JavaFX：按行和列获取Node

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

getChildren

Pane

getRowIndex(Nodechild)

getColumnIndex(Node child)

GridPane

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

以下是示例代码，可从中获取

Node

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()

getColumnIndex()

GridPane.getRowIndex(node)

GridPane.getColumnIndex(node)