我有以下python代码:
def split_arg(argv):
buildDescriptor = argv[1]
buildfile, target = buildDescriptor.split("#")
return buildfile, target
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它需要一个argv[1]表单的string()buildfile#target并将它们分成两个同名的变量.所以像" my-buildfile#some-target " 这样的字符串将分别被分解为my-buildfile和some-target.
有时候,不会有"#"和目标; 有时你只会有" my-buildfile ",在这种情况下我只想让目标成为""(空).
如何修改此函数以便它将处理"#"不存在的实例并返回带有空目标的buildfile?
目前,如果我只传递构建文件,它会抛出一个错误:
buildfile, target = buildDescriptor.split("#")
ValueError: need more than 1 value to unpack
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提前致谢!
ch3*_*3ka 10
我会用明显的方法:
buildfile, target = buildDescriptor.split("#") if \
"#" in buildDescriptor else \
(buildDescriptor, "")
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请注意,当buildDescriptor中存在多个"#"时,这也会抛出异常(这通常是一件好事!)
首先,将拆分结果放在一个列表中:
split_build_descriptor = buildDescriptor.split("#")
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然后检查它有多少元素:
if len(split_build_descriptor) == 1:
buildfile = split_build_descriptor[0]
target = ''
elif len(split_build_descriptor) == 2:
buildfile, target = split_build_descriptor
else:
pass # handle error; there's two #s
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>>> buildfile, _, target = "hello#world".partition("#")
>>> buildfile, target
('hello', 'world')
>>> buildfile, _, target = "hello".partition("#")
>>> buildfile, target
('hello', '')
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