x1 = "這是字串資料型態的範例"
x2 = '使用單引號的字串'
x3 = '123'            # 雖是數字, 但用單引號框起來，故也是字串

x4 = "This is Mary's house." #字串包含了單引號, 故用雙引號框起來
x5 = '小明說:"這是我的書。"'    # 字串如包含雙引號, 則用單引號框起來

x1 = "字串1"
x2 = '字串2'
x3 = x1 + x2  # 串接, 存入另一個變數

print(x3)
print(x1 + x2) # 也可直接執行「串接」運算

x4 = '詞彙'
x5 = '頻率'
x6 = x4 + "," + x5    # 多個字串串接
print(x6)

# 字串內容太長，超過一行
x1 = """清晨進入一個曲折像蛇般的峽灣中，兩邊都是陸地，
上午8點見到了英國國旗在一個小山的山頂上揚起。
很快的，船之就進入到一個較為寬廣的水域，
四周被小山圍繞著。同時在兩邊高崗上，
層層羅列著房舍，終於到達香港。
回到艙房中，感謝上帝的照顧以及美善。
登岸，遇到艾德博士，他問我「你是馬偕嗎？」""" 

str1 = '千山鳥飛絕 萬徑人蹤滅'  # 同一行
str2 = '千山鳥飛絕\n萬徑人蹤滅' # 加上換行字元\n
str3 = '千山鳥飛絕\t萬徑人蹤滅' # 加上Tab字元\t

str4 = 'This is John\'s pen.' # 單引號,雙引號的「逸出字元」的寫法

score = int(input('輸入分數:'))
str1 = '你輸入的分數是' + score      # 錯誤：字串, 整數無法相加

str2 = '你輸入的分數是' + str(score) # 正確：字串相加
print(str2)

words = ['學堂', '醫館', '講道']
counts = [123, 456, 145]

for i in range(len(words)):
    str1 = words[i] + ',' + str(counts[i])  # 3個字串串接
    print(str1)

str1 = '千山鳥飛絕'
print(str1[3])   # 印出'飛'

for i in range(len(str1)):
    print(str1[i])         # 逐行印出str1內容

str2 = '千山鳥飛絕 萬徑人蹤滅'
print('字串第1～3個字元', str2[0:3]) # 索引值0開始,2結束
print('字串第2～4個字元', str2[1:4]) # 索引值1開始,3結束
print('字串第2個字元到最後', str2[1:])  # 索引值1開始,到最後結束
print('字串第3,5,7字元', str2[2:7:2])  # 索引值2開始,6結束,每隔2取值

[開始值:結束值:間隔]

import re
...
手機規則 = re.compile(r'\d\d\d\d-\d\d\d-\d\d\d')
...

import re

字串 = '請明天13:00請到牛津藝術中心參加策展茶會, 可用0933-080-080聯絡我'
手機規則 = re.compile(r'\d\d\d\d-\d\d\d-\d\d\d')
比對結果 = 手機規則.search(字串)
if (比對結果 == None):
    print('找不到')
else:
    print(比對結果.group())

比對結果 = 手機規則.search(字串)

import re

def parseAllString(string):
    """解析字串中所有的電話號碼"""
    手機規則 = re.compile(r'\d\d\d\d-\d\d\d-\d\d\d')
    比對結果串列 = 手機規則.findall(string)
    return 比對結果串列

msg1 = 'Please call my secretary using 0930-919-919 or 0952-001-001'
msg2 = '請明天13:00請到牛津藝術中心參加策展茶會'
msg3 = '請明天13:00請到牛津藝術中心參加策展茶會, 可用0933-080-080聯絡我'
print(parseAllString(msg1))
print(parseAllString(msg2))
print(parseAllString(msg3))

import re
...
手機規則 = re.compile(r'\d\d\d\d-\d\d\d-\d\d\d')
...

import re
...
手機規則 = re.compile(r'\d{4}-\d{3}-\d{3}')
...

import re
...
市話規則 = re.compile(r'\d{2}-\d{8}')
...

import re

str1 = '台北校區 地址： 25103 新北市淡水區真理街32號 電話：02-26212121'
市話規則 = re.compile(r'\d{2}-\d{8}')
result = 市話規則.search(str1)
if (result != None):
    print('找到了 %s' % result.group())
else:
    print('找不到市話號碼')

import re

msg = '台北校區 地址： 25103 新北市淡水區真理街32號 電話：02-26212121'
pattern = r'(\d{2})-(\d{8})'  # \d{2}一組,,\d{8}一組
規則 = re.compile(pattern)
phoneNum = 規則.search(msg)           # 傳回搜尋結果

print("完整號碼是: %s" % phoneNum.group())   # 顯示完整號碼
print("完整號碼是: %s" % phoneNum.group(0))  # 顯示完整號碼
print("區域號碼是: %s" % phoneNum.group(1))  # 顯示區域號碼
print("電話號碼是: %s" % phoneNum.group(2))  # 顯示電話號碼

import re
...
市話規則 = re.compile(r'\(\d{2}\)-\d{8}'')
...

import re
# 測試1
msg = 'Johnson will attend my party tonight.'
pattern = 'John((na)?son)'
規則 = re.compile(pattern)
txt = 規則.search(msg)      # 傳回搜尋結果
print(txt.group())
# 測試2
msg = 'Johnnason will attend my party tonight.'
pattern = 'John((na)?son)'
規則 = re.compile(pattern)
txt = 規則.search(msg)      # 傳回搜尋結果
print(txt.group())

import re
...
pattern = 'John((na)?son)'  # 問號前的分組可有可無
規則 = re.compile(pattern)
...

# ch16_16.py
import re

# 測試1
msg = 'Please call my secretary using 02-26669999'
pattern = r'(\d\d-)?(\d{8})'                 # 增加?號
規則 = re.compile(pattern)
phoneNum = 規則.search(msg)           # 傳回搜尋結果
print("完整號碼是: %s" % phoneNum.group())   # 顯示完整號碼

# 測試2
msg = 'Please call my secretary using 26669999'
pattern = r'(\d\d-)?(\d{8})'                 # 增加?號
規則 = re.compile(pattern)
phoneNum = 規則.search(msg)           # 傳回搜尋結果
print("完整號碼是: %s" % phoneNum.group())   # 顯示完整號碼

import re
# 測試1
msg = 'Johnson will attend my party tonight.'
pattern = 'John((na)*son)'
規則 = re.compile(pattern)
txt = 規則.search(msg)      # 傳回搜尋結果
print(txt.group())
# 測試2
msg = 'Johnnason will attend my party tonight.'
pattern = 'John((na)*son)'
規則 = re.compile(pattern)
txt = 規則.search(msg)      # 傳回搜尋結果
print(txt.group())
# 測試3
msg = 'Johnnanason will attend my party tonight.'
pattern = 'John((na)*son)'
規則 = re.compile(pattern)
txt = 規則.search(msg)      # 傳回搜尋結果
print(txt.group())

import re
# 測試1
msg = 'Johnson will attend my party tonight.'
pattern = 'John((na)+son)'
規則 = re.compile(pattern)
txt = 規則.search(msg)      # 傳回搜尋結果
print(txt.group())
# 測試2
msg = 'Johnnason will attend my party tonight.'
pattern = 'John((na)+son)'
規則 = re.compile(pattern)
txt = 規則.search(msg)      # 傳回搜尋結果
print(txt.group())
# 測試3
msg = 'Johnnanason will attend my party tonight.'
pattern = 'John((na)+son)'
規則 = re.compile(pattern)
txt = 規則.search(msg)      # 傳回搜尋結果
print(txt.group())

msg = 'Johnson will attend my party.'
pattern = 'John((na)+son)'
txt = re.search(pattern, msg)      # 傳回搜尋結果

msg = 'John, and Johnson will attend my party.'
pattern = '\w+'                    # 不限長度的單字
txt = re.findall(pattern,msg)      # 傳回搜尋結果

import re
# 測試1將字串從句子分離
msg = 'John, Johnson, Johnnason and Johnnathan will attend my party.'
pattern = '\w+'                    # 不限長度的單字
txt = re.findall(pattern,msg)      # 傳回搜尋結果
print(txt)
# 測試2將John開始的字串分離
msg = 'John, Johnson, Johnnason and Johnnathan will attend my party.'
pattern = 'John\w*'                # John開頭的單字
txt = re.findall(pattern,msg)      # 傳回搜尋結果
print(txt)

import re

msg = '1 cat, 2 dogs, 3 pigs, 4 swans'
pattern = '\d+\s\w+'
txt = re.findall(pattern,msg)      # 傳回搜尋結果
print(txt)

import re
# 測試1搜尋[aeiouAEIOU]字元
msg = 'John, Johnson, Johnnason and Johnnathan will attend my party.'
pattern = '[aeiouAEIOU]'           
txt = re.findall(pattern,msg)      # 傳回搜尋結果
print(txt)
# 測試2搜尋[2-5.]字元
msg = '1. cat, 2. dogs, 3. pigs, 4. swans'
pattern = '[2-5.]'
txt = re.findall(pattern,msg)      # 傳回搜尋結果
print(txt)