非常新的Python，我刚刚创建了我的第一段代码，以帮助我完成一项非常平凡的工作任务。一切都很顺利，直到我尝试将while循环与CSV读取结合起来，我只是不确定该去哪里。
我正在努力的代码：
airport = input('Please input the airport ICAO code: ')
with open('airport-codes.csv', encoding='Latin-1') as f:
reader = csv.reader(f, delimiter=',')
for row in reader:
if airport.lower() == row[0].lower():
airportCode = row[2] + "/" + row[0]
print(airportCode)
else:
print('Sorry, I don\'t recognise that airport.')
print('Please try again.')
执行此代码会导致'else'连续打印，直到代码停止，无论输入是否与CSV文件中的输入匹配。我删除此语句的那一刻代码运行正常（如果输入不匹配，则不会打印任何内容）。
我的目标是尝试做的是问题循环，直到真实。所以我的尝试如下：
with open('airport-codes.csv', encoding='Latin-1') as f:
reader = csv.reader(f, delimiter=',')
for row in reader:
while True:
airport = input('Please input the airport ICAO code: ')
if airport.lower() == row[0].lower():
airportCode = row[2] + "/" + row[0]
print(airportCode)
break
else:
print('Sorry, I don\'t recognise that airport.')
print('Please try again.')
False
我很确定我的有限经验导致我监督一个明显的问题，但我找不到与我的搜索查询类似的东西，所以我的下一站就在这里。
先谢谢，我为任何格式问题道歉。
根据要求，CSV文件的几行：
EDQO small_airport Ottengrüner Heide Airport 50.22583389, 11.73166656
EDQP small_airport Rosenthal-Field Plössen Airport 49.86333466,
EDQR small_airport Ebern-Sendelbach Airport 50.03944397, 10.82277775
EDQS small_airport Suhl-Goldlauter Airport 50.63194275, 10.72749996
EDQT small_airport Haßfurt-Schweinfurt Airport 50.01805496,
EDQW small_airport Weiden in der Oberpfalz Airport 49.67890167,
答案 0 :(得分：0)
一个简单的解决方案是这样的：
airport = input('Please input the airport ICAO code: ')
airportCode = None
with open('airport-codes.csv', encoding='Latin-1') as f:
reader = csv.reader(f, delimiter=',')
for row in reader:
if airport.lower() == row[0].lower():
airportCode = row[2] + "/" + row[0]
print(airportCode)
break
if airportCode is not None:
print('Sorry, I don\'t recognise that airport.')
print('Please try again.')
答案 1 :(得分：0)
我有一个不同的建议使用函数：
for (Iterator i = c.iterator(); i.hasNext(); ) {
Element e = i.next(); //Get the element
System.out.println(e); //access or modify the element
}
答案 2 :(得分：0)
最好先将整个CSV数据读入Python字典，而不是尝试将CSV中的每一行与输入的输入相匹配。这样您就可以轻松“查找”您喜欢的任何机场代码，而无需继续重读整个文件：
import csv
airport_codes = {}
with open('airport-codes.csv', encoding='Latin-1', newline='') as f_input:
csv_input = csv.reader(f_input, delimiter='\t')
for row in csv_input:
if len(row) >= 2: # Make sure the CSV line has at least 2 columns
airport_codes[row[0].lower()] = row[2]
while True:
icao_code = input('Please input the airport ICAO code: ').lower().strip()
try:
print("{}/{}".format(row[0], airport_codes[icao_code]))
except KeyError:
print("Sorry, I don't recognise airport.")
print("Please try again.\n")
这样做会让你的脚本提示多个问题。
注意：您在问题中发布的数据似乎是制表符分隔的，如果是这种情况，则需要\t才能正确读取条目。如果newline=''语句将与documentation中显示的CSV阅读器一起使用，还建议您将open()添加到try语句中。
except和except是所谓的异常处理，在这种情况下，如果您在字典中查找的条目不存在，则代码跳转到A = LOAD 'file_location' USING PigStorage('\t') AS (name:chararray, age:int, gpa:float);
部分。这是Python中的首选方法。
答案 3 :(得分：0)
我认为你的逻辑流程有点偏差。如果要继续循环遍历csv文件，直到找到机场代码，然后如果代码不存在则打印错误消息，那么我们需要以不同的顺序执行操作。目前，代码循环遍历文件的每一行。对于每一行，如果输入代码与csv文件中的任何机场代码都不匹配，它会打印出不存在的消息。相反，我们想要打印错误消息，如果它不在任何行中。我们也只想在整个循环中通过csv文件提示一个提示符。而不是：
with open('airport-codes.csv', encoding='Latin-1') as f:
reader = csv.reader(f, delimiter=',')
for row in reader: # Loop over every row in the csv
while True: # This effectively keeps you in the loop until you 'break'
airport = input('Please input the airport ICAO code: ')
if airport.lower() == row[0].lower():
airportCode = row[2] + "/" + row[0]
print(airportCode)
break # break if this row matches the airport code
else:
print('Sorry, I don\'t recognise that airport.')
print('Please try again.') # Prints if this row doesn't match
False # This doesn't set anything to 'False'
试试这个：
with open('airport-codes.csv', encoding='Latin-1') as f:
reader = csv.reader(f, delimiter=',')
while True:
matches = False
airport = input('Please input the airport ICAO code:')
for row in reader:
if airport.lower() == row[0].lower():
airportCode = row[2] + "/" + row[0]
print(airportCode)
matches = True
break
if not matches:
print('Sorry, I don\'t recognise that airport.')
print('Please try again.')
False
使用此代码，流程是不同的。首先，要求机场代码。然后针对该文件进行检查，直到您到达文件末尾或找到匹配的代码。如果在文件中找到代码，请将其打印出来。如果找不到，请打印一条消息并再次询问用户新的机场代码。
答案 4 :(得分：0)
如果您想要按照自己的方式进行操作，只需确保当用户输入错误代码时，csv读者的指针会返回到文件的开头，例如：
with open('airport-codes.csv', encoding='Latin-1') as f:
reader = csv.reader(f, delimiter=',')
airportCode = None # store for our code
while not airportCode: # loop until valid code found
airport = input('Please input the airport ICAO code: ') # ask for the code
for row in reader: # loop through our CSV
if airport.lower() == row[0].lower(): # if code found in the first column...
airportCode = row[2] + "/" + row[0]
break # no need to search further
if not airportCode: # we didn't find the code, print a helpful message
print('Sorry, I don\'t recognise that airport.')
print('Please try again.')
f.seek(0) # reset the CSV pointer to the beginning for the next loop
print("Airport found: {}".format(airportCode))
但是我建议你只需将代码加载到内存中并在现场快速查找：
airportCode = None # store for our code
airport_codes = {} # our in-memory fast lookup
with open('airport-codes.csv', encoding='Latin-1') as f:
reader = csv.reader(f, delimiter=',')
for row in reader: # loop through our CSV
airport_codes[row[0].lower()] = row[2] + "/" + row[0] # store the lookup map
while not airportCode:
airport = input('Please input the airport ICAO code: ') # ask for the code
airportCode = airport_codes.get(airport.lower(), None)
if not airportCode:
print('Sorry, I don\'t recognise that airport.')
print('Please try again.')
print("Airport found: {}".format(airportCode))