質問編集履歴
2
実際のコードを記載しました
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以下エラーがでる箇所
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```
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#カテゴリ毎にURLを解析する
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base = "https://www.gucci.com/jp/ja/"
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実際のコードを記載させていただきました。
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test
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@@ -10,27 +10,103 @@
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コード
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プログラミング初心者のため。コードも汚く、見にくいものとなっております。申し訳ございません。
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```ここに言語を入力
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import requests
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from bs4 import BeautifulSoup
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import pprint
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import os
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from urllib.parse import urljoin
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import itertools
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sex_URL = "https://www.gucci.com/jp/ja/ca/men-c-men"
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soup = BeautifulSoup(requests.get(sex_URL).content,'lxml')
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for name in category:
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soup= BeautifulSoup(requests.get(dict[name]).content,'lxml')
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def make_folder(category_folder_name):
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os.makedirs("/Users/RP/Desktop/GUCCI/メンズ/" + category_folder_name,exist_ok=True)
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#カテゴリ毎のURLを取得し、相対パスを絶対パスに変換
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#item_category_listにURLを格納
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base = "https://www.gucci.com/jp/ja/"
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item_category_link = soup.find_all("a",class_="category-product")
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item_category_list = []
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for item_category_link in item_category_link:
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item_category_list.append(urljoin(base,item_category_link.get("href")))
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print(item_category_list)
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#商品カテゴリ名を取得し、item_category_name_listに格納
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item_category_name = soup.select("header > h2")
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item_category_name_list = []
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for item_category_name in item_category_name:
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item_category_name_list.append((item_category_name).getText())
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make_folder((item_category_name).getText())
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print(item_category_name_list)
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#item_category_listとitem_category_name_listのリストを1つのディクショナリに変換
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keys = item_category_name_list
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values = item_category_list
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category = dict(zip(keys,values))
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print(category)
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```
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91
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92
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93
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以下エラーがでる箇所
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94
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+
|
|
95
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+
```#カテゴリ毎にURLを解析する
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96
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97
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base = "https://www.gucci.com/jp/ja/"
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98
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99
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item_list = []
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for category_name in category:
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soup= BeautifulSoup(requests.get(category[category_name]).content,'lxml')
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item_link = soup.find_all("a",class_="product-tiles-grid-item-link")
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for item_link in item_link:
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item_list.append(urljoin(base,link).get("href"))
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item_list.append(urljoin(base,item_link).get("href"))
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110
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print(item_list)
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112
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