小编Pet*_*zov的帖子

我想为设置面板导航创建这样的按钮:

你能告诉我如何在图标上创建这种悬停效果吗？对我来说最困难的部分是创建看起来像图片的CSS代码.

在没有使用第三方解决方案的情况下,我可以使用默认JavaFX 8软件包中的日期选择器和时间选择器的实现吗？

我想创建这样的下拉菜单:

我希望当我将鼠标放在文本上以查看可用于选择值的组合框时.当我删除鼠标时,我想看到简单的标签.我怎么能这样做？

实时更新折线图

我想修改这个简单的折线图示例并添加实时更新.

import javafx.application.Application;
import javafx.scene.Scene;
import javafx.scene.chart.LineChart;
import javafx.scene.chart.NumberAxis;
import javafx.scene.chart.XYChart;
import javafx.stage.Stage;


public class LineChartSample extends Application {

    @Override public void start(Stage stage) {
        stage.setTitle("Line Chart Sample");
        //defining the axes
        final NumberAxis xAxis = new NumberAxis();
        final NumberAxis yAxis = new NumberAxis();
        xAxis.setLabel("Number of Month");
        //creating the chart
        final LineChart<Number,Number> lineChart = 
                new LineChart<Number,Number>(xAxis,yAxis);

        lineChart.setTitle("Stock Monitoring, 2010");
        //defining a series
        XYChart.Series series = new XYChart.Series();
        series.setName("My portfolio");
        //populating the series with data
        series.getData().add(new XYChart.Data(1, 23));
        series.getData().add(new XYChart.Data(2, 14)); …

我试图用它来从自定义组合框中选择一个值：

import java.util.List;
import javafx.application.Application;
import static javafx.application.Application.launch;
import static javafx.application.Application.launch;
import javafx.beans.value.ChangeListener;
import javafx.beans.value.ObservableValue;
import javafx.scene.Scene;
import javafx.scene.control.ComboBox;
import javafx.scene.control.ListCell;
import javafx.scene.control.ListView;
import javafx.scene.layout.StackPane;
import javafx.stage.Stage;
import javafx.util.Callback;
import javafx.util.StringConverter;

public class MainApp extends Application
{

    public static void main(String[] args)
    {
        launch(args);
    }

    @Override
    public void start(Stage stage)
    {

        final ComboBox<ListGroupsObj> listGroups = new ComboBox();

        listGroups.setButtonCell(new GroupListCell());
        listGroups.setCellFactory(new Callback<ListView<ListGroupsObj>, ListCell<ListGroupsObj>>()
        {
            @Override
            public ListCell<ListGroupsObj> call(ListView<ListGroupsObj> p)
            {
                return new GroupListCell();
            }
        });

        listGroups.setEditable(true);
        listGroups.setConverter..............

        // Insert Some data …

我需要帮助代码,用于将javax.servlet.http.Part转换为java.io.File

我发现这个有用的代码,但我需要帮助正确实现代码.

private void processFilePart(Part part, String filename) throws IOException
    {
        filename = filename.substring(filename.lastIndexOf('/') + 1).substring(filename.lastIndexOf('\\') + 1);
        String prefix = filename;
        String suffix = "";
        if (filename.contains("."))
        {
            prefix = filename.substring(0, filename.lastIndexOf('.'));
            suffix = filename.substring(filename.lastIndexOf('.'));
        }
        File file = File.createTempFile(prefix + "_", suffix, new File(location));
        if (multipartConfigured)
        {
            part.write(file.getName());
        }
        else
        {
            InputStream input = null;
            OutputStream output = null;
            try
            {
                input = new BufferedInputStream(part.getInputStream(), DEFAULT_BUFFER_SIZE);
                output = new BufferedOutputStream(new FileOutputStream(file), DEFAULT_BUFFER_SIZE);
                byte[] buffer = new byte[DEFAULT_BUFFER_SIZE];
                for …

我想使用Java API从GitHub获取所有提交。到目前为止，我设法创建了以下简单代码：

import java.io.IOException;
import java.util.List;
import org.eclipse.egit.github.core.Repository;
import org.eclipse.egit.github.core.client.GitHubClient;
import org.eclipse.egit.github.core.service.RepositoryService;
import org.junit.After;
import org.junit.AfterClass;
import org.junit.Before;
import org.junit.BeforeClass;
import org.junit.Test;

public class GithubImplTest
{
    public void testSomeMethod() throws IOException
    {
        GitHubClient client = new GitHubClient();
        client.setCredentials("sonratestw@gmail.com", "sono");

        RepositoryService service = new RepositoryService(client);

        List<Repository> repositories = service.getRepositories();

        for (int i = 0; i < repositories.size(); i++)
        {
            Repository get = repositories.get(i);
            System.out.println("Repository Name: " + get.getName());
        }
    }
}

如何从该帐户将所有提交提交到存储库中？

我想使用Java Google Drive API.我试过这段代码:

public Drive getDriveService() throws GeneralSecurityException, IOException, URISyntaxException
    {
        HttpTransport httpTransport = new NetHttpTransport();
        JacksonFactory JSON_FACTORY = new JacksonFactory();

        GoogleCredential credential = new GoogleCredential.Builder()
            .setTransport(httpTransport)
            .setJsonFactory(JSON_FACTORY)
            .setServiceAccountId("sonoratestw-226@sonora-project.iam.gserviceaccount.com")
            .setServiceAccountPrivateKeyFromP12File(new java.io.File("C:\\buffer\\sonora project-3256770463ed.p12"))
            .setServiceAccountScopes(Collections.singleton(SQLAdminScopes.SQLSERVICE_ADMIN))
            .setServiceAccountUser("sonoratestw@gmail.com")
            .build();

        Drive service = new Drive.Builder(httpTransport, JSON_FACTORY, null)
            .setApplicationName("FileListAccessProject")
            .setHttpRequestInitializer(credential).build();

        return service;
    }

但我得到这个错误:

An error occurred: com.google.api.client.auth.oauth2.TokenResponseException: 401 Unauthorized

我用这个配置:

你能告诉我如何解决这个问题吗？

我想格式化包含Angular 6应用程序中的XML数据的表行.我试过这个:

<div class="list">
    <table class="table table-hover">
      <thead>
      <tr>
        <th scope="col">#</th>
        ..............
      </tr>
      </thead>
      <tbody>
      <tr *ngFor="let processingLog of processingLogs">
        ...............
        <td>{{processingLog.title}}</td>
        <td>{{processingLog.message}}</td>
      </tr>
      </tbody>
    </table>
  </div>

零件:

@Component({
  ..........
})
export class TransactionDetailsComponent implements OnInit {

  processingLogs: ProcessingLogs = new ProcessingLogs(null, null, null, null, null);
  transaction: Transaction;

  constructor(......) { }

  ngOnInit() {
    this.route.params.pipe(
      flatMap(params => {
        ............
      })
    ).subscribe(value => {
      if(value != null) {
        this.transaction = value;
        this.route.params.pipe(
          flatMap(params => {
            if (value.unique_id) {
              return this.processingLogsService.getProcessingLogsList(value.unique_id);
            } else …

我有MariaDB的SQL查询.

select @ref:=id as id, unique_id, reference_id
from mytable
join (select @ref:=id from mytable WHERE unique_id = 55544)tmp
where reference_id=@ref

https://www.db-fiddle.com/f/jKJodfVfvw65aMaVDyFySd/0

如何在HQL查询中实现此查询？我想在JPA中使用它？