本文整理汇总了Python中DatabaseController.getAccountPasswordBasedOnPersonIdAndType方法的典型用法代码示例。如果您正苦于以下问题：Python DatabaseController.getAccountPasswordBasedOnPersonIdAndType方法的具体用法？Python DatabaseController.getAccountPasswordBasedOnPersonIdAndType怎么用？Python DatabaseController.getAccountPasswordBasedOnPersonIdAndType使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类DatabaseController的用法示例。
在下文中一共展示了DatabaseController.getAccountPasswordBasedOnPersonIdAndType方法的2个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: checkIfCurrentPersonPasswordIsCorrect
# 需要导入模块: import DatabaseController [as 别名]
# 或者: from DatabaseController import getAccountPasswordBasedOnPersonIdAndType [as 别名]
def checkIfCurrentPersonPasswordIsCorrect(personId, accountType, currentPassword):
accountPasswordRecord = DatabaseController.getAccountPasswordBasedOnPersonIdAndType(personId, accountType)
accountPassword = accountPasswordRecord[0][0]
if currentPassword == accountPassword:
return True
else:
return False示例2: getStudentAccountsList
# 需要导入模块: import DatabaseController [as 别名]
# 或者: from DatabaseController import getAccountPasswordBasedOnPersonIdAndType [as 别名]
def getStudentAccountsList(schoolId, firstName, surname):
listOfStudentAccounts = []
listOfStudentRecords = DatabaseController.getStudentAccountsList(schoolId, firstName, surname)
for eachRecord in listOfStudentRecords:
for index in range(0, eachRecord.__len__()):
listOfStudentAccounts.append(eachRecord[index])
currentPersonId = int(eachRecord[0])
accountPasswordRecord = DatabaseController.getAccountPasswordBasedOnPersonIdAndType(currentPersonId, accountTypes[2])
if accountPasswordRecord.__len__() > 0:
listOfStudentAccounts.append(accountPasswordRecord[0][0])
listOfStudentAccounts.append(True)
else:
listOfStudentAccounts.append("")
listOfStudentAccounts.append(False)
return listOfStudentAccounts