Question

我正在使用Jinja模板引擎在Flask中创建一个网站。我有一份公告清单，但没有好办法对它们进行排序。

他们现在正在由id调用的表中。我想先按最新排序（这意味着最高的id）

像：

P1
- 5
- 4
- 3
P2
- 2
- 1

在我的HTML中，我已经为Jinja提出了这条指令：

<div id="main"> 

{% autoescape false %}
{% for announcement in announcements.items|sort(attribute='id', reverse = True) %}

  <a name={{announcement.title}}></a>
  <h1>{{announcement.title}} - Posted on: {{announcement.date}}</h1>
  {{announcement.body}}
  <br/>
  <hr/>

{% endfor %}
{% endautoescape %}

 <p class = "footer">{% if announcements.has_prev %}<a href="{{ url_for('index', page = announcements.prev_num) }}"><< Newer posts</a>{% else %}<< Newer posts{% endif %} | 
 {% if announcements.has_next %}<a href="{{ url_for('index', page = announcements.next_num) }}">Older posts >></a>{% else %}Older posts >>{% endif %} </p>

<br/>
</div>

蟒蛇：

from flask import render_template, Markup
from app import app
from config import POSTS_PER_PAGE
from models import Announcement, VideoAnnouncement

@app.route('/')
@app.route('/index')
@app.route('/index/<int:page>')
def index(page = 1):
    loggedOut = True
    announcements = Announcement.query.paginate(page, POSTS_PER_PAGE, True)
    videoAnnounce = VideoAnnouncement.query.all()
    return render_template("index.html", announcements = announcements, videoAnnouncements = videoAnnounce , loggedOut = loggedOut)

但是那样的话：

P1
- 3
- 2
- 1
P2
- 5
- 4

无论如何要按降序对它进行排序，不会被网页弄乱？

（我希望我的问题有道理）

Answer 1

采取gnibbler的建议。我在分页之前重新排序了查询。

更新的代码：

from flask import render_template, Markup
from app import app
from config import POSTS_PER_PAGE
from models import Announcement, VideoAnnouncement

@app.route('/')
@app.route('/index')
@app.route('/index/<int:page>')
def index(page = 1):
    loggedOut = True
    query = Announcement.query.order_by(Announcement.id.desc())
    announcements = query.paginate(page, POSTS_PER_PAGE, True)
    videoAnnounce = VideoAnnouncement.query.all()
    return render_template("index.html", announcements = announcements, videoAnnouncements = videoAnnounce , loggedOut = loggedOut)

按降序Jinja排序列表

1 个答案: