spinbox input add rows ??

n-2204

yes, thanks using this works
but what happening is i added 10 rows and when i change value in spinbox to 1 then no. of rows in table also changing to 1

so i think using loop this prblm will be solved

int noofrows = ui.spinBox->value();
   for (int i = 0;i < noofrows;i++)
{
   model->insertRows(i,noofrows);
}

if use this loop when i enter 1 in spinbox it adds two rows ?

JonB

@n-2204
I already told you about that earlier.

And finally I doubt you really intend

model->insertRows(i, model->rowCount(QModelIndex()));

Why rowCount() as the number of rows to add?? I would have thought given your loop you would want 1 there. You are confusing insertRows() to add multiple rows without a loop versus a loop to add one row at a time....

EITHER a for loop with insertRow(), OR no for loop and insertRows(). You really should be able to understand this.

Put some qDebug() statements in to see for yourself what is going on. That is what debugging is about.

n-2204

ok understood
thank you all

n-2204

 tooll::tool(QWidget* parent)
: QMainWindow(parent)
  {
  ui.setupUi(this);
 model = new QStandardItemModel(10, 13, this)
 ui.tableView->setModel(model);
 connect(ui.spinBox, SIGNAL(valueChanged(int)), model, 
 SLOT(setTableRows())); 
}
void tool::setTable2Rows()
   { 
 int noofrows2 = ui.spinBox_2->value();
 for (int i = 0;i < noofrows2;++i)
  {
      model2->insertRow(i);
    //qDebug() << i;
}
}

when i am using new signal/slot why i am getting error
(Error (active) E0304 no instance of overloaded function)

 connect(ui.spinBox, &QSpinBox::valueChanged, this,&tool::setTableRows);
connect( sender, &Sender::valueChanged,    receiver, &Receiver::updateValue);

jsulm

@n-2204 Please post whole error message. Does your tooll class have setTableRows slot? I'm asking because I only see setTable2Rows()

n-2204

yes slot setTablerow() is also there

 void tool::setTableRows()
{ 
int noofrows = ui.spinBox->value();
for (int i = 0;i < noofrows;++i)
{
  model->insertRow(i);
//qDebug() << i;
}

c++ no instance of overloaded function matches the argument list argument types are: (QSpinBox *, <unknown-type>, tooll , void (tool::)())
Severity Code Description Project File Line Suppression State Detail Description
Error (active) E0304 no instance of overloaded function "tool::connect" matches the argument list tool C:\User\source\repos\tool\tooll.cpp 265 argument types are: (QSpinBox *, <unknown-type>, tool , void (tool::)())

VRonin

@n-2204 said in spinbox input add rows ??:

tooll::tool(QWidget* parent)

????

What's the name of the class? tooll or tool? looks like a big mess here

Christian Ehrlicher

@n-2204 said in spinbox input add rows ??:

c++ no instance of overloaded function matches the argument list argument types are: (QSpinBox *, <unknown-type>, tooll , void (tool::)())

How about taking a single look into the documentation? The problem is exactly explained there.

Simply writing something without understanding what you're doing doesn't help you at all.

n-2204

thanks for sharing ref doc, its done

 connect(ui.spinBox, QOverload<int>::of(&QSpinBox::valueChanged), 
         this, &tool::setTableRows);

n-2204

what changes need to do to end the row at end of row, now when i enter value in spinbox row is adding in top but i need to add the row at end how to do ?

 void tool::setTableRows()
{
int noofrows = ui.spinBox->value();
for (int i = 0;i < noofrows;++i)
{
   model->insertRow(i);
}
 }
connect(ui.spinBox, QOverload<int>::of(&QSpinBox::valueChanged), 
         this, &tool::setTableRows);

jsulm

@n-2204 https://doc.qt.io/qt-5/qabstractitemmodel.html#insertRow
"Inserts a single row before the given row in the child items of the parent specified".
So, what do you think should you pass as row to insertRow to add at the end of the table?

JonB

@n-2204

@jsulm said in spinbox input add rows ??:

@n-2204 https://doc.qt.io/qt-5/qabstractitemmodel.html#insertRow
"Inserts a single row before the given row in the child items of the parent specified".
So, what do you think should you pass as row to insertRow to add at the end of the table?

Further to @jsulm. insertRow() says

Note: This function calls the virtual method insertRows.

And in the docs for insertRows, immediately adjacent to insertRow, if you read it it actually tells you what to pass so that

the rows are appended to any existing rows

Please take the time to read docs....

n-2204

how can i set value of a spinbox and also edittable for user ?

 void tool::setTable2Rows()
 {
QAbstractItemModel* table2 = ui.tableView_2->model();
int rows2 = ui.spinBox_2->value();
  ui.spinBox->setValue(10);//this is not working
  model2->setRowCount(rows2);
}

if i setvalue 10 then that no. of rows should be there

JonB

@n-2204 said in spinbox input add rows ??:

ui.spinBox->setValue(10);//this is not working

Don't what you mean. "Not working" could indicate anything. I imagine this works perfectly well.

how can i set value of a spinbox and also edittable for user ?

if i setvalue 10 then that no. of rows should be there

Don't know what these mean either.

n-2204

not working means value is not set to 10

and when i setvalue as 10 , 10 rows should be there in my table
also when user change the spinbox value to 11 or 8 no. of rows in table should update

JonB

@n-2204 said in spinbox input add rows ??:

not working means value is not set to 10

What value is not set to 10? You can see what it sets:

ui.spinBox->setValue(10);

I would bet my bottom $ that it does exactly that.

and when i setvalue as 10 , 10 rows should be there in my table

Your code has

model2->setRowCount(rows2);

so that is what it does.

also when user change the spinbox value to 11 or 8 no. of rows in table should update

Which spinbox? Only if you call the code to set the rows when the spin value changes, and only if the code is correct. Which i have to say does not look like the case for the 4 lines of code you have shown...

n-2204

code:

 tool::tool(QWidget* parent)
: QMainWindow(parent)
{
ui.setupUi(this);
model = new QStandardItemModel(0, 13, this);//row*col
model2 = new QStandardItemModel(0, 19, this)
ui.spinBox->setValue(10);
ui.spinBox_2->setValue(10);

using this in spinbox value is 10 but the same no. of rows is not updated in my table
so when i set value in spinbox same no. of rows should be added in my table
and when user change value in spinbox row should update (that is happening)

 connect(ui.spinBox, QOverload<int>::of(&QSpinBox::valueChanged), 
         this, &tool::setTableRows);
 connect(ui.spinBox_2, QOverload<int>::of(&QSpinBox::valueChanged),
         this, &tool::setTable2Rows);
 } 
void tool::setTableRows()
{
QAbstractItemModel* table1 = ui.tableView->model();
int rows = ui.spinBox->value();
model->setRowCount(rows);
 } 

 void tool::setTable2Rows()
{
QAbstractItemModel* table2 = ui.tableView_2->model();
int rows2 = ui.spinBox_2->value();
model2->setRowCount(rows2);
}

JonB

@n-2204 said in spinbox input add rows ??:

using this in spinbox value is 10 but the same no. of rows is not updated in my table

If I understand correctly.

At the time you call

ui.spinBox->setValue(10);

you have not yet done the

 connect(ui.spinBox, QOverload<int>::of(&QSpinBox::valueChanged), 
         this, &tool::setTableRows);

So if you are expecting the slot setTableRows() to be called to actually create rows during the setValue(), it won't be. Why should it?

Either connect the signals before setting the values, or call the slots explicitly first time before you start out.

Is that what you mean?

n-2204

@JonB said in spinbox input add rows ??:

@n-2204 said in spinbox input add rows ??:

using this in spinbox value is 10 but the same no. of rows is not updated in my table

If I understand correctly.

At the time you call
ui.spinBox->setValue(10);
you have not yet done the
 connect(ui.spinBox, QOverload<int>::of(&QSpinBox::valueChanged), 
         this, &tool::setTableRows);
So if you are expecting the slot setTableRows() to be called to actually create rows during the setValue(), it won't be. Why should it?

Either connect the signals before setting the values, or call the slots explicitly first time before you start out.
ok understood
Is that what you mean?
yes
thanks