尝试重现(以最简单的方式)Maciej Cegłowski 的 http://pinboard.in ;我有书籍和标签,而不是链接和标签。每本Book
都可以用任意数量的Tag
进行标记,并且一个Tag
与许多Book
相关联。 s。
class Book(db.Model):
__tablename__ = 'books'
book_id = db.Column(db.Integer, primary_key=True)
title = db.Column(db.String(120), unique=True)
auth = db.Column(db.String(120), unique=True)
comment = db.Column(db.String(120), unique=True)
date_read = db.Column(db.DateTime)
era = db.Column(db.String(36))
url = db.Column(db.String(120))
notable = db.Column(db.String(1))
tagged = db.relationship('Tag', secondary=assoc, backref=db.backref('thebooks',lazy='dynamic'))
class Tag(db.Model):
__tablename__ = 'tags'
tag_id = db.Column(db.Integer, primary_key=True)
tag_name = db.Column(db.String(120))
def construct_dict(query):
books_dict = {}
for each in query: # query is {<Book object>, <Tag object>} in the style of assoc table - therefore, must make a dictionary bc of the multiple tags per Book object
book_data = books_dict.setdefault(each[0].book_id, {'bookkey':each[0], 'tagkey':[]}) # query is a list like this {index-book_id, {<Book object>}, {<Tag object #1>, <Tag object #2>, ... }}
book_data['tagkey'].append(each[1])
return books_dict
@app.route('/query')
def query():
query = db.session.query(Book, Tag).outerjoin('tagged') # query to get all books and their tags
books_dict = construct_dict(query)
return render_template("query.html", query=query, books_dict=books_dict)
这是我开始有点迷失的地方;也就是说,构建正确的逻辑来处理我想要做的事情,这将在下面详细描述。
{% for i in books_dict %}
<a href="{{books_dict[i].bookkey.url}}" target="_blank">{{books_dict[i].bookkey.title}}</a>
{% for i in books_dict[i].tagkey %} # tagkey is a list of Tag objects; for each Book's tagkey, print each Object's tag_name
<a href="/tag/{{i.tag_name}}" class="tag-link">{{i.tag_name}}</a>
{% endfor %}
<a href="" class="edit">edit</a> # eventually, edit link will display the form
<form method="add_tag_to_book">
{% for j in books_dict[i].tagkey %}
<input type="text" name="tag" value="{{j.tag_name}}" />
{% endfor %}
<input type="submit" value="save">
</form>
{% endfor %}
对于任何一本书,用户(目前只有我)应该能够:
更新
assoc
表,以便在 Book 实例和 Tag 实例之间创建新的关联;- 或者,如果 Tag 不存在,则创建一个新的 Tag 实例(当然,还要
UPDATE
assoc
表,以便 Book 实例与新标签)
我认为这个任务对我来说很复杂,因为我仍在为 Jinja 循环的范围而苦苦挣扎。然而,我意识到我需要做这样的事情:
获取用户的输入;检查
__tablename__ = "tags"
中是否已存在如果
tag_name
已存在,则获取其tag_id
以及 Book 实例的book_id
并将行添加到assoc
表(即 book_id|tag_id)如果
tag_name
不存在,则创建新的Tag()
实例,然后执行步骤2
tag_name
最佳答案
一、构建book_dict
字典:
def construct_dict(query):
books_dict = {}
for each in query: # query is {<Book object>, <Tag object>} in the style of assoc table - therefore, must make a dictionary bc of the multiple tags per Book object
book_data = books_dict.setdefault(each[0].book_id, {'bookkey':each[0], 'tagkey':[]}) # query is a list like this {index-book_id, {<Book object>}, {<Tag object #1>, <Tag object #2>, ... }}
book_data['tagkey'].append(each[1])
return books_dict
@app.route('/query')
def query():
query = db.session.query(Book, Tag).outerjoin('tagged') # query to get all books and their tags
books_dict = construct_dict(query)
return render_template("query.html", query=query, books_dict=books_dict)
然后,除了打印每个 Book
实例位于 book_dict
(并列出本书的相关 Tag
对象),我们为每个 Book
创建一个表单允许用户关联新的 Tag
的实例是 Book
:
{% for i in books_dict %}
<a href="{{books_dict[i].bookkey.url}}">{{books_dict[i].bookkey.title}}</a>,
{{books_dict[i].bookkey.auth}}
{% for i in books_dict[i].tagkey %}
<a href="/tag/{{i.tag_name}}" class="tag-link">{{i.tag_name}}</a>
{% endfor %}
<form action="{{ url_for('add_tag_to_book') }}" method=post class="edit-form">
<input type="hidden" name="book_id" value={{books_dict[i].bookkey.book_id}} />
<input type="text" name="tag_name" value="" />
<input type="submit" value="save">
</form>
{% endfor %}
...可见的<input>
将采用用户输入的值,其中有 name="tag_name"
;提交表单后,路线 /add_tag_to_book
路线被称为。从表格中,我们抓取 book_id
(以表单打印但不可见,即 <input type="hidden" name="book_id" value={{books_dict[i].bookkey.book_id}} />
);我们还获取 <input>
的值元素为 name="tag_name"
):
@app.route('/add_tag_to_book', methods=['POST'])
def add_tag_to_book():
b = request.form['book_id']
t = request.form['tag_name']
接下来,我们应该检查是否 tag_name
用户提交的已经是Tag()
; Python 返回 None
如果在 Tag
中未找到 tag_name table ;否则,它将返回 Tag
具有 tag_name=t
的对象(即tag_name
由用户提交);如果tag_object == None
,我们需要创建 Tag()
的新实例使用tag_name
用户提供:
tag_object = Tag.query.filter_by(tag_name=t).first()
if tag_object == None:
new_tag = Tag(tag_name=t)
db.session.add(new_tag)
db.session.commit()
tag_object = Tag.query.filter_by(tag_name=t).first()
tag_object_id = tag_object.tag_id
此时,我们将有一个 tag_object
(无论是新创建的还是之前在我们的 Tag
表中创建的)其 tag_id
我们可以抓取并插入到我们的关联表中,以及 book_id
对于Book
目的。接下来,我们创建一个数据库连接,插入book_id
& tag_id
,提交到数据库,然后将用户返回到query
页面:
conn = db.session.connection()
ins = assoc.insert().values(book_id=b,tag_id=tag_object_id)
result = conn.execute(ins)
db.session.commit()
return redirect(url_for('query'))
把它们放在一起,完整的 @app.route('/add_tag_to_book')
看起来像这样:
@app.route('/add_tag_to_book', methods=['POST'])
def add_tag_to_book():
b = request.form['book_id']
t = request.form['tag_name']
tag_object = Tag.query.filter_by(tag_name=t).first()
if tag_object == None:
new_tag = Tag(tag_name=t)
db.session.add(new_tag)
db.session.commit()
tag_object = Tag.query.filter_by(tag_name=t).first()
tag_object_id = tag_object.tag_id
conn = db.session.connection()
ins = assoc.insert().values(book_id=b,tag_id=tag_object_id)
result = conn.execute(ins)
db.session.commit()
return redirect(url_for('query'))
关于python - 如何通过HTML表单插入关联表?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42840641/