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我在制作动态URI的时候需要传递参数。当我尝试传递一个参数时,出现了类型错误消息。它一直说我传递了0个参数。我找不到问题。Flask:TypeError,无法传递参数
这是views.py
@app.route('/delete_todo/<int:todo_id>', methods=['POST'])
@login_required
def delete_todo(todo_id):
if request.form['todo_id']:
g.db.execute('''delete * from todo where todo_id = ?''', request.form['todo_id'])
g.db.commit()
flash('Your message was deleted')
return redirect(url_for('index'))
范本中应该经过论证
{% for todo in todo %}
<form action="{{ url_for('delete_todo', todo_id=todo.todo_id) }}" method="post">
<input type="hidden" name="todo_id" value="{{ todo.todo_id }}">
<input type="submit" value="Delete">
</form>
{% else %}
{% endfor %}
这是错误消息。
TypeError
TypeError: _wrapped_view() takes at least 1 argument (0 given)
Traceback (most recent call last)
File "C:\Python27\lib\site-packages\flask\app.py", line 1994, in __call__
return self.wsgi_app(environ, start_response)
File "C:\Python27\lib\site-packages\flask\app.py", line 1985, in wsgi_app
response = self.handle_exception(e)
File "C:\Python27\lib\site-packages\flask\app.py", line 1540, in handle_exception
reraise(exc_type, exc_value, tb)
File "C:\Python27\lib\site-packages\flask\app.py", line 1982, in wsgi_app
response = self.full_dispatch_request()
File "C:\Python27\lib\site-packages\flask\app.py", line 1614, in full_dispatch_request
rv = self.handle_user_exception(e)
File "C:\Python27\lib\site-packages\flask\app.py", line 1517, in handle_user_exception
reraise(exc_type, exc_value, tb)
File "C:\Python27\lib\site-packages\flask\app.py", line 1612, in full_dispatch_request
rv = self.dispatch_request()
File "C:\Python27\lib\site-packages\flask\app.py", line 1598, in dispatch_request
return self.view_functions[rule.endpoint](**req.view_args)
TypeError: _wrapped_view() takes at least 1 argument (0 given)
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为什么在使用'POST'方法时在URL中传递参数? – stamaimer
已经可以从request.form ['todo_id']''todo_id''为什么使用动态url? – metmirr