如何在python中检查空白输入
我对编程和python非常陌生。我正在编写脚本,如果客户键入空白,我想退出脚本。 问题是我怎么做才对? 这是我的尝试,但我认为是错误的 比如说如何在python中检查空白输入,python,file-io,Python,File Io,我对编程和python非常陌生。我正在编写脚本,如果客户键入空白,我想退出脚本。 问题是我怎么做才对? 这是我的尝试,但我认为是错误的 比如说 userType = raw_input('Please enter the phrase to look: ') userType = userType.strip() line = inf.readline() while (userType == raw_input) print "userType\n" if (userTyp
userType = raw_input('Please enter the phrase to look: ')
userType = userType.strip()
line = inf.readline()
while (userType == raw_input)
print "userType\n"
if (userType == "")
print "invalid entry, the program will terminate"
# some code to close the app
您提供的程序不是有效的python程序。因为您是初学者,所以对您的程序进行一些小的更改。这应该运行,并且做我理解的事情 这只是一个起点:结构不明确,你必须根据需要进行更改
userType = raw_input('Please enter the phrase to look: ')
userType = userType.strip()
#line = inf.readline() <-- never used??
while True:
userType = raw_input()
print("userType [%s]" % userType)
if userType.isspace():
print "invalid entry, the program will terminate"
# some code to close the app
break
userType=raw\u input('请输入要查看的短语:')
userType=userType.strip()
#line=inf.readline()您提供的程序不是有效的python程序。因为您是初学者,所以对您的程序进行一些小的更改。这应该运行,并且做我理解的事情
这只是一个起点:结构不明确,你必须根据需要进行更改
userType = raw_input('Please enter the phrase to look: ')
userType = userType.strip()
#line = inf.readline() <-- never used??
while True:
userType = raw_input()
print("userType [%s]" % userType)
if userType.isspace():
print "invalid entry, the program will terminate"
# some code to close the app
break
userType=raw\u input('请输入要查看的短语:')
userType=userType.strip()
#line=inf.readline()应用strip删除空白后,请改用此选项:
if not len(userType):
# do something with userType
else:
# nothing was entered
在应用strip删除空白后,请改用此选项:
if not len(userType):
# do something with userType
else:
# nothing was entered
您可以输入并检查是否还有剩余内容
import string
userType = raw_input('Please enter the phrase to look: ')
if not userType.translate(string.maketrans('',''),string.whitespace).strip():
# proceed with your program
# Your userType is unchanged.
else:
# just whitespace, you could exit.
您可以输入并检查是否还有剩余内容
import string
userType = raw_input('Please enter the phrase to look: ')
if not userType.translate(string.maketrans('',''),string.whitespace).strip():
# proceed with your program
# Your userType is unchanged.
else:
# just whitespace, you could exit.
我知道这已经很老了,但这可能对将来的人有所帮助。我想出了如何使用正则表达式来实现这一点。这是我的密码:
import re
command = raw_input("Enter command :")
if re.search(r'[\s]', command):
print "No spaces please."
else:
print "Do your thing!"
我知道这已经很老了,但这可能对将来的人有所帮助。我想出了如何使用正则表达式来实现这一点。这是我的密码:
import re
command = raw_input("Enter command :")
if re.search(r'[\s]', command):
print "No spaces please."
else:
print "Do your thing!"