Python skipping my if and else statement in a for loop, but the for loop is running-CodePudding

Well im still learning and im now into pyqt5 designer but i have a problem with a for and if statement where python dont do neither of the else or if, i use the print (empleados[i]["Nombre"]), to check if the else is running but none of the if or else run

        for i in range(0,len(empleados)):
            print(empleados[i]["Numero"])
            if self.ui.num.text() == empleados[i]["Numero"]:
                QtWidgets.QMessageBox.information(self,"Advertencia","No se pueden registrar dos numeros de empleado iguales")    
            else:
                
                empleados.append({"Nombre":self.ui.nom.text(),
                                  "Numero":self.ui.num.text(),
                                  "Salario":self.ui.salario.text(),
                                  "Dep":self.ui.dep.currentText(),
                                  "FechaC":self.ui.dateC.date(),
                                  "FechaN":self.ui.dateN.date()})
                print (empleados[i]["Nombre"])
                break

CodePudding user response：

Is the intent to append a new element to empleados if and only if there isn't any with a matching Numero? If so I think this'll do the trick:

if any(e["Numero"] == self.ui.num.text() for e in empleados):
    QtWidgets.QMessageBox.information(
        self,
        "Advertencia",
        "No se pueden registrar dos numeros de empleado iguales"
    )
else:
    empleados.append({
        "Nombre":self.ui.nom.text(),
        "Numero":self.ui.num.text(),
        "Salario":self.ui.salario.text(),
        "Dep":self.ui.dep.currentText(),
        "FechaC":self.ui.dateC.date(),
        "FechaN":self.ui.dateN.date()
    })

In general, modifying a list as you're iterating over it doesn't work well, but in this case I don't think you want to do the append inside the loop anyway, since you only want to append one element.

If Numero is unique, you might consider making empleados a dictionary keyed on that field, e.g.:

emp_by_num = {e["Numero"]: e for e in empleados}

so that you can do quick checks with in rather than having to use any:

if self.ui.num.text() in emp_by_num: