我尝试从QGridLayout中的指定行中删除小部件,如下所示：
void delete_grid_row(QGridLayout *layout,int row)
{
if (!layout || row < 0) return;
for (int i = 0; i < layout->columnCount(); ++i) {
QLayoutItem* item = layout->itemAtPosition(row,i);
if (!item) continue;
if (item->widget()) {
layout->removeWidget(item->widget());
} else {
layout->removeItem(item);
}
delete item;
}
}
但是当我打电话给它时,应用程序会在第一次迭代中使用SIGSEGV删除项目.有任何想法吗？
解决方法
注意：如果您只是寻找一个快速解决这个问题,那么只需使用下面提供的代码.如果您对网格布局中行列处理的背景和困难感兴趣,请继续阅读.
关于QGridLayout行和列
首先,请注意rowCount()和columnCount()返回内部分配的行和列数.例如,如果在新构造的网格布局上调用addWidget(widget,5,7),则行计数将为6,列计数将为8,并且除了行索引5之外的单元格的网格布局的所有单元格并且列索引7将为空,因此在GUI内不可见.
请注意,不可能从网格布局中删除这样的内部行或列.换句话说,网格布局的行和列计数因此总是只能增长,但不会缩小.
您可以做的是删除特定行或列的单元格的内容,这将有效地与删除行或列本身具有相同的效果.但是请注意,这不会更改任何行或列计数,也不会更改任何行或列索引.
清除行或列
那么一行或一列的内容怎么能被清除？这不幸也不像看起来那么容易.
首先,您需要考虑如果您只想从布局中删除小部件,或者您还希望将其删除.如果您仅从布局中删除小部件,则可以将它们重新放入不同的布局,或者手动给它们一个合理的几何体. (请注意,如果不这样做,这些小部件将只会在当前位置保持可见,这几乎绝对不是您想要的.)如果小部件也被删除,它们将从GUI中消失.下面提供的代码使用一个布尔参数来控制小部件删除.
接下来,我们必须考虑一个布局单元格不能只包含一个窗口小部件,而且还包含嵌套布局,其本身可以包含嵌套布局等等.我们还需要处理跨越多行和列的布局项.最后,还有一些行和列属性,如最小宽度和高度,不依赖于单元格内容,但仍然必须重置.
把它放在一起 – 代码
使用下面的removeRow()和removeColumn()函数从网格布局中删除一行或一列.
/**
* Helper function. Removes all layout items within the given @a layout
* which either span the given @a row or @a column. If @a deleteWidgets
* is true,all concerned child widgets become not only removed from the
* layout,but also deleted.
*/
void remove(QGridLayout *layout,int row,int column,bool deleteWidgets) {
// We avoid usage of QGridLayout::itemAtPosition() here to improve performance.
for (int i = layout->count() - 1; i >= 0; i--) {
int r,c,rs,cs;
layout->getItemPosition(i,&r,&c,&rs,&cs);
if ((r <= row && r + rs - 1 >= row) || (c <= column && c + cs - 1 >= column)) {
// This layout item is subject to deletion.
QLayoutItem *item = layout->takeAt(i);
if (deleteWidgets) {
deleteChildWidgets(item);
}
delete item;
}
}
}
/**
* Helper function. Deletes all child widgets of the given layout @a item.
*/
void deleteChildWidgets(QLayoutItem *item) {
if (item->layout()) {
// Process all child items recursively.
for (int i = 0; i < item->layout()->count(); i++) {
deleteChildWidgets(item->layout()->itemAt(i));
}
}
delete item->widget();
}
/**
* Removes all layout items on the given @a row from the given grid
* @a layout. If @a deleteWidgets is true,all concerned child widgets
* become not only removed from the layout,but also deleted. Note that
* this function doesn't actually remove the row itself from the grid
* layout,as this isn't possible (i.e. the rowCount() and row indices
* will stay the same after this function has been called).
*/
void removeRow(QGridLayout *layout,bool deleteWidgets) {
remove(layout,row,-1,deleteWidgets);
layout->setRowMinimumHeight(row,0);
layout->setRowStretch(row,0);
}
/**
* Removes all layout items on the given @a column from the given grid
* @a layout. If @a deleteWidgets is true,but also deleted. Note that
* this function doesn't actually remove the column itself from the grid
* layout,as this isn't possible (i.e. the columnCount() and column
* indices will stay the same after this function has been called).
*/
void removeColumn(QGridLayout *layout,column,deleteWidgets);
layout->setColumnMinimumWidth(column,0);
layout->setColumnStretch(column,0);
}
