最近在处理数据的时候,想把一个字符串开头的“)”符号去掉,所以使用targetStr.lstrip(")"),发现在
将处理完的数据插入到数据库时会出现编码报错,于是在网上搜到了这个帖子。出现上述编码错误问题的原因
是我对lstrip函数的理解错误,权威的解释如下:
str.lstrip([chars])
Return a copy of the string with leading characters removed. The chars argument is a string
specifying the set of characters to be removed. If omitted or None, the chars argument
defaults to removing whitespace. The chars argument is not a prefix; rather, all combinations
of its values are stripped:
>>> ' spacious '.lstrip() 'spacious ' >>> 'www.example.com'.lstrip('cmowz.') 'example.com'
strip, lstrip, rstrip的情况都一样,最重要的是要注意:
参数chars是一个集合,而不是prefix(chars集合中字符的所有组合都会被删除)
所以分析我出错的情况:
>>> a = ")" >>> b = "(" >>> a 'xefxbcx89' >>> b 'xefxbcx88' >>> len(a) 3 >>> a.lstrip(b) 'x89' >>> b.lstrip(a) 'x88' >>> targetStr = ")求教python高手:(一)" >>> print targetStr.lstrip(a) 求教python高手:(一) >>> print targetStr.lstrip(b) �求教python高手:(一) >>> targetStr = "(1)求教python高手:(一)" >>> print targetStr.lstrip(a) �1)求教python高手:(一) >>> print targetStr.lstrip(b) 1)求教python高手:(一)
所以我采用的解决方法如下,
方法1:
>>> targetStr = ")求教python高手:(一)" >>> targetStr = targetStr[3:] if targetStr.find(")") == 0 else targetStr >>> print targetStr 求教python高手:(一) >>> targetStr = "(1)求教python高手:(一)" >>> targetStr = targetStr[3:] if targetStr.find(")") == 0 else targetStr >>> print targetStr (1)求教python高手:(一)
方法2:
>>> len(u")") 1 >>> targetStr = u")求教python高手:(一)" >>> print targetStr )求教python高手:(一) >>> targetStr u'uff09u6c42u6559pythonu9ad8u624buff1auff08u4e00uff09' >>> targetStr = targetStr.lstrip(u")") >>> targetStr u'u6c42u6559pythonu9ad8u624buff1auff08u4e00uff09' >>> print targetStr 求教python高手:(一) >>> targetStr = u"(1)求教python高手:(一)" >>> print targetStr (1)求教python高手:(一) >>> targetStr u'uff081uff09u6c42u6559pythonu9ad8u624buff1auff08u4e00uff09' >>> targetStr = targetStr.lstrip(u")") >>> targetStr u'uff081uff09u6c42u6559pythonu9ad8u624buff1auff08u4e00uff09' >>> print targetStr (1)求教python高手:(一)
如果各位有更好的方法,一定要告诉我 : )
此外,从这个帖子我还学习到了另外一点:
>>> c = int(" 22 ") >>> c 22
References:
求教python高手:一个简单的问题,lstrip函数切割错误