我也在尝试发送异常，但它给了我语法错误

Question

错误：

File "/home/soubhagya/Desktop/carrier-circle/backend/details/urls.py", line 2, in <module>
from .views import *
File "/home/soubhagya/Desktop/carrier-circle/backend/details/views.py", line 298
return {"code":402,"error":except}
                                ^
SyntaxError: invalid syntax

代码：

   try:
        api = Details.get_important_links(
         "OdishaGovtJobs",
         "OdishaGovtJobDetails",
         "state-govt-jobs/odisha-govt-jobs"
         )
        return JsonResponse(api,safe=False)
    except:
        return {"code":402,"error":except}

我正在尝试发送异常以及状态代码，但这给了我语法错误

Answer 1

我假设您在函数内

您可以试试吗？这对我有用

def my_def():
    try:
        api = Details.get_important_links(
            "OdishaGovtJobs",
            "OdishaGovtJobDetails",
            "state-govt-jobs/odisha-govt-jobs"
        )
        print ('Im a print in the try')
        return JsonResponse(api,safe=False)

    except Exception  as e :
        print ('Im in the except' )
        return {"code":402,"error":e}


print (my_def())