如何在python中将变量值传递给字符串

我需要从json文件中读取一个字符串，并为其赋值，使其成为一个可配置的文件名

我从json读取的字符串：

data_file_format =  "sourcesystem_batchid_extractname_loadtype"

我有一个变量，它保存代码中的值，如

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

所以我的数据文件格式应该是

data_file_format = "xyz_101_abc_Delta"

---

您可以使用+

---

这将为您提供xyz_101_abc_Delta

，您可以使用+

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)

---

这将为您提供xyz_101_abc_Delta

有多种方法可以做到这一点：

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)

fstring

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

或者使用.format

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

---

有多种方法可以做到这一点：

fstring

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

或者使用.format

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

---

或者你可以选择基本的，然后像

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

---

前面的答案也不遥远。它只是忘记了下划线

或者你可以选择基本的，然后继续这样做

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

---

前面的答案也不遥远。它只是忘记了下划线

，因此，您需要根据输入数据文件格式动态生成文件名。你能将数据存储在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_三角洲

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)

---

因此，您需要根据输入数据文件格式动态生成文件名。你能将数据存储在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_三角洲

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)

.join将是此场景中的最佳答案。答案由bitto提供。但对于未来，一些好方法是：

>>> 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

.join将是此场景中的最佳答案。答案由bitto提供。但对于未来，一些好方法是：

>>> 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

---

看这个，特别是这个。看这个，特别是这个。我想动态生成这个。。由于我的文件格式因文件而异，我希望动态生成此文件。。因为我的文件格式因文件而异