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1.解决办法

crontab要设置正确，因为1.9.1之后订单邮件是任务计划队列发出

SSH
sudo crontab -e
#add
 * * * * * /bin/sh /www/wwwroot/path/magento/cron.sh

按照你的路径来修改上方，推荐Aoe_Scheduler 插件

https://github.com/AOEpeople/Aoe_Scheduler

此外，stackoverflow 有人给出直接修改成不用队列发送的，如果修改不了crontab也可以考虑

连接如下

1. 修改email template （推荐）
2. 修改order
3. BSS
4. 官方论坛的帖子1 帖子2

1. 上网问题

使用的CentOS-7-x86_64-Minimal-1908，输入ifconfig的时候提示，没有此命令。

vi /etc/sysconfig/network-scripts ifcfg-ens33

把 ONBOOT属性更改为yes ，NAT模式下（我使用的）就获取到IP 了

2.install 问题 could not resolve host

网络通了以后换源，我使用的是aliyunhttps://developer.aliyun.com/mirror/centos

官方的知道方案，比较详细

步骤一，生成CSR 和私钥

https://www.chinassl.net/ssltools/generator-csr.html

当然有其他家都可以，这一步得到CSR和密钥，就是宝塔证书的左栏（ 密钥(KEY) ）

步骤二，拿着CSR去godaddy 申请证书（如果你域名在godaddy上就不用这样，直接就生成了）

步骤三，下载证书crt文件，选择是一串数字和字母的，不要包含bundle字样的，编辑代码，获取就是 宝塔右栏（ 证书(PEM格式) ）

最后，保存生效就可以了

参看链接：

https://www.bt.cn/bbs/forum.php?mod=viewthread&tid=704&highlight=ssl%E8%AF%81%E4%B9%A6

统计学理论学完后，就到了技术部分，包含Python和SQL（使用的是Postgre）

今天先回顾学过的Python部分：

1.Data Types And Operators数据类型和操作符

首先学的是操作符，简单的加减乘除就和普通一样，比较特殊的是

** 求幂运算，简单来说就是次方，3的6次方 3**6

// 向下整数除法 7//2 =3   -7//2=-4

% 求余数

负值是等号，==是判断是否相等

+= -= *=

整数和浮点数

type() 检查参数类型

int(49.7) =49 直接舍去

报错会有两种 Exception 和error

exception是运行时报错

error是运行前，检查出来的错误。

布尔值true false、逻辑运算and or not

接下来是比较重要的难点

字符串string、列表list、元组tuples、set集合、字典 dictionary

字符串：有序，不可修改

单引号来包含双引号

\ 来转义

+ 来合并字符串，*来复制，没有减法和除法

len()/format()

a.count(b)/a.find(b)

list列表 可排序可改变，使用广泛

names = [“Carol”, “Albert”, “Ben”, “Donna”]

names[0]第一个 ；names[-1]最后一个，[len(string)-1]

month[6:9]  7-9月份，不包含上限。

[:6]前6个元素，不包含7

in 和not in判断

max() 最大的int，字符串首字母靠后的，多重type无法比较-min()

sorted()从下到大排序

sorted(list,reverse=True)从大到小排序

“-“.join(list)连接字符串

b.append(‘one’)添加到列表末尾 直接使用

b=names.pop(1) 提出第二个元素

元组Tuples 包含不可变 有序的数据结构，括号可选

没有append和pop方法

tup1[0] 方括号索引查找

内置4个函数，cmp比较两个元组，max,min和tuple(listname)转化

location = (13.4125, 103.866667)
print("Latitude:", location[0])
print("Longitude:", location[1])

元组解包，就是把元组内的参数赋值，下方第二行就是解包

dimensions = 52, 40, 100
length, width, height = dimensions
print("The dimensions are {} x {} x {}".format(length, width, height))

也可以写成

length, width, height = 52, 40, 100
print("The dimensions are {} x {} x {}".format(length, width, height))

set 集合 包含唯一元素 且 顺序不定，可以修改

scores=[‘4′,’5′,’6′,’9’]

可以由list生成 set(listname)

set.add(‘newone’)添加元素 没有append() 随机添加到一个位置

set.pop()随机删除一个元素

dictionary 字典 存储的是元素对 key:value，数组的概念

key可以是int string float 不可变的值且统一

elements = {“hydrogen”: 1, “helium”: 2, “carbon”: 6}

print(elements[“helium”])

添加就直接添加 elements[“lithium”] = 3

“some” in elements 检查是否存在于字典

elements .get(“some”) 比直接方括号elements.[“some”]要安全，不存咋的话，也只返回None而不是报错

还可以设置成elements .get(“some”，”There\`s no such key”)来自定义返回语句

Compound Data Structures 应该是python的多维数组

elements = {"hydrogen": {"number": 1,
                         "weight": 1.00794,
                         "symbol": "H"},
              "helium": {"number": 2,
                         "weight": 4.002602,
                         "symbol": "He"}}

hydrogen_weight = elements[“hydrogen”][“weight”]

控制流 control flow

IF ELIF ELSE

注意冒号和缩进。

points = 174

points = 174  # use this input when submitting your answer

# set prize to default value of None
prize = None

# use the value of points to assign prize to the correct prize name
if points <= 50:
    prize = "wooden rabbit"
elif 151 <= points <= 180:
    prize = "wafer-thin mint"
elif points >= 181:
    prize = "penguin"

# use the truth value of prize to assign result to the correct message
if prize:
    result = "Congratulations! You won a {}!".format(prize)
else:
    result = "Oh dear, no prize this time."

print(result)

循环 loop

for和while的使用

usernames = ["Joey Tribbiani", "Monica Geller", "Chandler Bing", "Phoebe Buffay"]

for i in range(len(usernames)):
    usernames[i] = usernames[i].lower().replace(" ", "_")

print(usernames)

while求阶乘

# Start with a sample number for first test - change this when testing your code more!
number = 6   
# We'll always start with our product equal to the number
product = number

# Write while loop header line - how will you tell it when to stop looping?
while  number > 1:
    # Each time through the loop, what do we want to do to our number?
    number -= 1
    # Each time, what do we want to multiply the current product by?
    product *= number
# Print out final product (how do we indicate this should happen after loop ends?)
print(product)

for的方式

# This is the number we'll find the factorial of - change it to test your code!
number = 6
# We'll start with the product equal to the number
product = number

# Write a for loop that calculates the factorial of our number 
for num in range(1, number):
    product *= num

# print the factorial of your number
print(product)

range函数，使用此函数得到list的索引

range(start,stop,step) start 和step默认是1

range(5)只有一个整数时，是stop的值 ，0-4

range(2,10) 是start =2 和stop=10 step默认为1， 2~8

迭代字典

fruit_count, not_fruit_count = 0, 0
basket_items = {'apples': 4, 'oranges': 19, 'kites': 3, 'sandwiches': 8}
fruits = ['apples', 'oranges', 'pears', 'peaches', 'grapes', 'bananas']

#Iterate through the dictionary
for fruit, count in basket_items.items():
    if fruit in fruits:
       fruit_count += count
    else:
        not_fruit_count += count

print("The number of fruits is {}.  There are {} items that are not fruits.".format(fruit_count, not_fruit_count))

while 是条件为真时继续运行

card_deck = [4, 11, 8, 5, 13, 2, 8, 10]
hand = []

# adds the last element of the card_deck list to the hand list
# until the values in hand add up to 17 or more
while sum(hand)  < 17:
    hand.append(card_deck.pop())

limit = 40

num = 0
while (num+1)**2 < limit:
    num += 1
nearest_square = num**2

print(nearest_square)

break 和 continue

break终止并跳出循环

continue 跳过本次迭代，而不完全跳出循环

headlines = ["Local Bear Eaten by Man",
             "Legislature Announces New Laws",
             "Peasant Discovers Violence Inherent in System",
             "Cat Rescues Fireman Stuck in Tree",
             "Brave Knight Runs Away",
             "Papperbok Review: Totally Triffic"]

news_ticker = ""
for headline in headlines:
    news_ticker += headline + " "
    if len(news_ticker) >= 140:
        news_ticker = news_ticker[:140]
        break

print(news_ticker)

zip and enumerate 打包和枚举

打包可简单理解成多个list拼成字典

x_coord = [23, 53, 2, -12, 95, 103, 14, -5]
y_coord = [677, 233, 405, 433, 905, 376, 432, 445]
z_coord = [4, 16, -6, -42, 3, -6, 23, -1]
labels = ["F", "J", "A", "Q", "Y", "B", "W", "X"]

points = []
for point in zip(labels, x_coord, y_coord, z_coord):
    points.append("{}: {}, {}, {}".format(*point))

for point in points:
    print(point)

cast_names = ["Barney", "Robin", "Ted", "Lily", "Marshall"]
cast_heights = [72, 68, 72, 66, 76]

cast = dict(zip(cast_names, cast_heights))
print(cast)

{'Lily': 66, 'Barney': 72, 'Marshall': 76, 'Ted': 72, 'Robin': 68}

枚举 Enumerate

cast = ["Barney Stinson", "Robin Scherbatsky", "Ted Mosby", "Lily Aldrin", "Marshall Eriksen"]
heights = [72, 68, 72, 66, 76]

for i, character in enumerate(cast):
    cast[i] = character + " " + str(heights[i])

print(cast)

['Barney Stinson 72', 'Robin Scherbatsky 68', 'Ted Mosby 72', 'Lily Aldrin 66', 'Marshall Eriksen 76']

List Comprehensions 列表推导

应该理解成对list的筛选

multiples_3 = [x * 3 for x in range(1, 21)]
print(multiples_3)

[3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60]

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