Python中出现ValueError: Invalid format specifier错误该如何解决？

错误原因：

• 代码里的"是HTML转义的双引号，直接放到Python f-string里会导致语法混乱
• f-string中要输出字面大括号必须用{{和}}转义，原代码里部分大括号未正确转义，同时{i}前后的引号拼接方式错误，让Python误将部分内容识别为格式说明符，触发报错

修正后的代码：

for i in range(0, 2):
    payload = f'{{"requests": [{{"indexName": "products", "params": "query=&hitsPerPage=20&maxValuesPerFacet=1000&page={i}&facets=%5B%22family.name%22%2C%22wine_type.name%22%2C%22vina.name%22%2C%22cepa.name%22%2C%22linea.name%22%2C%22procedencia.name%22%2C%22wine_harvest_year%22%2C%22price%22%2C%22score%22%5D&tagFilters=&facetFilters=%5B%22family.name%3AVinos%22%5D"}}]}}'
    print(payload)

关键修改点：

• 把所有"替换为实际的双引号"
• 确保所有需要作为字面量输出的大括号都用{{和}}转义（比如最外层的{{"requests": ...}}、数组内的{{"indexName": ...}}）
• 直接在page=后插入{i}，无需额外引号包裹（params参数本身是字符串，{i}直接拼接即可）

更推荐的写法（用字典生成JSON，避免手动拼接的繁琐）：

import json

for i in range(0, 2):
    payload_dict = {
        "requests": [
            {
                "indexName": "products",
                "params": f"query=&hitsPerPage=20&maxValuesPerFacet=1000&page={i}&facets=%5B%22family.name%22%2C%22wine_type.name%22%2C%22vina.name%22%2C%22cepa.name%22%2C%22linea.name%22%2C%22procedencia.name%22%2C%22wine_harvest_year%22%2C%22price%22%2C%22score%22%5D&tagFilters=&facetFilters=%5B%22family.name%3AVinos%22%5D"
            }
        ]
    }
    payload = json.dumps(payload_dict)
    print(payload)

这种方式无需手动转义大括号，可读性更强且不易出错。

内容的提问来源于stack exchange，提问作者John Doe