flask-pagination : 未返回的行数 |分页不起作用答案

【问题标题】：flask-pagination : Number of rows not returned | Pagination doesn't workflask-pagination : 未返回的行数 |分页不起作用
【发布时间】：2016-06-02 06:09:30
【问题描述】：

我无法成功地对记录进行分页。 This is the link from where I got the flask-pagination help.

我的视图函数是这样的（注意：我只展示了相关代码）：

from flask import Blueprint
from flask.ext.paginate import Pagination

mod = Blueprint('runserver', __name__)

@app.route('/home/all-puppies/<int:shelter_id>/<q>')
def showAllPuppies():
    search = False
    q = request.args.get('q')
    if q:
        search = True
    try:
        page = int(request.args.get('page', 1))
    except ValueError:
        page = 1

    puppies = session.query(Puppy, Shelter).join(Shelter).filter(Shelter.id == shelter_id).all()

    pagination = Pagination(page=page, total=puppies.count(puppies), search=search, record_name='puppies')
    return render_template('allpuppies.html',
                           puppies=puppies,
                           pagination=pagination,
                           )

我的 allpuppies.html 看起来像这样：

{% extends "master.html" %}

{% block title %}All Puppies{% endblock %}

{% block body %}
{{ pagination.info|safe }}
{{ pagination.links }}
<table class="table table-condensed" align:"center">
  <thead>
     <tr>
      <th>#</th>
      <th>Name</th>
      <th>Gender</th>
      <th>Weight</th>
      <th>D.O.B</th>
      <th>Shelter</th>
      <th>Location</th>
      <th>Details</th>
    </tr>
  </thead>
  <tbody>
    {% for puppy in puppies %}
      <tr>
        <td>1</td>
         <td>{{ puppy.Puppy.name }}</td>
            <td>{{ puppy.Puppy.gender }}</td>
            <td>{{ puppy.Puppy.weight }}</td>
            <td><b>{{ puppy.Puppy.dateOfBirth }}<b></td>
            <td>{{ puppy.Shelter.name }}</td>
            <td>{{ puppy.Shelter.city }},{{ puppy.Shelter.state}}</td>
            <td><a href='{{ url_for('showPuppyDetails', puppy_id = puppy.Puppy.id ) }}'> Details </a></td>
      </tr>
    {% endfor %}
  </tbody>
</table>
{{ pagination.links }}
{% endblock %}

编辑：1 我更改了查询以仅在特定避难所中过滤掉puppies。为此，我将@app.route() 更改为将shelter_id 也用作查询字符串参数。我还在路由中添加了<q>，因为该视图函数的第一行是q = request.args.get('q')。但是当我做这个改变时，它给了我一个 BuildError 如下：

werkzeug.routing.BuildError

BuildError: ('showAllPuppies', {'shelter_id': 2}, None)

【问题讨论】：

标签： python pagination flask-sqlalchemy

【解决方案1】：

答案有点晚了，但总数是一个单独的查询，将模块用作关键字。我会尝试单独查询。

total = puppies = session.query(Puppy, Shelter).join(Shelter).filter(Shelter.id == shelter_id).count()

然后将 total 作为 total=total 传递给 Pagination

【讨论】：