如何通过 Flask 从 python 脚本中 return 确切错误

Question

我正在调用 python 函数来调用脚本来创建目录。

app = Flask(__name__)

 

@app.route('/', methods=['POST'])

def share_drive():

    try:
    
        parentdir = request.json.get("parentdir")

        dirname = request.json.get("dirname")

        #parentdir = request.values.get("parentdir")

        #dirname = request.values.get("dirname")

                path = os.path.join(parentdir, dirname)

    # makedirs create directory recursively

    try:
    
        os.makedirs(path)

        #return ("Success Fileshare created: {} ".format(dirname))
        resp = make_resopnse('{} successfully created.'.format(dirname))
        resp.status_code = 200
        return resp
    
    except OSError as error:
    
        #return ("Fileshare creation failed: {} ".format(dirname))       
        resp = make_resopnse('Failed to create fileshare {}'.format(dirname))
        resp.status_code = 400
        return resp

我正在通过 post man 调用它，return 语句正在运行。但是我正在从 return 做出回应并传递 resp.status 代码，那部分失败了。

post man

出错

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<title>500 Internal Server Error</title>
<h1>Internal Server Error</h1>
<p>The server encountered an internal error and was unable to complete your request. Either the server is overloaded or there is an error in the application.</p>

Answer 1

因为你的except有错别字:

return ("Fileshare creation failed: {} ". formart(dirname))

应该是：

return ("Fileshare creation failed: {} ".format(dirname))

并且，如果 dirname 未定义（它应该始终是，但显示确切错误的良好编码形式）：

print(error)
return ("Fileshare creation failed: {} ".format(dirname))

Answer 2

    os.makedirs(path)

    #return ("Success Fileshare created: {} ".format(dirname))
    resp = Response('{} successfully created.'.format(dirname))
    print (resp)
    resp.status_code = 200
    return resp

except OSError as error:

    #print(error)
    resp = Response('{} fileshare creation failed.{} filename  already exists'.format(dirname, error))
    print (resp)
    resp.status_code = 200
    return resp 
    #return ("Fileshare creation failed: {} ".format(dirname))

这已修复。

如何通过 Flask 从 python 脚本中 return 确切错误

How to return exact error from the python script through flask

python

flask

flask-restful