Question

我要从json文件中读取一个字符串，并为其赋值以使其成为可配置文件名。

我正在从json中读取字符串：

data_file_format =  "sourcesystem_batchid_extractname_loadtype"

我有一个变量，可以保存代码中的值，例如

sourcesystem ="xyz"
  batchid = "101"
  extractname = "abc"
  loadtype = "Delta"

所以我的data_file_format应该像这样

data_file_format = "xyz_101_abc_Delta"

Answer 1

sourcesystem ="xyz"
batchid = "101"
extractname = "abc"
loadtype = "Delta"
data_file_format="_".join([sourcesystem,batchid,extractname,loadtype])
#or
#data_file_format=sourcesystem+'_'+batchid +'_'+extractname+'_'+loadtype
print(data_file_format)

Answer 2

有多种方法可以做到这一点。

fstring

data_file_format = f'{sourcesystem}_{batchid}_{extractname}_{loadtype}'

或使用.format

data_file_format = '{}_{}_{}_{}'.format(sourcetype,batchid,extractname,loadtype)

Answer 3

因此，您需要根据输入data_file_format动态生成文件名。您可以将数据存储在dict中而不是单独的变量中吗？

data_file_format =  "sourcesystem_batchid_extractname_loadtype"

data = {
  "sourcesystem": "xyz",
  "batchid": "101",
  "extractname": "abc",
  "loadtype": "Delta"
}

filename = '_'.join([data[key] for key in data_file_format.split('_')])

print(filename)

xyz_101_abc_Delta

Answer 4

或者您也可以像这样基本走下去

data_file_format = datasourcesystem + "_" + batchid + "_" + extractname + "_" + loadtype

前面的答案也不远。它只是忘记了下划线

Answer 5

.join是这种情况下的最佳答案（按位排列）。但是，将来有一些好的方法：

>>> a = 1
>>> b = 2
>>> c = "three"
>>> "%d and %d then %s" % (a, b, c)
1 and 2 then three
>>> "{} and {} then {}".format(a, b, c)
1 and 2 then three
>>> f"{a} and {b} then {c}"
1 and 2 then three

Answer 6

您可以使用+在python中连接字符串

data_file_format =  sourcesystem + "_" + batchid + "_" + extractname + "_" + loadtype

这应该给您“ xyz_101_abc_Delta”

如何将变量值传递给python中的字符串

6 个答案: