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Basic Communications with QSerialPort

I am trying to rig up some basic serial communications in QT I am getting the port COM19 from QSerialPortInfo, and I speaking successfully to the port via Arduino. However, I cannot get anything back via QT.

#include "mainwindow.h"
#include <QApplication>
#include <QDebug>
#include <QTextStream>
#include <QFile>
#include <QtSerialPort/QSerialPort>
#include <QtSerialPort/QSerialPortInfo>

int main(int argc, char *argv[])
{
    QApplication a(argc, argv);

    foreach(const QSerialPortInfo &info, QSerialPortInfo::availablePorts()){
        qDebug() << "Name  :" << info.portName();
        qDebug() << "Description  :" << info.description();
        qDebug() << "Manufactuer :"  << info.manufacturer();

        QSerialPort serial;
        serial.setPort(info);

        if(serial.open(QIODevice::ReadWrite))
            qDebug() << serial.errorString();

            serial.write("M114 \n");

            qDebug() << serial.readAll();

            serial.close();
            // Now we need to send and receive commands

            serial.setPortName("COM19");
            serial.setBaudRate(QSerialPort::Baud57600);
            serial.setDataBits(QSerialPort::Data8);
            serial.setParity(QSerialPort::NoParity);
            serial.setStopBits(QSerialPort::OneStop);
            serial.setFlowControl(QSerialPort::NoFlowControl);

            if(serial.open(QIODevice::ReadWrite)){
                qDebug() << "opened";
              }else{
                qDebug() << "Not opened";
            }
           qDebug() << serial.errorString();

            serial.write("M114 \n");
            qDebug() << serial.readAll();

            serial.close();


    }





    MainWindow w;
    w.show();

    return a.exec();
}

As you can see, I am trying a simple connection along the lines of the documentation, and one where I write out all the baud rate information. They throw two different errors.

Like I said, I am connecting via arduino to this same port and having success. Any ideas what's wrong?

Name  : "COM19" 
Description  : "USB Serial (Communication Class, Abstract Control Model)" 
Manufactuer : "PJRC.COM, LLC." 
"Unknown error" 
"" 
opened 
"The handle is invalid." 
"" 

Any ideas for what I am doing wrong?

My idea is to send commands to the device, and read them back to the console.

like image 977
MrSynAckSter Avatar asked Feb 19 '26 11:02

MrSynAckSter


1 Answers

the code looks a bit confuse. You open all port available and then you try to do something wrong.

NOTE: You use a GUI application like a shell application. It is wrong.

Try:

#include "mainwindow.h"
#include <QApplication>
#include <QDebug>
#include <QTextStream>
#include <QFile>
#include <QtSerialPort/QSerialPort>
#include <QtSerialPort/QSerialPortInfo>

int main(int argc, char *argv[])
{
    QApplication a(argc, argv);

    QSerialPort serial;
    serial.setPortName("COM19");

    if(serial.open(QIODevice::ReadWrite)){
        //Now the serial port is open try to set configuration
        if(!serial.setBaudRate(QSerialPort::Baud57600))
            qDebug()<<serial.errorString();

        if(!serial.setDataBits(QSerialPort::Data8))
            qDebug()<<serial.errorString();

        if(!serial.setParity(QSerialPort::NoParity))
            qDebug()<<serial.errorString();

        if(!serial.setStopBits(QSerialPort::OneStop))
            qDebug()<<serial.errorString();

        if(!serial.setFlowControl(QSerialPort::NoFlowControl))
            qDebug()<<serial.errorString();

        //If any error was returned the serial il corrctly configured

        serial.write("M114 \n");
        //the serial must remain opened

        if(serial.waitForReadyRead(100)){
            //Data was returned
            qDebug()<<"Response: "<<serial.readAll();
        }else{
            //No data
            qDebug()<<"Time out";
        }

        //I have finish alla operation
        serial.close();
    }else{
        qDebug()<<"Serial COM19 not opened. Error: "<<serial.errorString();
    }

    MainWindow w;
    w.show();

    return a.exec();
}
like image 138
Elia Avatar answered Feb 22 '26 01:02

Elia



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