TypeError:预期的str,字节或os.PathLike对象,而不是_io.TextIOWrapper [英] TypeError: expected str, bytes or os.PathLike object, not _io.TextIOWrapper
问题描述
我正在尝试使用此处的示例打开,读取,修改和关闭json文件:
I am trying to open, read, modify, and close a json file using the example here:
如何在使用Python从文件中检索的JSON数据中添加键值?
import os
import json
path = '/m/shared/Suyash/testdata/BIDS/sub-165/ses-1a/func'
os.chdir(path)
string_filename = "sub-165_ses-1a_task-cue_run-02_bold.json"
with open ("sub-165_ses-1a_task-cue_run-02_bold.json", "r") as jsonFile:
json_decoded = json.load(jsonFile)
json_decoded["TaskName"] = "CUEEEE"
with open(jsonFile, 'w') as jsonFIle:
json.dump(json_decoded,jsonFile) ######## error here that open() won't work with _io.TextIOWrapper
我总是在末尾出现错误(由于open(jsonFile...)
我不能将jsonFile
变量与open()
一起使用.我使用了确切的格式作为上面链接中提供的示例,所以我没有确定为什么它不起作用.最终将使用更大的脚本,所以我想避免使用硬编码/使用字符串作为json文件名.
I keep getting an error at the end (with open(jsonFile...)
that I can't use the jsonFile
variable with open()
. I used the exact format as the example provided in the link above so I'm not sure why it's not working. This is eventually going in a larger script so I want to stay away from hard coding/ using strings for the json file name.
推荐答案
这个问题有点老了,但是对于有同样问题的任何人:
This Question is a bit old, but for anyone with the same issue:
是的,您无法打开jsonFile变量.它指向另一个文件连接并打开的指针需要一个字符串或类似的东西.值得注意的是,一旦退出"with"块,也应关闭jsonFile,因此不应在其之外对其进行引用.
You're right you can't open the jsonFile variable. Its a pointer to another file connection and open wants a string or something similar. Its worth noting that jsonFile should also be closed once you exit the 'with' block so it should not be referenced outside of that.
但是要回答这个问题:
with open(jsonFile, 'w') as jsonFIle:
json.dump(json_decoded,jsonFile)
应该是
with open(string_filename, 'w') as jsonFIle:
json.dump(json_decoded,jsonFile)
您可以看到我们只需要使用相同的字符串来打开新连接,然后就可以根据需要为它赋予与读取文件时相同的别名.就个人而言,我更喜欢in_file和out_file只是为了明确我的意图.
You can see we just need to use the same string to open a new connection and then we can give it the same alias we used to read the file if we want. Personally I prefer in_file and out_file just to be explicit about my intent.
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