当异常发生在 except 块中时，为什么最终块不返回最新的异常？

Question

我写了一段代码，其中我使用回溯返回异常堆栈。但我忘了导入回溯。

文件名-trial_main.py

from fastapi import FastAPI
from pydantic import BaseModel
import uvicorn

app = FastAPI()
class RequestJson(BaseModel):
data: str

@app.post("/trial", status_code=200)
async def trial(req:RequestJson):
try:
status = "default"
resp = {
"status": status,
"reason": "default reason"
}

# error point
a = 10/0

status = "complete"
reason = "Finished converting from base64 to image"

resp = {
"status": status,
"reason": reason
}
except Exception as e:
status = "incomplete"
resp =  {
"status": status,
"reason": traceback.format_exc(),
}
finally:
return resp

if __name__ == '__main__':
uvicorn.run("trial_main:app", host="0.0.0.0", port=5001, log_level="info")

让我困惑的是，既然找不到回溯模块，为什么代码没有异常退出。相反，它返回先前设置的默认响应。

这是我得到的API响应-

{
"status": "default",
"reason": "default reason"
}

这是我运行uvicorn服务器的终端上的输出

INFO:     Started server process [14744]
INFO:     Waiting for application startup.
INFO:     Application startup complete.
INFO:     Uvicorn running on http://0.0.0.0:5001 (Press CTRL+C to quit)
INFO:     127.0.0.1:61809 - "POST /trial HTTP/1.1" 200 OK

Api终点-

[POST] http://127.0.0.1:5001/trial

输入以重新创建情况-

{"data": "randomString"}

Answer 1

这里没有什么神秘之处。

在try块中，您成功地将值分配给resp变量，然后引发Exception并执行到except块。在这里，您试图将新值分配给resp，但ImportError发生在语句的右侧，所以resp仍然包含旧值，该值将在finally块中返回。异常不会进一步传播，因为finally块中有return，它只是抑制异常。

为了不被fastapi样板分散注意力，所有这些都可以用更简单的示例来说明

def always_return():
"""return 1"""
try:
res = 1
raise Exception()
except Exception:
res = 1 / 0
finally:
return res

def never_return():
"""raises ZeroDivisionError"""
try:
res = 1
raise Exception()
except Exception:
res = 1 / 0
finally:
print("finally block exists, but no return here")
return res

>>> print(always_return())
1
>>> print(never_return())
finally block exists, but no return here
ZeroDivisionError: division by zero... traceback