-active kubelet
,错误是output: System has not been booted with systemd as init system (PID 1). Can't operate.
: running command: sudo systemctl is-active kubelet
如何在 WSL 中启用 systemd 功能?有什么办法摆脱System has not been booted with systemd
ubuntu kubernetes windows-subsystem-for-linux minikube ubuntu-18.04
当我尝试使用命令启动弹性搜索5.0并获得以下错误时,我./elasticsearch
得到以下错误.
[2016-11-23T13:44:09,507][WARN ][o.e.b.ElasticsearchUncaughtExceptionHandler] [] uncaught exception in thread [main]
org.elasticsearch.bootstrap.StartupException: java.lang.RuntimeException: can not run elasticsearch as root
at org.elasticsearch.bootstrap.Elasticsearch.init(Elasticsearch.java:116) ~[elasticsearch-5.0.1.jar:5.0.1]
at org.elasticsearch.bootstrap.Elasticsearch.execute(Elasticsearch.java:103) ~[elasticsearch-5.0.1.jar:5.0.1]
at org.elasticsearch.cli.SettingCommand.execute(SettingCommand.java:54) ~[elasticsearch-5.0.1.jar:5.0.1]
at org.elasticsearch.cli.Command.mainWithoutErrorHandling(Command.java:96) ~[elasticsearch-5.0.1.jar:5.0.1]
at org.elasticsearch.cli.Command.main(Command.java:62) ~[elasticsearch-5.0.1.jar:5.0.1]
at org.elasticsearch.bootstrap.Elasticsearch.main(Elasticsearch.java:80) ~[elasticsearch-5.0.1.jar:5.0.1]
at org.elasticsearch.bootstrap.Elasticsearch.main(Elasticsearch.java:73) ~[elasticsearch-5.0.1.jar:5.0.1]
Caused by: java.lang.RuntimeException: can not run elasticsearch as root
at org.elasticsearch.bootstrap.Bootstrap.initializeNatives(Bootstrap.java:96) ~[elasticsearch-5.0.1.jar:5.0.1]
at org.elasticsearch.bootstrap.Bootstrap.setup(Bootstrap.java:155) ~[elasticsearch-5.0.1.jar:5.0.1]
at org.elasticsearch.bootstrap.Bootstrap.init(Bootstrap.java:286) ~[elasticsearch-5.0.1.jar:5.0.1]
at org.elasticsearch.bootstrap.Elasticsearch.init(Elasticsearch.java:112) ~[elasticsearch-5.0.1.jar:5.0.1]
... 6 more
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我试图通过切换到其他用户开始尝试sudo ./elasticsearch
并得到相同的错误.如何以root用户身份开始弹性搜索?
我尝试匹配TITLE列中的文本配置从 ASA5505 8.2 迁移到 ASA5516。
我的程序看起来像这样。
Directory directory = FSDirectory.open(indexDir);
MultiFieldQueryParser queryParser = new MultiFieldQueryParser(Version.LUCENE_35,new String[] {"TITLE"}, new StandardAnalyzer(Version.LUCENE_35));
IndexReader reader = IndexReader.open(directory);
IndexSearcher searcher = new IndexSearcher(reader);
queryParser.setPhraseSlop(0);
queryParser.setLowercaseExpandedTerms(true);
Query query = queryParser.parse("TITLE:Config migration from ASA5505 8.2 to ASA5516");
System.out.println(queryStr);
TopDocs topDocs = searcher.search(query,100);
System.out.println(topDocs.totalHits);
ScoreDoc[] hits = topDocs.scoreDocs;
System.out.println(hits.length + " Record(s) Found");
for (int i = 0; i < hits.length; i++) {
int docId = hits[i].doc;
Document d = searcher.doc(docId);
System.out.println("\"Title :\" " …
Run Code Online (Sandbox Code Playgroud) 我正在进行登录检查.一旦成功,我想将用户重定向到同一应用程序中的另一个页面.我有以下代码来做到这一点.
import { Component } from '@angular/core';
import { FormGroup, FormControl, Validators, FormBuilder } from '@angular/forms';
import { DataService } from '../services/data.service';
import {Location, LocationStrategy, PathLocationStrategy} from '@angular/common';
@Component({
selector: 'app-login',
templateUrl: './login.component.html',
providers: [Location, {provide: LocationStrategy, useClass: PathLocationStrategy}],
styleUrls: ['./login.component.css'],
})
export class LoginComponent {
static location:Location;
constructor(private formBuilder: FormBuilder,
private dataService: DataService,location: Location) {
LoginComponent.location = location;
this.loginForm = this.formBuilder.group({
username: this.username,
password: this.password
});
}
loginForm : FormGroup;
username = new FormControl('', Validators.required);
password = new FormControl('', …
Run Code Online (Sandbox Code Playgroud) 我有测试微服务的场景。场景大纲:运行对象查询
Scenario Outline: Run Object Queries
Given url homeLinks.queryUrl
And header Content-Type = 'application/x-www-form-urlencoded'
When method post
Then status 200
Then print response
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打印语句未将任何内容记录到 InteliJ 编辑器中的控制台。我也尝试在调试模式下运行。
我提到了空手道:在空手道的控制台上看不到打印语句,这没有帮助。
打印日志所需的任何配置更改?
如何window.onload
在控制器内部使用功能。
window.onload= function() {
console.log("Hi Page loaded")
};
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该控制器不适用于整页。我想在控制器加载至少之后执行一个函数。
我想将通知发送到使用Node.Js代码中的Ionic t开发的Android应用程序。我尝试了以下代码并得到Exactly one of topic, token or condition is required.
如何无条件发送所有用户通知?
var serviceAccount = require("/path/to/config.json");
admin.initializeApp({
credential: admin.credential.cert(serviceAccount),
databaseURL: "https://myApp.firebaseio.com"
});
var message = {
notification: {
title: '$GOOG up 1.43% on the day',
body: '$GOOG gained 11.80 points to close at 835.67, up 1.43% on the day.'
}
};
admin.messaging().send(message).then(res=>{
console.log("Success",res)
}).catch(err=>{
console.log("Error:",err)
})
Run Code Online (Sandbox Code Playgroud) firebase firebase-authentication firebase-cloud-messaging firebase-notifications
我从 REST Api 有 2 个不同的响应,如下所示
1. {"type":null,"title":"Request is invalid","status":"BAD_REQUEST","detail":"Please correct request and try again.","invalidParams":[{"name":"create.arg0.name","reason":"Name is required"}]}
2. {"type":null,"title":"Unicode char u0000 is not allowed","status":"BAD_REQUEST","detail":"Unicode char u0000 is not allowed"}
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我想写一个条件,如果invalidParams
存在于响应中,然后比较数组的内容。如果不是,invalidParams
应该为空/未定义。
Scenario Outline: Create Asset Model with missing name and status
Given url modelsUrl
And request somedata
When method POST
Then status 400
Then assert response.type == null
Then match response.status == 'BAD_REQUEST'
Then match (response.invalidParams != undefined && response.invalidParams[0].reason contains <reason>) OR (response.invalidParams == undefined)
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但与 …
我正在使用 Caffeine 缓存来存储从外部系统接收到的数据。
LoadingCache<String, String> clientCache = Caffeine.newBuilder().expireAfterWrite(1, TimeUnit.MINUTES).build(id -> {
System.out.println("Generating new value for " + id);
if (id.equalsIgnoreCase("1")) {
return null;
}
return new Date().toString();
});
System.out.println(clientCache.get("1"));
System.out.println(clientCache.get("1"));
System.out.println(clientCache.get("2"));
System.out.println(clientCache.get("3"));
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这导致,
Generating new value for 1
null
Generating new value for 1
null
Generating new value for 2
Wed May 20 17:11:01 IST 2020
Generating new value for 3
Wed May 20 17:11:01 IST 2020
Wed May 20 17:11:01 IST 2020
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咖啡因不会在缓存中保存空值。如何在咖啡因中存储空值?
我正在开发一个带有 3 个选项卡的 Android 应用程序。一旦我完成从服务器获取数据,我将在第一个选项卡中设置文本Tab1 数据集成功,并且新文本在视图中可见。然后我切换到tab3并切换回tab1。现在文本是在 tab1.xml 中定义的Tab 1 。那么当我在选项卡之间切换时如何保留以前的视图内容。
我的安卓代码如下
public class secondPage extends AppCompatActivity {
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main_screen);
Intent intent = getIntent();
String dateStrig = intent.getStringExtra("Date");
String rData = null;
TabLayout tabLayout = (TabLayout) findViewById(R.id.tab_layout);
tabLayout.addTab(tabLayout.newTab().setText("TAB 1"));
tabLayout.addTab(tabLayout.newTab().setText("TAB 2"));
tabLayout.addTab(tabLayout.newTab().setText("TAB 3"));
tabLayout.setTabGravity(TabLayout.GRAVITY_FILL);
final ViewPager viewPager = (ViewPager) findViewById(R.id.pager);
final PagerAdapter adapter = new PagerAdapter
(getSupportFragmentManager(), tabLayout.getTabCount());
viewPager.setAdapter(adapter);
new fetchData().execute(dateStrig);
viewPager.addOnPageChangeListener(new TabLayout.TabLayoutOnPageChangeListener(tabLayout));
tabLayout.setOnTabSelectedListener(new TabLayout.OnTabSelectedListener() {
@Override
public void onTabSelected(TabLayout.Tab tab) { …
Run Code Online (Sandbox Code Playgroud) 我尝试在 Oracle 中添加一个表到缓存
alter table Test_Table cache;
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那么现在我怎样才能将其从缓存中删除呢?
尝试使用我构建的另一个自定义映像 (Y) 构建 Docker 映像 (X)。在这里,我尝试将文件夹从基本图像 Y 复制到 X,如下所示。
FROM some-image as base
COPY --from=base /root/.m2 /root
RUN mkdir -p /build
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但这会引发错误invalid from flag value base: pull access denied for base, repository does not exist or may require 'docker login': denied: requested access to the resource is denied
我怎样才能做到这一点?
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