Setting same instance of widget not working

Question

I have a pyqt application and when using the same instance of the QLabel class on the gridlayout it is not working. I see only one instance being displayed.

import sys

from PyQt5.QtWidgets import QApplication, QWidget, QToolTip, QPushButton, QMessageBox, QAction, qApp, QMenu, QTextEdit, QToolBar, QMdiArea, QGridLayout, QLabel, QDialog
from PyQt5.QtGui import QFont, QIcon, QPainter, QVector2D
from PyQt5.Qt import QDesktopWidget, QMainWindow, Qt, QHBoxLayout, QVBoxLayout,\
    QLineEdit
from PyQt5.QtCore import QPoint


class Example(QDialog):
    count = 0
    def __init__(self, *args, **kwargs):
        super(Example, self).__init__(*args, **kwargs)            
        self.initUI()


    def initUI(self):
        grid = QGridLayout()
        a1 = QLabel('alphanumeric characters')
        a2 = QLabel('alphanumeric characters')

        grid.addWidget(QLabel('Name'), 1, 0)
        grid.addWidget(QLineEdit(), 1, 1)
        grid.addWidget(QLabel('only alphanumeric'), 1, 2)
        grid.addWidget(QLabel('Street1'), 2, 0)
        grid.addWidget(QLineEdit(), 2, 1)
        grid.addWidget(QLabel('only alphanumeric'), 2, 2)
        grid.addWidget(QLabel('Street2'), 3, 0)
        grid.addWidget(QLineEdit(), 3, 1)
        grid.addWidget(QLabel('only alphanumeric'), 3, 2)
        grid.addWidget(QLabel('City'), 3, 0)
        grid.addWidget(QLineEdit(), 3, 1)
        grid.addWidget(QLabel('only alphanumeric'), 3, 2)

        self.setLayout(grid)
        self.setGeometry(500, 500, 500, 500)
        self.setWindowTitle('Lines')
        self.show()


if __name__ == '__main__':

    app = QApplication(sys.argv)
    ex = Example()
#     ex.show()
    sys.exit(app.exec_())

typo: change `grid.addWidget(a1, 1, 0) grid.addWidget(a1, 1, 1)` to `grid.addWidget(a1, 1, 0) grid.addWidget(a2, 1, 1)` — eyllanesc, Jul 21 '18 at 20:55
Can't I use the same instance - a1? My grid is going to have 3 columns and multiple rows..Most of the 3rd column will just contain informational msg - 'only alphanumeric characters" — user1050619, Jul 21 '18 at 20:57
You could explain yourself better, try changing both lines to: grid.addWidget(a1, 1, 0, 2, 1) — eyllanesc, Jul 21 '18 at 21:00
I have edited the code, if you notice I have create multiple QLabel instance just to display 'only alphanumeric' text on column 3 of the grid. — user1050619, Jul 21 '18 at 21:13
okay, that's another thing, your initial example was very confusing that made me think he was a typo. — eyllanesc, Jul 21 '18 at 21:14

score 1 · Accepted Answer · answered Jul 21 '18 at 21:21

First the coordinates start from (0, 0), so that part I have corrected.

Going to the problem, you do not have to add one to one, the addWidget() method is overloaded, so there is a 4 and 5 argument that indicates the span in the row and column, respectively.

void addWidget(QWidget *widget, int row, int column, Qt::Alignment alignment = ...)

void addWidget(QWidget *widget, int fromRow, int fromColumn, int rowSpan, int columnSpan, Qt::Alignment alignment = ...)

import sys

from PyQt5.QtWidgets import QApplication, QDialog, QGridLayout, QLabel, QLineEdit


class Example(QDialog):
    def __init__(self, *args, **kwargs):
        super(Example, self).__init__(*args, **kwargs)            
        self.initUI()

    def initUI(self):
        grid = QGridLayout(self)
        a1 = QLabel('alphanumeric characters')
        a2 = QLabel('alphanumeric characters')

        grid.addWidget(QLabel('Name'), 0, 0)
        grid.addWidget(QLineEdit(), 0, 1)
        grid.addWidget(QLabel('Street1'), 1, 0)
        grid.addWidget(QLineEdit(), 1, 1)
        grid.addWidget(QLabel('Street2'), 2, 0)
        grid.addWidget(QLineEdit(), 2, 1)
        grid.addWidget(QLabel('City'), 3, 0)
        grid.addWidget(QLineEdit(), 3, 1)

        grid.addWidget(QLabel('only alphanumeric'), 0, 2, 4, 1)

        self.setGeometry(500, 500, 500, 500)
        self.setWindowTitle('Lines')
        self.show()


if __name__ == '__main__':

    app = QApplication(sys.argv)
    ex = Example()
#     ex.show()
    sys.exit(app.exec_())

Setting same instance of widget not working

1 Answers1