已解决

Question

我有一个鲜花（属，种，名）的已连接SQL数据库，其中填充了一个仅显示该信息的表。我只能让属列正确填充，而不能获取种类/名称列。

工作代码-正确显示一列：

update.html

#!/bin/python
import psutil
def status(service):
for proc in psutil.process_iter():
    if proc.name() == service:
       result=True
       return result

尝试对其他人使用for循环会中断页面（不确定原因）：

<!--this is for the table -->
   <div class="container">
       <div class="row">
           <div class="col">
               <table id="table" border="1">
                   <tr>
                       <th class="cell">Genus</th>
                       <th class="cell">Species</th>
                       <th class="cell">Comname</th>
                   </tr>
                   <!--the next line with the python code works as long as you only want the genus information-->
                   {% for g in genus_update %}
                   <tr>
                       <td class="cell">{{g}}</td>
                       <!--<td class="cell">{{g}}</td>-->
                       <!--<td class="cell">{{c}}</td>-->
                   </tr>
                   {% endfor %}
               </table>
           </div>
           <div class="col">
               <!--the right column so that everything is lined up on the left side-->
           </div>
       </div>
   </div>

Python.py：

{% for s in species_update %}
  <tr>
    <td class="cell">{{s}}</td>
  </tr>
{% endfor %}

{% for c in comname_update %}
  <tr>
    <td class="cell">{{c}}</td>
  </tr>
{% endfor %}

已解决

解决方案

html代码：

from flask import Flask, render_template, request, g
import sqlite3

app = Flask (__name__)

# conn = sqlite3.connect('flowers.db')
# c = conn.cursor()

DATABASE = 'flowers.db'
def get_db():
   db = getattr(g, '_database', None)
   if db is None:
       db = g._database = sqlite3.connect(DATABASE)
   return db

@app.teardown_appcontext
def close_connection(exception):
   db = getattr(g, '_database', None)
   if db is not None:
       db.close()

@app.route('/')
def index():
   c = get_db().cursor()
   c.execute('SELECT COMNAME FROM FLOWERS')
   all_flowers = c.fetchall()
   return render_template("index.html", all_flowers=all_flowers)

@app.route('/update')
def update():
   c = get_db().cursor()
   # this just gets the data from the db
   c.execute('SELECT COMNAME FROM FLOWERS')
   comname_update = c.fetchall()
   c.execute('SELECT GENUS FROM FLOWERS')
   genus_update = c.fetchall()
   c.execute('SELECT SPECIES FROM FLOWERS')
   species_update = c.fetchall()
   zipped = zip(genus_update, species_update)
   return render_template("update.html", comname_update=comname_update, genus_update=genus_update, species_update=species_update, zipped=zipped)

@app.route('/profile/<name>')
def profile(name):
   return render_template("profile.html", name=name)


if __name__ == "__main__":
   app.run(debug=True)

python代码：

{% for g, s, c in genus_flowers%}
                   <tr>
                       <td class="cell">{{g}}</td>
                       <td class="cell">{{s}}</td>
                       <td class="cell">{{c}}</td>

                   </tr>
{% endfor %}

Answer 1

我知道在Django（另一个适用于python的网络框架）中，您必须引用对象中的字段，而不仅仅是对象本身。因此，如果您执行Select *，而不是Select'field'：

@app.route('/update')
def update():
   c = get_db().cursor()
   # this just gets the data from the db
   c.execute('SELECT * FROM FLOWERS')
   flowers = c.fetchall()
   zipped = zip(genus_update, species_update)
   return render_template("update.html", flowers=flowers, zipped=zipped)

然后，您可以执行以下操作：

<!--this is for the table -->
   <div class="container">
       <div class="row">
           <div class="col">
               <table id="table" border="1">
                   <tr>
                       <th class="cell">Genus</th>
                       <th class="cell">Species</th>
                       <th class="cell">Comname</th>
                   </tr>
                   <!--the next line with the python code works as long as you only want the genus information-->
                   {% for f in flowers %}
                   <tr>
                       <td class="cell">{{ f.genus }}</td>
                       <td class="cell">{{ f.species }}</td>
                       <td class="cell">{{ f.comname }}</td>
                   </tr>
                   {% endfor %}
               </table>
           </div>
           <div class="col">
           </div>
       </div>
   </div>

Answer 2

我不确定这里到底会发生什么，但是由于我处于这种情况，所以我建议您先测试是否从数据库中获取数据，而不是直接检查Flask所在的部分渲染它。确保也存在要传递的查询数据。

希望这会有所帮助。

Python / SQL不显示连接的数据库到表

已解决

2 个答案: