重複輸出

Question

非常非常新的Python的，我瞭解的情況是我剛剛建在最後幾個小時：

from sys import argv 
import requests 
import xml.etree.ElementTree as ET 

script, gameid = argv 

game = requests.get("http://boardgamegeek.com/xmlapi/game/" + (gameid)) 
r = game.text 
root = ET.fromstring(r) 
boardgame = root.findall('boardgame') 
for b in boardgame: 
    name = b.find('name').text 
    year = b.find('yearpublished').text 
    mech = b.find('boardgamemechanic').text 
    cat = b.find('boardgamecategory').text 
    print (name,",",year,",",mech,",",cat) 

這是做什麼我希望它這是根據用戶輸入給我的上述四條信息。

我想知道的是，如果可以運行此腳本n次，其中n是在cmd行輸入的參數數量？

輸出類似於如下：

C:\Python>bgg.py 822 
Carcassonne , 2000 , Area Control/Area Influence , City Building 

C:\Python>bgg.py 25417 
BattleLore , 2006 , Campaign/Battle Card Driven , Fantasy 

我想是如果我能得到類似下面：

C:\Python>bgg.py 822 25417 
Carcassonne , 2000 , Area Control/Area Influence , City Building 
BattleLore , 2006 , Campaign/Battle Card Driven , Fantasy 

乾杯

Answer 1

關鍵目標是重複需要多次執行的代碼。爲此，從argv列表中提取您的遊戲ID（您似乎知道這是如何工作的）。現在，遍歷每個ID併爲每個ID做，你在原始代碼中做了什麼。現在

from sys import argv 
import requests 
import xml.etree.ElementTree as ET 

gameids = argv[1:] 

for gameid in gameids: 
    game = requests.get("http://boardgamegeek.com/xmlapi/game/" + (gameid)) 
    r = game.text 
    root = ET.fromstring(r) 
    boardgame = root.findall('boardgame') 
    for b in boardgame: 
     name = b.find('name').text 
     year = b.find('yearpublished').text 
     mech = b.find('boardgamemechanic').text 
     cat = b.find('boardgamecategory').text 

     print (name, ",", year, ",", mech, ",", cat) 

，與python bgg.py 822 25417調用你的程序，你會看到你正在尋找的結果。

Answer 2

for gameid in argv[1:]: 
    ... 

[1:]手段沒有第一個元素（腳本）的列表的副本，它被稱爲切片。