当 csv 文件中出现 HTTP 错误时跳过图像 url

Question

我有一个项目试图从网站抓取图像。我使用包含所有 url 的 csv 文件。有些 url 我没有权限打开（或者它们不存在）。我在 phyton 中收到了一个 Http 错误 403。我只想尝试 csv 文件中的下一个 url 并忽略错误。

import urllib.request
import csv

with open ('urls_01.csv') as images:
    images = csv.reader(images)
    img_count = 1
    for image in images:
        urllib.request.urlretrieve(image[0],
                'images/image_{0}.jpg'.format(img_count)) 
        img_count += 1

这是错误

Traceback (most recent call last):
  File "c:\Users\Heigre\Documents\Phyton\img_test.py", line 8, in <module>
    urllib.request.urlretrieve(image[0],
  File "C:\Program Files\Python310\lib\urllib\request.py", line 241, in urlretrieve
    with contextlib.closing(urlopen(url, data)) as fp:
  File "C:\Program Files\Python310\lib\urllib\request.py", line 216, in urlopen
    return opener.open(url, data, timeout)
  File "C:\Program Files\Python310\lib\urllib\request.py", line 525, in open
    response = meth(req, response)
  File "C:\Program Files\Python310\lib\urllib\request.py", line 634, in http_response
    response = self.parent.error(
  File "C:\Program Files\Python310\lib\urllib\request.py", line 563, in error
    return self._call_chain(*args)
  File "C:\Program Files\Python310\lib\urllib\request.py", line 496, in _call_chain
    result = func(*args)
  File "C:\Program Files\Python310\lib\urllib\request.py", line 643, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 403: Forbidden

Answer 1

使用 try 块捕获 http 错误：

import urllib.request
import csv

with open ('urls_01.csv') as images:
    images = csv.reader(images)
    img_count = 1
    for image in images:
        try:
            urllib.request.urlretrieve(image[0],
                    'images/image_{0}.jpg'.format(img_count)) 
            img_count += 1
        except HTTPError:
            pass

Answer 2

使用 try 块捕获 http 错误：


    import urllib.request
    import csv

    with open ('urls_01.csv') as images:
        images = csv.reader(images)
        img_count = 1
        for image in images:
            try:
                urllib.request.urlretrieve(image[0],
                        'images/image_{0}.jpg'.format(img_count)) 
                img_count += 1
            except HTTPError:
                pass

Answer 3

不确定是否需要导入，但我根据您的错误输出提供了它。还提供了打印语句的想法，如果您只需要传递特定错误，它可能会有所帮助……

import urllib.request
from urllib.error import HTTPError
import csv

with open ('urls_01.csv') as images:
    images = csv.reader(images)
    img_count = 1
    for image in images:
        try:
            urllib.request.urlretrieve(
                image[0],
                'images/image_{0}.jpg'.format(img_count)
            ) 
            img_count += 1
        except HTTPError as ex:
            # This will catch and hide any HTTPError the below print not tested...
            # print(ex, ex.code)
            pass

当 csv 文件中出现 HTTP 错误时跳过图像 url

Skip image url when HTTP Error appears in csv file

python

urllib

http-error