W3Schools Python环境中如何避免URL里的&not被替换为¬

在W3Schools的Python环境中运行以下构建URL的代码时，输出的URL里&notify片段中的&not被自动替换成了¬符号：

man_ids = ['1215', '1216', '1217']
a = 'https://DOMAINURL/cgi/editor.cgi?article='
z = '&context=STRUCTURELABEL&notify_authors=0&window=abstract&withdraw_for_author=Withdraw+submission'

for x in man_ids:
    print(a + x + z)

实际输出示例：

https://DOMAINURL/cgi/editor.cgi?article=1215&context=STRUCTURELABEL¬ify_authors=0&window=abstract&withdraw_for_author=Withdraw+submission

尝试使用原始字符串字面量无效，需要解决如何让代码输出完整的&notify部分。

方法1：对&进行转义 #

在字符串z中，把每个&替换成\&，避免环境将&not解析为特殊字符：

man_ids = ['1215', '1216', '1217']
a = 'https://DOMAINURL/cgi/editor.cgi?article='
z = '\&context=STRUCTURELABEL\&notify_authors=0\&window=abstract\&withdraw_for_author=Withdraw+submission'

for x in man_ids:
    print(a + x + z)

方法2：使用URL编码的&（即%26） #

URL中&的标准编码形式是%26，直接替换字符串中的&为%26，输出的URL既符合规范，也不会被错误解析：

man_ids = ['1215', '1216', '1217']
a = 'https://DOMAINURL/cgi/editor.cgi?article='
z = '%26context=STRUCTURELABEL%26notify_authors=0%26window=abstract%26withdraw_for_author=Withdraw+submission'

for x in man_ids:
    print(a + x + z)

方法3：拆分参数拼接 #

将URL参数拆分为单独的列表项，再通过&拼接，从根源避免&not连续字符出现：

man_ids = ['1215', '1216', '1217']
base_url = 'https://DOMAINURL/cgi/editor.cgi?article={}'
params = [
    'context=STRUCTURELABEL',
    'notify_authors=0',
    'window=abstract',
    'withdraw_for_author=Withdraw+submission'
]

for man_id in man_ids:
    full_url = f"{base_url.format(man_id)}&{'&'.join(params)}"
    print(full_url)

内容的提问来源于stack exchange，提问作者Sasha Hoffman