模型 Django 不可迭代

Posted 2023-02-23

【中文标题】模型 Django 不可迭代

【英文标题】：Models Django not iterable

【发布时间】：2021-12-09 06:48:04

【问题描述】：

我对 Django 模型有一些问题，我想迭代从基于模型的表单的文本字段中获取的数据。但我不能它仍然是一个错误：

TypeError at /kod/start/
'Passwords' object is not iterable
Request Method: POST
Request URL:    http://127.0.0.1:8000/kod/start/
Django Version: 3.2.8
Exception Type: TypeError
Exception Value:    
'Passwords' object is not iterable
Exception Location: /home/natalia/PycharmProjects/django_project/kodszyfru/views.py, line 52, in start
Python Executable:  /home/natalia/PycharmProjects/django_project/venv/bin/python3
Python Version: 3.8.10
Python Path:    
['/home/natalia/PycharmProjects/django_project',
 '/usr/lib/python38.zip',
 '/usr/lib/python3.8',
 '/usr/lib/python3.8/lib-dynload',
 '/home/natalia/PycharmProjects/django_project/venv/lib/python3.8/site-packages']
Server time:    Fri, 22 Oct 2021 16:03:53 +0000

代码模型.py

from django.db import models

# Create your models here.
class Passwords(models.Model):
    text = models.TextField(max_length=2000,null=True)
    cipher = models.CharField(max_length=64)

    def __str__(self):
        return self.text

views.py

def start(request):
    password_form = PasswordsForm(request.POST or None)
    if password_form.is_valid():
        text = password_form.save()
        try:
            for i in text:
                value = (alphabet[i])
                klucz = 10
                new_value = int(value) + klucz
                if new_value > 70:
                    new_value = new_value - 70
                for key, value in alphabet.items():
                    if value == str(new_value):
                        print(key, end='')
            return ''
        except KeyError:
            return ('Nie znaleziono wartości' + i + ' w słowniku!')

            return render(request, 'kod.html', 'kod': kod)

    return render(request, 'start.html', 'password_form': password_form)

我将不胜感激！

娜塔莉亚

【参考方案1】：

text 是 Password 对象，而不是字符串。您应该遍历密码：

password_form = PasswordsForm(request.POST or None)
if password_form.is_valid():
    password = password_form.save()
    try:
        for i in password.text:
            # …
    except KeyError:
        # …

【讨论】：

谢谢你，Willem，你能不能再看看另一个问题？以下？提前谢谢你:)

@NataliaOsysko：请提出一个新问题。通常解决一个问题每个问题。