【发布时间】:2015-12-06 22:49:12
【问题描述】:
我有一个简单的 Flask Web 应用程序,它使用 peewee 来管理 MySQL 连接。不幸的是,我似乎错过了我的代码中的一些重要内容,因此在使用该网站 10 到 20 分钟后,我收到了 pymysql.err.OperationalError: (2006, "MySQL server has gone away.. 错误。如果我重新启动应用程序,它会再次正常工作,所以我认为我处理连接/连接错误。
我有一些基本查询用于在我的 UI 上显示列表。由于我是 Python 新手,因此我的整个逻辑可能是错误的,所以如果有人能解释在我的情况下使用 peewee 管理(连接-关闭连接)查询的正确方法是什么(简单的网站没有任何用户函数)。
你可以在下面的代码中看到我现在是怎么做的。在我连接之前很好,但是在我连接到数据库之后,我只是调用查询而不处理连接。我假设我应该根据已调用的查询关闭连接,但找不到正确的方法。调用 myDB.close() 是不够的,或者我用错了。
# this is how I connect to the MySQL
import peewee as pw
from flask import Flask, request, session, g, redirect, url_for, abort, render_template, flash
myDB = pw.MySQLDatabase("", host="", user="", passwd="")
class MySQLModel(pw.Model):
"""A base model that will use our MySQL database"""
class Meta:
database = myDB
class city(MySQLModel):
...
class building(MySQLModel):
...
myDB.connect()
# this is how I manage the homepage
cityList = []
cityList = city.select().order_by(city.objectId).limit(15)
buildings = []
buildings = building.select().order_by(building.buildingName).limit(180)
def getSearchResult (content_from_url):
searchResultList = []
searchResultList = city.select().where(city.objTitle.contains(content_from_url))
return searchResultList
@app.route('/', methods=['GET', 'POST'])
def main_page():
search = request.args.get('search')
if search:
return render_template("results.html", search = getSearchResult(search), str = search)
else:
return render_template("home.html", name=cityList, buildList=buildings)
# and this is how the subpage
relatedCityList = []
slugObj = []
buildings2 = []
buildings2 = building.select().order_by(building.buildingName).limit(180)
def getLink (title_for_link):
slugObj = city.get(city.urlSlug == title_for_link)
return slugObj
def displayRelatedCity (city_to_filter):
resultCity = city.get(city.urlSlug == city_to_filter)
relatedCityList = city.select().where(city.objTitle == resultCity.objTitle).limit(20)
return relatedCityList
@app.route('/city-list/<content>', methods=['GET', 'POST'])
def city_page(content):
linkText = getLink(content)
return render_template("details.html", relatedCity = displayRelatedCity(content), buildingName = linkText.buildingName, buildz = buildings2)
myDB.close()
【问题讨论】: