Question

我有一个代码，每次出现错误时都会停止运行。有没有办法在脚本中添加代码，忽略所有错误并继续运行脚本直到完成？

以下是代码：

import sys
import tldextract

def main(argv):

        in_file = argv[1]
        f = open(in_file,'r')
        urlList = f.readlines()
        f.close()
        destList = []

        for i in urlList:
            print i
            str0 = i
            for ch in ['\n','\r']:
                    if ch in str0:
                        str0 = str0.replace(ch,'')
            str1 = str(tldextract.extract(str0))

            str2 = i.replace('\n','') + str1.replace("ExtractResult",":")+'\n'
            destList.append(str2)

        f = open('destFile.txt','w')
        for i in destList:
                f.write(i)

        f.close()

        print "Completed successfully:"


if __name__== "__main__":
    main(sys.argv)

非常感谢

Answer 1

你应该总是尝试＆＃39;打开文件。这样，如果文件不存在，您可以管理异常。在Python Tutorial Exeption Handling

处获取战利品

import sys

try:
    f = open('myfile.txt')
    s = f.readline()
    i = int(s.strip())
except IOError as e:
    print "I/O error({0}): {1}".format(e.errno, e.strerror)
except ValueError:
    print "Could not convert data to an integer."
except:
    print "Unexpected error:", sys.exc_info()[0]
    raise

或

for arg in sys.argv[1:]:
    try:
        f = open(arg, 'r')
    except IOError:
        print 'cannot open', arg
    else:
        print arg, 'has', len(f.readlines()), 'lines'
        f.close()

不要（！）只是通过＆＃39;在异常块中。这会（！）让你更加沮丧。

Answer 2

如果您的错误发生在哪里，您可以将其包装在try / except块

中

for i in loop:
    try:
        code goes here...
    except:
        pass

Python代码忽略错误

2 个答案: