Question

学习“利用django学习网站开发”。在第3章,在建立图书标识数据模型时,我遵循了除“外”以外的指示和代码。

from django.contrib.auth.models import User
class Bookmark(models.Model):
title = models.CharField(maxlength=200)
user = models.ForeignKey(User)
link = models.ForeignKey(Link)

一、导言

title = models.CharField(maxlength=200)

页: 1

title = models.CharField(max_length=200)

我正在收到错误信息。之后,我管理着 p。 py syncdb, 然后,python管理。 py sql bookmarks. 当我检查。我收到这一错误信息:

Request Method:     GET
Request URL:    http://127.0.0.1:8000/user/j/
Django Version:     1.3
Exception Type:     TemplateSyntaxError
Exception Value:    

**Caught DatabaseError while rendering: no such column: bookmarks_bookmark.title**

Looking this up, I learned that sqlite3, the database i was using, cannot find bookmarks_bookmark.title. I went back 页: 1the book to make sure I have everything copied correctly, and I did except for that part that I changed (max_length). When I ran python manage.py sql bookmarks, it gave me--

BEGIN;
CREATE TABLE "bookmarks_bookmark" (
"id" integer NOT NULL PRIMARY KEY,
"title" varchar(200) NOT NULL,
"user_id" integer NOT NULL REFERENCES
"auth_user" ("id"),
"link_id" integer NOT NULL REFERENCES
"bookmarks_link" ("id"),
);
CREATE TABLE "bookmarks_link" (
"id" integer NOT NULL PRIMARY KEY,
"url" varchar(200) NOT NULL UNIQUE
);
COMMIT;

我如何确定这一点? 感谢!

Answer 1

手册删除工作。

sqlite3 yourdb.db 
> drop table bookmarks_bookmark;
> .quit
cd yourpythonproj
python2 manage.py syncdb
python2 manage.py runserver

bash 写

#!/bin/bash
sqlite3 ../yourdb.db  drop table yourtable 
python2 manage.py syncdb

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