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Python 构造函数无法获取输入参数_Python_Linux_Class_Error Handling_Constructor - Fatal编程技术网

Python 构造函数无法获取输入参数

Python 构造函数无法获取输入参数,python,linux,class,error-handling,constructor,Python,Linux,Class,Error Handling,Constructor,我是Python新手。我一直在关注有关python类的在线教程,但我遇到了一个奇怪的错误 我不知道我到底做了什么 下面是我的代码: class StudentData: "Contains information of all students" studentNumber = 0; def _init_(self,name,age,marks): self.name = name; self.age = age; self.marsk = mar

我是Python新手。我一直在关注有关python类的在线教程,但我遇到了一个奇怪的错误

我不知道我到底做了什么

下面是我的代码:

class StudentData:
  "Contains information of all students"
  studentNumber = 0;
  def _init_(self,name,age,marks):
      self.name = name;
      self.age = age;
      self.marsk = marks;
  def displayStudentNUmber(self):
      print 'Total Number of students = ',studentNumber;
  def displayinfo(self):
      print 'Name of the Student: ',name;
      print 'Age of the Student: '.age;
      print 'Marks of the Student: '.marks;
student1 = StudentData('Ayesha',12,90)
student2 = StudentData('Sarah',13,89)
print "Student number in case of student 1",student1.displayStudentNumber();
print "Information of the Student",student1.dispalyinfo();
print "Student number in case of student 1",student2.displayStudentNumber();
print "Information of the Student",student2.dispalyinfo();
下面是错误

回溯(最近一次调用last):文件“main.py”,第14行,在 student1=StudentData('Ali',12,90)类型错误:此构造函数采用 没有争论

有人能解释我为什么会犯这个错误吗

很抱歉提出了一个蹩脚的问题:(

\uuuu init\uuuu()
应该在两边都有两个下划线,因此python将您的函数视为类中的普通函数

更改:

def _init_(self,name,age,marks):

\uuuu init\uuuu()
的两边应该有两个下划线,因此python将您的函数视为类中的普通函数

更改:

def _init_(self,name,age,marks):