playwright也是可以做接口测试的，但个人觉得还是没有requests库强大，但和selenium相比的话，略胜一筹，毕竟支持API登录，也就是说可以不用交互直接调用接口操作了。
怎么用
既然是API的测试了，那肯定就别搞UI自动化那套，搞什么浏览器交互，那叫啥API测试，纯属扯淡。
也不像有些博主更懒，直接贴的官方例子，难道我用你再帮我复制一次？
来下面，说明下使用playwright如何做API测试？
实例化request对象
示例代码如下：
python
playwright.request.new_context()没错，实例化后，就是调API，看吧，其实也不是很难是不是？
实战举栗
这里用我自己写的学生管理系统的部分接口来做演示，并对部分常用api做以说明，代码示例都是用同步的写法。
1、GET请求
示例如下：
python
def testQueryStudent(playwright: Playwright):
"""
查询学生
"""
url = 'http://localhost:8090/studentFindById'
param = {
'id': 105
}
request_context = playwright.request.new_context()
response = request_context.get(url=url, params=param)
assert response.ok
assert response.json()
print('\n', response.json())效果：
2、POST请求
示例代码：
python
def testAddStudent(playwright: Playwright):
"""
新增学生
:return:
"""
url = 'http://localhost:8090/studentAdd'
request_body = {
"className": "banji",
"courseName": "wuli",
"email": "ales@qq.com",
"name": "ales",
"score": 70,
"sex": "boy",
"studentId": "92908290"
}
header = {"Content-Type": "application/json"}
request_context = playwright.request.new_context()
response = request_context.post(url=url, headers=header, data=request_body)
assert response.ok
assert response.json()
print('\n', response.json())效果：
3、PUT请求
示例代码：
python
def testUpdateStudents(playwright: Playwright):
"""
修改学生
"""
url = 'http://localhost:8090/studentUpdate/100'
param = {
'studentId': "id" + str(100),
'name': "name" + str(100),
'score': 100,
"sex": "girl",
"className": "class" + str(100),
"courseName": "course" + str(100),
"email": str(100) + "email@qq.com"
}
request_context = playwright.request.new_context()
response = request_context.put(url=url, form=param)
assert response.ok
assert response.json()
print('\n', response.json())效果：
4、DELETE请求
示例代码：
python
def testDeleteStudents(playwright: Playwright):
"""
删除学生
"""
url = 'http://localhost:8090/studentDelete/' + str(105)
request_context = playwright.request.new_context()
response = request_context.delete(url=url)
assert response.ok
assert response.json()
print('\n', response.json())效果：
5、上传文件
这个是特例吧，按照官方给的方法，我真的是死活也不能成功，一直都是提示上上传文件不能为空，也不到为啥，结果我用了一个替代方案，就是抓包模拟的构造入参，才成功，也是曲折呀。
示例代码：
python
def test_upload_file(playwright: Playwright):
'''
上传文件
:param playwright:
:return:
'''
# 创建请求上下文
request_context = playwright.request.new_context()
# 定义上传文件的URL
upload_url = "http://localhost:8090/fileUpload"
# 文件路径
file_path = "d:/demo.txt"
# 获取文件名和MIME类型
filename = file_path.split('/')[-1]
mime_type, _ = mimetypes.guess_type(file_path)
if not mime_type:
mime_type = 'application/octet-stream'
# 读取文件内容
with open(file_path, 'rb') as file:
file_content = file.read()
# 构造multipart/form-data的边界字符串
boundary = '---------------------' + str(random.randint(1e28, 1e29 - 1))
# 构造请求体
body = (
f'--{boundary}\r\n'
f'Content-Disposition: form-data; name="file"; filename="{filename}"\r\n'
f'Content-Type: {mime_type}\r\n\r\n'
f'{file_content.decode("utf-8") if mime_type.startswith("text/") else file_content.hex()}'
f'\r\n--{boundary}--\r\n'
).encode('utf-8')
# 设置请求头
headers = {
'Content-Type': f'multipart/form-data; boundary={boundary}',
}
# 发起POST请求
response = request_context.post(upload_url, data=body, headers=headers)
# 检查响应
assert response.status == 200, f"Upload failed with status: {response.status}"
assert response.ok
assert response.json()
print('\n', response.json())效果：
官方写法：
python
# 读取文件内容
with open(file_path, 'rb') as file:
file_content = file.read()
response = request_context.post(upload_url, multipart={
"fileField": {
"name": "demo.txt",
"mimeType": "text/plain",
"buffer": file_content,
}
})
print('\n', response.json())效果：
官方写法，我不知道为啥，有大侠知道，还请帮忙给个例子，小弟不胜感激呀！
写在最后
我还是觉得微软很强呀，这套框架确实比selenium略胜一筹，综合来看。
终于有时间了，来更新一篇，感觉文章对你有用，转发留言都可，谢谢！
对了，那个上传文件的为啥不行，还请前辈们帮看一下呀！