相关疑难解决方法(0)

如果语句SyntaxError:语法无效

我想计算中位数我不知道这里有什么问题更新:

def median(a,b,c):
    if a>b:
        if b>c:
        return b
        else:
        if a>c:
            return c
        else:
            return a
    else:
        if b<c:
            return b
        else:
        if a>c:
            return a
        else: 
            return c
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我的错误是:文件"prog.py",第4行返回b ^ IndentationError:预期缩进块

python

-2
推荐指数
1
解决办法
3585
查看次数

Python错误:"预期缩进块"

我有"通常"预期的缩进块.所有的缩进都是正确的.我已经在各种编辑器中打开了我的脚本,并且没有任何问题可能会导致错误,或隐藏由标签引起的空白.

如果有人能对这个问题有所了解,真的很感激.

这是导致问题的脚本的一部分:

def findCSVs():
'''
looks into a 'tocom_data' subdirectory, finds 'tocomxxx.csv' files,
retuns a sorted list of filenames that conform: begins with TOCOM, ends in .csv
'''
    csvlist = []
    datadir=os.path.join('.','tocom_data')
    flist = os.listdir(datadir)
    for fname in flist:
        fsplit = fname.split('.')
        if len(fsplit)>1:
            if fsplit[1]=="csv" and fname[0:5]=="TOCOM":
                completeFname= os.path.join(datadir,fname)
                csvlist.append(completeFname)
                csvlist.sort()
    return csvlist
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Python期望该行有一个缩进块 if len(fsplit)>1:

非常感激

何塞

python syntax-error

-2
推荐指数
1
解决办法
5259
查看次数

python中的"缩进错误:意外缩进"

所以我正在学习Django教程,在页面的中间部分,我必须在mysite\polls\views.py中进行一些更改.

这就是我如何进行更改class IndexViewclass DetailView按照要求:

mysite的\调查\ views.py:

class IndexView(generic.ListView):
    template_name = 'polls/index.html'
    context_object_name = 'latest_question_list'

    def get_queryset(self):
        """
        Return the last five published questions (not including those set to be
        published in the future).
        """
        return Question.objects.filter(
        pub_date__lte=timezone.now()
        ).order_by('-pub_date')[:5]

class DetailView(generic.DetailView):
    model = Question
    template_name = 'polls/detail.html'

    def get_queryset(self): #The Error Points Here
        """
        Excludes any questions that aren't published yet.
        """
        return Question.objects.filter(pub_date__lte=timezone.now())
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mysite的\调查\ urls.py:

from django.conf.urls import url

from . import …
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python django

-2
推荐指数
1
解决办法
4234
查看次数

运行脚本时出现Python错误 - "IndentationError:unindent与任何外部缩进都不匹配"

我尝试运行脚本时收到错误

Error:"IndentationError: unindent does not match any outer indentation"
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抛出错误的代码snipet:

def update():
    try:
        lines = open("vbvuln.txt", "r").readlines()
    except(IOError): 
         print "[-] Error: Check your phpvuln.txt path and permissions"
         print "[-] Update Failed\n" 
        sys.exit(1)
    try:
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这是出现错误的实际行:

print "[-] Update Failed\n" 
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python syntax indentation

-4
推荐指数
1
解决办法
3118
查看次数

USB RFID阅读器+覆盆子pi

我有一个连接到覆盆子pi的RFID阅读器RFID阅读器是中国品牌,它像键盘一样,它读取前十位数

所以我在这里尝试做的是从卡中读取并与我的代码中存储的数字进行比较

#!/usr/bin/env python
import time
import sys

card = '0019171125'        # Stored good card number consider using a list or a file.
def main():                # define a main function.
while True:                # loop until the program encounters an error.
sys.stdin = open('/dev/tty0', 'r')
RFID_input = input()            
if RFID_input == card:      # coppare the stored number to the input and if True execute code.
  print "Access Granted" 
  print "Read code from RFID reader:{0}".format(RFID_input)
else:                    # and if the condition is …
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python

-4
推荐指数
1
解决办法
2万
查看次数

try和catch方法,python,缩进错误无法搞清楚

 #setBirthday sets their Birthday to a date
    def setBirthday(self):
        while True:
            try:
                day = int(raw_input('Please enter the date of the month (1-31) you were born on here ->'))
                if day <= 0 or day >= 32:
                    print "better try again..."
            except ValueError:

                continue
            else:
                break

                month = int(raw_input('Please enter the month of the year (1-12) you were born on here ->'))
                if month <= 0 or day >= 13:
                    print "better try again..."
            except ValueError:

                continue
            else:
                break
                year = …
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python try-catch

-5
推荐指数
1
解决办法
267
查看次数

这个错误是什么意思 ?"预计缩进块"Python

我的代码如下:

    def value(one,two):
if one < two:
return two
else:
return one
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每次我尝试运行它时,都会出现以下错误:

IndentationError:预期缩进块

我尝试重写代码,但仍然没有发生任何事情.

python

-6
推荐指数
1
解决办法
245
查看次数

标签 统计

python ×7

django ×1

indentation ×1

syntax ×1

syntax-error ×1

try-catch ×1