即使不滿意，我的IF也不會移動到ELIF

這是我的代碼，我遇到的問題是它沒有從IF語句排序到ELIF有什麼方法可以解決這個問題嗎？基本上它沒有「或」部分，但它不會在您添加或部分時使用。這與其他問題不同，因爲它使用的是字符串，而另一個問題使用整數。即使不滿意，我的IF也不會移動到ELIF

import time 
print "Welcome to the Troubleshooting Program" 
time.sleep(0.2) 
query=raw_input("What is your query? ") 
querystr=str(query) 
if "screen" or "Screen" in querystr: 
    in_file=open("Screen.txt", "r") 
    screen_answer=in_file.read() 
    print "We have identified the keyword Screen, here are your possible solutions:" 
    time.sleep(0.5) 
    print screen_answer 

elif "wet" or "Wet" or "water" or "Water" in querystr: 
    in_file2=open("Wet.txt", "r") 
    screen_answer=in_file.read() 
    print "We have identified that the device has been in contact with water, here are your possible solutions:" 
    time.sleep(0.5) 
    print screen_answer 

elif "Bath" or "bath" or "sink" or "Sink" or "toilet" or "Toilet" in querystr: 
    in_file3=open("Wet.txt", "r") 
    screen_answer=in_file.read() 
    print "We have identified that the device has been in contact with water, here are your possible solutions:" 
    time.sleep(0.5) 
    print screen_answer 

elif "battery" or "Batttery" or "charge" in querystr: 
    in_file=open("Battery.txt", "r") 
    screen_answer=in_file.read() 
    print "We have identified that there is an issue with the devices battery, here are your possible solutions:" 
    time.sleep(0.5) 
    print screen_answer 

elif "speaker" or "Speaker" or "Sound" or "sound" in querystr: 
    in_file=open("Speaker.txt", "r") 
    screen_answer=in_file.read() 
    print "We have identified that there is an issue with the devices speakers, here are your possible solutions:" 
    time.sleep(0.5) 
    print screen_answer 

elif "port" or "Port" in querystr: 
    in_file=open("Port.txt", "r") 
    screen_answer=in_file.read() 
    print "We have identified that there is an issue with the devices ports, here are your possible solutions:" 
    time.sleep(0.5) 
    print screen_answer 

else: 
    repeat=raw_input("Do you have another query? Y/N") 
    if repeat =="Y" or "y": 
     queryloop() 
    else: 
     print "Thank you!"

謝謝你的時間。

來源

2016-11-17 Alex Mantle

「屏幕」爲真，那麼第一個條件總是得到滿足。你需要在querystr或querystr中的「Screen」中設置「screen」。或者在querystr.lower（）'中輸入「screen」。 –

非常感謝，這已經修復了我的代碼。 –

每個OR之間是一個條件。一個條件必須評估爲真或假，但你擁有的只是一個字符串，它總是會評估爲真，因爲這就是python的工作原理。貌似你試圖檢查，如果事情是等於在這種情況下，你應該使用事情的另一件事==「另一回事」或者你可以只用通過所有的手術室，並使用返回功能一個布爾值。這就是我會做的。該函數可以獲取一個字符串列表並查看它們中的每一個，並檢查它們是否在查詢中 –

問題在這裏：

您的條件：

if "screen" or "Screen" in querystr:

永遠是True因爲：

"screen" or "Screen" in querystr

總是會返回"screen"和你if會將其視爲True。

怎麼辦？

如果你想檢查一些詞是否在querystr或不。你可能不喜歡它：

if any(word in querystr for word in ["screen", "Screen"])

在這種特定情況下，由於只是要檢查的條件忽略的情況下，可以寫成：

if "screen" in querystr.lower():

來源

2016-11-17 12:12:59

即使不滿意，我的IF也不會移動到ELIF

回答

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