如何在谓词中对 FilteredList 结果进行优先级排序/排名?

Zep*_*hyr 4 java javafx predicate

我的应用程序包含 aTextField和 a ListView。允许TextField用户输入搜索词,这些搜索词将在他们键入时过滤内容ListView

过滤过程将匹配每个DataItem字段中的多个字段ListView,如果其中任何一个匹配,则返回结果。

然而,我想做的是让这些结果优先考虑与一个特定字段匹配的项目而不是其他字段。

例如,在下面的 MCVE 中,我有两项:ComputerPaper。该Computer项目有一个keyword“paper”,因此搜索“paper”应该返回Computer结果。

但是,由于我还有一个名为 的项目Paper,因此搜索应返回Paper到列表顶部。但在 MCVE 中,结果仍然按字母顺序排列:

截屏

问题:我如何确保DataItem.name上面列出的任何匹配项都与 a 匹配DataItem.keywords

编辑:在搜索字段中输入“pap”也应该在顶部返回“Paper”,然后是其余的匹配项,因为部分搜索词与名称部分匹配DataItem


MCVE


数据项.java:

import java.util.List;

public class DataItem {

    // Instance properties
    private final IntegerProperty id = new SimpleIntegerProperty();
    private final StringProperty name = new SimpleStringProperty();
    private final StringProperty description = new SimpleStringProperty();

    // List of search keywords
    private final ObjectProperty<List<String>> keywords = new SimpleObjectProperty<>();

    public DataItem(int id, String name, String description, List<String> keywords) {
        this.id.set(id);
        this.name.set(name);
        this.description.set(description);
        this.keywords.set(keywords);
    }

    /**
     * Creates a space-separated String of all the keywords; used for filtering later
     */
    public String getKeywordsString() {
        StringBuilder sb = new StringBuilder();

        for (String keyword : keywords.get()) {
            sb.append(keyword).append(" ");
        }

        return sb.toString();

    }

    public int getId() {
        return id.get();
    }

    public IntegerProperty idProperty() {
        return id;
    }

    public String getName() {
        return name.get();
    }

    public StringProperty nameProperty() {
        return name;
    }

    public String getDescription() {
        return description.get();
    }

    public StringProperty descriptionProperty() {
        return description;
    }

    public List<String> getKeywords() {
        return keywords.get();
    }

    public ObjectProperty<List<String>> keywordsProperty() {
        return keywords;
    }

    @Override
    public String toString() {
        return name.get();
    }
}
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主要.java:

import javafx.application.Application;
import javafx.collections.FXCollections;
import javafx.collections.ObservableList;
import javafx.collections.transformation.FilteredList;
import javafx.collections.transformation.SortedList;
import javafx.geometry.Insets;
import javafx.geometry.Pos;
import javafx.scene.Scene;
import javafx.scene.control.ListView;
import javafx.scene.control.TextField;
import javafx.scene.layout.VBox;
import javafx.stage.Stage;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;

public class Main extends Application {

    // TextField used for filtering the ListView
    TextField txtSearch = new TextField();

    // ListView to hold our DataItems
    ListView<DataItem> dataItemListView = new ListView<>();

    // The ObservableList of DataItems
    ObservableList<DataItem> dataItems;

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

    @Override
    public void start(Stage primaryStage) {

        // Simple Interface
        VBox root = new VBox(10);
        root.setAlignment(Pos.CENTER);
        root.setPadding(new Insets(10));

        // Add the search field and ListView to the layout
        root.getChildren().addAll(txtSearch, dataItemListView);

        // Build the dataItems List
        dataItems = FXCollections.observableArrayList(buildDataItems());

        // Add the filter logic
        addSearchFilter();

        // Show the stage
        primaryStage.setScene(new Scene(root));
        primaryStage.setTitle("Sample");
        primaryStage.show();
    }

    /**
     * Adds the functionality to filter the list dynamically as search terms are entered
     */
    private void addSearchFilter() {

        // Wrap the dataItems list in a filtered list, initially showing all items, alphabetically
        FilteredList<DataItem> filteredList = new FilteredList<>(
                dataItems.sorted(Comparator.comparing(DataItem::getName)));

        // Add the predicate to filter the list whenever the search field changes
        txtSearch.textProperty().addListener((observable, oldValue, newValue) ->
                filteredList.setPredicate(dataItem -> {

                    // Clear any selection already present
                    dataItemListView.getSelectionModel().clearSelection();

                    // If the search field is empty, show all DataItems
                    if (newValue == null || newValue.isEmpty()) {
                        return true;
                    }

                    // Compare the DataItem's name and keywords with the search query (ignoring case)
                    String query = newValue.toLowerCase();

                    if (dataItem.getName().toLowerCase().contains(query)) {
                        // DataItem's name contains the search query
                        return true;
                    } else {

                        // Otherwise check if any of the search terms match those in the DataItem's keywords
                        // We split the query by space so we can match DataItems with multiple keywords
                        String[] searchTerms = query.split(" ");
                        boolean match = false;
                        for (String searchTerm : searchTerms) {
                            match = dataItem.getKeywordsString().toLowerCase().contains(searchTerm);
                        }
                        return match;
                    }
                }));

        // Wrap the filtered list in a SortedList
        SortedList<DataItem> sortedList = new SortedList<>(filteredList);

        // Update the ListView
        dataItemListView.setItems(sortedList);
    }

    /**
     * Generates a list of sample products
     */
    private List<DataItem> buildDataItems() {

        List<DataItem> dataItems = new ArrayList<>();

        dataItems.add(new DataItem(
                1, "School Supplies", "Learn things.",
                Arrays.asList("pens", "pencils", "paper", "eraser")));
        dataItems.add(new DataItem(
                2, "Computer", "Do some things",
                Arrays.asList("paper", "cpu", "keyboard", "monitor")));
        dataItems.add(new DataItem(
                3, "Keyboard", "Type things",
                Arrays.asList("keys", "numpad", "input")));
        dataItems.add(new DataItem(
                4, "Printer", "Print things.",
                Arrays.asList("paper", "ink", "computer")));
        dataItems.add(new DataItem(
                5, "Paper", "Report things.",
                Arrays.asList("write", "printer", "notebook")));

        return dataItems;
    }
}
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JKo*_*dis 5

如果没有记错的话,您只需要找到一种方法来正确对过滤结果进行排序。为了简单起见,我将使用这个比较器而不是你的:

Comparator<DataItem> byName = new Comparator<DataItem>() {
            @Override
            public int compare(DataItem o1, DataItem o2) {
                String searchKey = txtSearch.getText().toLowerCase();
                int item1Score = findScore(o1.getName().toLowerCase(), searchKey);
                int item2Score = findScore(o2.getName().toLowerCase(), searchKey);

                if (item1Score > item2Score) {
                    return -1;
                }

                if (item2Score > item1Score) {
                    return 1;
                }

                return 0;
            }

            private int findScore(String itemName, String searchKey) {
                int sum = 0;
                if (itemName.startsWith(searchKey)) {
                    sum += 2;
                }

                if (itemName.contains(searchKey)) {
                    sum += 1;
                }
                return sum;
            }
        };
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在上面的代码中,我比较了两个 DataItem。每个人都会有一个“分数”,这取决于他们的名字与我们的搜索关键字的相似程度。searchKey为了简单起见,假设如果出现在我们的项目名称中,我们给 1 分,如果项目名称以 开头,我们给 2 分searchKey,所以现在我们可以比较这两个并对它们进行排序。如果我们返回-1,则 item1 将被放置在前面,如果我们返回 1,则 item2 将被放置在前面,否则返回 0。

这是addSearchFilter()我在您的示例中使用的方法:

private void addSearchFilter() {
        FilteredList<DataItem> filteredList = new FilteredList<>(dataItems);

        txtSearch.textProperty().addListener((observable, oldValue, newValue) -> filteredList.setPredicate(dataItem -> {

            dataItemListView.getSelectionModel().clearSelection();

            if (newValue == null || newValue.isEmpty()) {
                return true;
            }

            String query = newValue.toLowerCase();

            if (dataItem.getName().toLowerCase().contains(query)) {
                return true;
            } else {
                String[] searchTerms = query.split(" ");
                boolean match = false;
                for (String searchTerm : searchTerms) {
                    match = dataItem.getKeywordsString().toLowerCase().contains(searchTerm);
                }
                return match;
            }
        }));

        SortedList<DataItem> sortedList = new SortedList<>(filteredList);

        Comparator<DataItem> byName = new Comparator<DataItem>() {
            @Override
            public int compare(DataItem o1, DataItem o2) {
                String searchKey = txtSearch.getText().toLowerCase();
                int item1Score = findScore(o1.getName().toLowerCase(), searchKey);
                int item2Score = findScore(o2.getName().toLowerCase(), searchKey);

                if (item1Score > item2Score) {
                    return -1;
                }

                if (item2Score > item1Score) {
                    return 1;
                }

                return 0;
            }

            private int findScore(String itemName, String searchKey) {
                int sum = 0;
                if (itemName.startsWith(searchKey)) {
                    sum += 2;
                }

                if (itemName.contains(searchKey)) {
                    sum += 1;
                }
                return sum;
            }
        };

        sortedList.setComparator(byName);

        dataItemListView.setItems(sortedList);
    }
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当然,findScore()如果您想创建更复杂的评分系统(例如检查大小写字母,根据项目名称中找到的关键字的位置给出更多分数等),则可能会更复杂。