如何把所有的單詞列表中的元素作爲變量

Question

我有以下程序，在其中我想傳遞一個元素，以連續的谷歌搜索的列表：

search_terms = ['Telejob (ETH)', 'Luisa da Silva','The CERN Recruitment Services'] 
for el in search_terms: 
    webpage = 'http://google.com/search?q='+el) 
    print('xxxxxxxxxxxxxxxxxxx') 
    print(webpage) 

不幸的是我的計劃是不把所有的單詞每個列表項，但只取第一個，給我這樣的輸出：

http://google.com/search?q=Telejob (ETH) 
xxxxxxxxxxxxxxxxxxx 
http://google.com/search?q=Luisa da Silva 
xxxxxxxxxxxxxxxxxxx 
http://google.com/search?q=The CERN Recruitment Services 
xxxxxxxxxxxxxxxxxxx 
http://google.com/search?q=The Swiss National Science Foundation 

Altough你可以看到每一個字整個項目被添加到上面的搜索，當我驗證鏈接，它將僅作爲元素連接每個項目的第一個單詞，例如：

http://google.com/search?q=Telejob 
xxxxxxxxxxxxxxxxxxx 
http://google.com/search?q=Luisa 
xxxxxxxxxxxxxxxxxxx 
http://google.com/search?q=The 
xxxxxxxxxxxxxxxxxxx 
http://google.com/search?q=The 

我在做什麼錯，有什麼解決方案將每個列表項中的所有單詞連接到谷歌搜索？

謝謝

Answer 1

我相信你的問題是與網址編碼。

要允許他們有地方的URL空間由 '％20'

試着改變你的鏈接要像

https://www.google.com/search?q=The%20CERN%20Recruitment%20Services

Answer 2

這條線：

webpage = 'http://google.com/search?q='+el) 

應拆分並加入％20木匠：

webpage = 'http://google.com/search?q='+'%20'.join(el.split())) 

Answer 3

您可以在python3中使用urllib.parse.urlencode。對於python2，你可以使用urllib.urlencode。

import urllib 

search_terms = ['Telejob (ETH)', 'Luisa da Silva','The CERN Recruitment Services'] 
for el in search_terms: 
    query = urllib.parse.urlencode({'q': el}) # urllib.urlencode({'q': el}) 
    webpage = 'http://google.com/search?{}'.format(query) 
    print('xxxxxxxxxxxxxxxxxxx') 
    print(webpage) 

Answer 4

這些答案都不能解決基本問題：您需要將整個字符串編碼爲url。

我選擇urllib.quote()：

>>> import urllib 
>>> for term in search_terms: 
    print urllib.quote(term) 
Telejob%20%28ETH%29 
Luisa%20da%20Silva 
The%20CERN%20Recruitment%20Services 

通知的()也被編碼，如將任何其他奇怪的字符可能博克您的查詢。

在你的情況，這將是：

webpage = 'http://google.com/search?q=' + urllib.quote(el)) 

等值PY3：

from urllib import parse 
for term in search_terms: 
    print(parse.quote(term)) 

所以

webpage = 'http://google.com/search?q=' + parse.quote(el)) 

Answer 5

事情就是URL必須百分比編碼，網址中有特殊含義的字符，例如：

• #：去到某個位置在頁面
• /：我想你知道這個做什麼？

您應該使用quote()來解決這個問題，而只記得：

• urllib.quote()是Python2
• url.parse.quote()是Python3

下面是Python3一些例子：

from urllib.parse import quote 


quote('/bars/will/stay/intact') 
#'/bars/will/stay/intact' 

quote('/bars/wont/stay/intact', safe='') 
#'%2Fbars%2Fwont%2Fstay%2Fintact' #Actually, everything will be encoded here 

quote('()ñ´ ç') 
#'%28%29%C3%B1%C2%B4%20%C3%A7' 

所以，你的代碼是現在：

search_terms = ['Telejob (ETH)', 'Luisa da Silva','The CERN Recruitment Services'] 
for el in search_terms: 
    webpage = 'http://google.com/search?q='+quote(el) 
    print('xxxxxxxxxxxxxxxxxxx') 
    print(webpage) 

由於search_terms可能包括不會被quote('something')轉義其他字符，你將不得不使用它的安全參數：

search_terms = ['Telejob (ETH)', 'Luisa da Silva','The CERN Recruitment Services'] 
for el in search_terms: 
    webpage = 'http://google.com/search?q='+quote(el, safe='') 
    print('xxxxxxxxxxxxxxxxxxx') 
    print(webpage) 

這最後一個，輸出：

xxxxxxxxxxxxxxxxxxx 
http://google.com/search?q=Telejob%20%28ETH%29 
xxxxxxxxxxxxxxxxxxx 
http://google.com/search?q=Luisa%20da%20Silva 
xxxxxxxxxxxxxxxxxxx 
http://google.com/search?q=The%20CERN%20Recruitment%20Services 

我建議你看看：https://docs.python.org/3/library/urllib.parse.html#url-quoting進一步的信息（見？ ！一個#字符）

Answer 6

谷歌查詢的格式爲https://www.google.com/search?q=keyword_1+...+keyword_N所以你應該格式化你的查詢，像這樣：

search_terms = ["Telejob (ETH)", "Luisa da Silva","The CERN Recruitment Services"] 
for search_term in search_terms: 
    query = "+".join(search_term.split()) 
    url = "http://google.com/search?q=" + query