I have a SQLite-Database and I did it into a QSqlTableModel.
To show the Database, I put that Model into a QTableView.
Now I want to create
Quark's answer (the selected one) is good for pointing people in the right direction, but his algorithm is entirely incorrect. In addition to an off by one error and incorrect assignment, its not even syntactically correct. Below is a working version that I just wrote and tested.
Let's assume our example table looks like so:
A | B | C
D | E | F
The problem with Quark's algorithm is the following:
If we replace his \t separator with a ' | ', it will produce this output:
B | C | D
E | F |
The off by one error is that D appears in the first row. The incorrect assignment is evidenced by the omission of A
The following algorithm corrects these two problems with correct syntax.
QString clipboardString;
QModelIndexList selectedIndexes = view->selectionModel()->selectedIndexes();
for (int i = 0; i < selectedIndexes.count(); ++i)
{
QModelIndex current = selectedIndexes[i];
QString displayText = current.data(Qt::DisplayRole).toString();
// If there exists another column beyond this one.
if (i + 1 < selectedIndexes.count())
{
QModelIndex next = selectedIndexes[i+1];
// If the column is on different row, the clipboard should take note.
if (next.row() != current.row())
{
displayText.append("\n");
}
else
{
// Otherwise append a column separator.
displayText.append(" | ");
}
}
clipboardString.append(displayText);
}
QApplication::clipboard()->setText(clipboardString);
The reason I chose to use a counter instead of an iterator is just because it is easier to test if there exists another index by checking against the count. With an iterator, I suppose maybe you could just increment it and store it in a weak pointer to test if it is valid but just use a counter like I did above.
We need to check if the next line will be on on a new row. If we are on a new row and we check the previous row as Quark's algorithm does, its already too late to append. We could prepend, but then we have to keep track of the last string size. The above code will produce the following output from the example table:
A | B | C
D | E | F