Question

我对python（以及一般的编程）非常陌生。下面粘贴的代码用于编辑现有的csv。但是我收到了一个错误。错误是

Traceback (most recent call last):
  File "C:\Users\Samsung Pc\Desktop\hungrypy1\hungry_data.py", line 34, in <module>
    writer.writeheader()
  File "C:\Users\Samsung Pc\AppData\Local\Programs\Python\Python36-32\lib\csv.py", line 144, in writeheader
    self.writerow(header)
  File "C:\Users\Samsung Pc\AppData\Local\Programs\Python\Python36-32\lib\csv.py", line 155, in writerow
    return self.writer.writerow(self._dict_to_list(rowdict))
  File "C:\Users\Samsung Pc\AppData\Local\Programs\Python\Python36-32\lib\tempfile.py", line 483, in func_wrapper
    return func(*args, **kwargs)
TypeError: a bytes-like object is required, not 'str'
[Finished in 0.5s]

我试图寻找答案，它似乎是从二进制到字符串的类型不匹配。我试图将文件打开为＆＃34; r＆＃34;而不是＆＃34; rb＆＃34;然而是同样的错误。请帮忙。代码如下。

import csv
import shutil
from tempfile import NamedTemporaryFile

def get_length(file_path):
    with open("data.csv", "r") as csvfile:
        reader = csv.reader(csvfile)
        reader_list = list(reader)
        print(reader_list)
        return len(reader_list)

def append_data(file_path, name, email):
    fieldnames = ['id', 'name', 'email'
    next_id = get_length(file_path)
    with open(file_path, "a", newline = '') as csvfile:
        writer = csv.DictWriter(csvfile, fieldnames=fieldnames)
        writer.writeheader() 
        writer.writerow({
                "id": next_id,
                "name": name,
                "email": email,
            })

append_data("data.csv", "Niraj", "test@test.com")

filename = "data.csv"
temp_file = NamedTemporaryFile(delete = False)

with open(filename, "r") as csvfile, temp_file:
    reader = csv.DictReader(csvfile)
    fieldnames = ['id','name','email','amount','sent']
    writer = csv.DictWriter(temp_file, fieldnames = fieldnames)
    writer.writeheader()

    for row in reader:
        writer.writerow({
            "id" : row["id"],
            "name" : row["name"],
            "email": row["email"],
            "amount": "1234.34",
            "sent": "",
            })

#shutil.move(temp_file.name. filename)

Answer 1

您不应同时读取和写入同一文件：

with open(filename, "r") as csvfile, temp_file:

即使您为此文件指定了两个名称，它们也指向您磁盘中的同一文件，将其更改为：

 with open(filename, "r") as csvfile, open('temp_file.csv', 'w') as temp_file:

Answer 2

你正在做什么：

temp_file = NamedTemporaryFile(delete = False)
with open(filename, "r") as csvfile, temp_file:

, temp_file部分严格禁止。你必须打开文件写...

之后您使用temp_file作为输出文件句柄，但它实际上是字符串，因此您将获得该消息。

你想这样做：

with open(filename, "r") as csvfile, open(temp_file,"w",newline='') as outfile:

然后使用outfile，而不是temp_file：

writer = csv.DictWriter(outfile, fieldnames = fieldnames)

除此之外：不要使用writer.writeheader()：无论如何都会写出标题，你会得到它两次。

读/写csv文件时出错。 str和二进制之间的类型不匹配

2 个答案: