JavaFX:按行和列获取节点

Question

如果我知道它的位置(行和列)或从gridPane获取节点的任何其他方式,有没有办法从gridPane获取特定节点？

Answer 1

我看不出有任何直接的API来获取由行列索引节点,但可以使用getChildrenAPI从Pane,并getRowIndex(Node child)与getColumnIndex(Node child)来自GridPane

//Gets the list of children of this Parent. 
public ObservableList<Node> getChildren() 
//Returns the child's column index constraint if set
public static java.lang.Integer getColumnIndex(Node child)
//Returns the child's row index constraint if set.
public static java.lang.Integer getRowIndex(Node child)

以下是从中获取Node行和列索引的示例代码GridPane

public Node getNodeByRowColumnIndex (final int row, final int column, GridPane gridPane) {
    Node result = null;
    ObservableList<Node> childrens = gridPane.getChildren();

    for (Node node : childrens) {
        if(gridPane.getRowIndex(node) == row && gridPane.getColumnIndex(node) == column) {
            result = node;
            break;
        }
    }

    return result;
}

重要更新: getRowIndex()和getColumnIndex()现在的静态方法和应改为GridPane.getRowIndex(node)和GridPane.getColumnIndex(node).