GtkTreePath *path;
GtkTreeIter iter;
/* With NULL as iter, we get the number of toplevel nodes. */
gint rows = gtk_tree_model_iter_n_children(GTK_TREE_MODEL(model), NULL);
/* Now get a path from the index. */
path = gtk_tree_path_new_from_indices(rows - 1, -1);
/* Ask the model for an iter to the node identified by the path. */
gtk_tree_model_get_iter(GTK_TREE_MODEL(model), &iter, path);
/* Drop the path, we're done with it, iter is final output. */
gtk_tree_path_free(path);
def model_get_iter_last( model, parent=None ):
"""Returns a gtk.TreeIter to the last row or None if there aren't any rows.
If parent is None, returns a gtk.TreeIter to the last root row."""
n = model.iter_n_children( parent )
return n and model.iter_nth_child( parent, n - 1 )
def get_last_iter (self, model):
n = model.iter_n_children (None)
# list empty
if n == 0:
return None
else:
iter = model.iter_nth_child (None, n - 1)
n = model.iter_n_children (iter)
# last top-level item has no children, so select that last item
if n == 0:
return iter
else:
# traverse thru every last child
while 1:
iter = model.iter_nth_child (iter, n - 1)
n = model.iter_n_children (iter)
if n == 0:
return iter
您应该能够通过GtkTreePath,可能是一个黑客,但仍然在API中:
注意:我在编写上面的代码时还没有测试过,但是我很确定我在实际代码中做了一些非常类似的事情。在
尽管unwind的答案是正确的,足以让我走上正轨,但我还是想为了完整起见添加我的最终解决方案:
Python:
C:
^{pr2}$感谢前面所有的答案,但是我认为最终的解决方案应该包括TreeModel(带TreeStore)可以存储具有多个嵌套级别的项(例如path(2,0,4,2))。“fresh”发布的解决方案对我不起作用,因为我需要一个能够正确遍历所有最后一个子级的函数,直到它真正找到最后一行。以下是我的最终解决方案:
或者更简短的形式:
^{pr2}$相关问题 更多 >
编程相关推荐