如何从输入的搜索值中获得特定对象的详细信息?

Question

我正在做一个大学项目，遇到了一个障碍。我希望程序能够在"employee.txt"中搜索特定对象。根据已输入的搜索值列表。我不允许使用SQL，只有文本文件。我的导师不知道如何解决这个问题，所以我来请求您的帮助!

我试过了:

empList = open("employee.txt")
def add_emp():
empSurname = txtSurname.get()
empForename = txtForename.get()
empPhone = txtPhone.get()
empGender = comboGender.get()
empDOB = txtDOB.get()
empEmail = txtEmail.get()
empHouseNum = txtHouseNum.get()
empStreetName = txtStreetName.get()
empTC = txtTc.get()
empCounty = txtCounty.get()
empPostcode = txtPostcode.get()
Employee(empSurname, empForename, empGender, empPhone, empDOB, empEmail, empHouseNum, empStreetName, empPostcode, empTC, empCounty)
for i in empList:
if i.surname == empSurname:
Details = str(i)
print(Details)

然而，我得到了一个属性错误，它指出"str对象不包含属性'姓'"这是我的员工文件:

class Employee():
def __init__(self, surname, forename, gender, phoneNum, DOB, email, houseNum, streetName, postcode,     tc, county):
self.surname = surname
self.forename = forename
self.phoneNum = phoneNum
self.DOB = DOB
self.email = email
self.houseNum = houseNum
self.streetName = streetName
self.postcode = postcode
self.tc = tc
self.county = county

CAPL = self.surname.upper()
CAPF = self.forename.upper()

count = 1
for i in employeeList:
if i[:4] == CAPL[:3] + CAPF[:1]:
count+= 1

empID = CAPL[:3] + CAPF[:1] + str("{:02d}".format(count))
self.empID = empID

employeeA = open("employee.txt", "a")
employeeA.write(empID + " " + surname + " " + forename + " " + gender + " " + phoneNum + " " + DOB + " " + email + " " + houseNum + " " + streetName + " " + postcode + " " + tc + " " + county + "n")
employeeA.close()

非常感谢你的帮助!

Answer 1

您需要将文件每行的第二个字段与输入的姓氏进行比较。

如果您遍历整个文件而没有找到匹配项，那么您可以创建新员工，并将其添加到文件中。

def add_emp():
empSurname = txtSurname.get()
empForename = txtForename.get()
empPhone = txtPhone.get()
empGender = comboGender.get()
empDOB = txtDOB.get()
empEmail = txtEmail.get()
empHouseNum = txtHouseNum.get()
empStreetName = txtStreetName.get()
empTC = txtTc.get()
empCounty = txtCounty.get()
empPostcode = txtPostcode.get()
with open("employee.txt") as empList:
for i in empList:
fields = i.split()
if len(fields) < 2:
continue # skip lines that don't have enough fields
oldSurname = i[1]
if oldSurname == empSurname:
print(f"Employee {empSurname} already exists")
print(Details)
break
else:
new_emp = Employee(empSurname, empForename, empGender, empPhone, empDOB, empEmail, empHouseNum, empStreetName, empPostcode, empTC, empCounty)
return new_emp