如何使用Python版Playwright发送PUT请求上传multipart/form-data？

要复现你给出的curl命令效果，关键是在Playwright的multipart参数中，将文件字段配置为包含路径、文件名和MIME类型的嵌套字典，而非拆分多个独立键。

同步版本代码 #

from playwright.sync_api import sync_playwright

with sync_playwright() as p:
    browser = p.chromium.launch()
    context = browser.new_context()
    # 发送PUT请求，配置multipart文件字段
    response = context.request.put(
        url="localhost:8080",
        multipart={
            "file": {
                "path": "./test.zip",  # 对应curl中的@test.zip
                "filename": "test.zip",  # 显式指定文件名
                "content_type": "application/x-zip-compressed"  # 设置文件MIME类型
            }
        }
    )
    print(f"响应状态码: {response.status}")
    browser.close()

异步版本代码 #

如果你使用异步API，写法如下：

import asyncio
from playwright.async_api import async_playwright

async def main():
    async with async_playwright() as p:
        browser = await p.chromium.launch()
        context = await browser.new_context()
        response = await context.request.put(
            url="localhost:8080",
            multipart={
                "file": {
                    "path": "./test.zip",
                    "filename": "test.zip",
                    "content_type": "application/x-zip-compressed"
                }
            }
        )
        print(f"响应状态码: {response.status}")
        await browser.close()

asyncio.run(main())

关键说明 #

• multipart字典的顶层键file对应curl命令中的表单字段名file
• 嵌套字典的path指定本地文件的路径，等价于curl中的@test.zip
• filename和content_type分别对应curl里的filename=test.zip和type=application/x-zip-compressed，这两个参数是可选的，省略时Playwright会自动从文件路径推断文件名和MIME类型
• 不要将这些属性拆分为独立的顶层键（比如单独设置filename键），否则会生成多个独立的表单部分，导致请求格式错误

内容的提问来源于stack exchange，提问作者daviewales