为什么Apache服务器在python中抛出雪花摄取模块语法错误

Question

我有一个flask应用程序在Linux box（ec2实例）中的apache服务器上运行，它调用python脚本，将文件从s3移动到snowflake。如果直接执行该脚本，则效果良好。当试图使用virtualhost和mod_wsgi在apache上托管snowflake的ingest python模块时，它抛出以下错误。该模块安装在默认站点包中

Apache服务器版本：2.4.41

Python版本：3.7

雪花：雪花_连接器_python-2.1.3-py3.7-nspkg.pth

操作系统：亚马逊linux 2

下面是我尝试重新启动时的apache错误日志：

[Fri Jan 10 15:47:53.703083 2020][：error][pid 19755][remote 67.79.202.36:20]mod_wsgi（pid=19755）：无法将目标wsgi脚本'/var/www/FLASKAPPS/snowflakeingestapp/snowflakeingestapp.wsgi'作为Python模块加载

[Fri Jan 10 15:47:53.703113 2020][：error][pid 19755][remote 67.79.202.36:20]mod_wsgi（pid=19755）：处理wsgi脚本'/var/www/FLASKAPPS/snowflakeingestapp/snowflakeingestapp.wsgi'时发生异常

[Fri Jan 10 15:47:53.703129 2020][：error][pid 19755][remote 67.79.202.36:20]回溯（最近一次呼叫最后一次）：

[Fri Jan 10 15:47:53.703142 2020][：error][pid 19755][remote 67.79.202.36:20]文件“/var/www/FLASKAPPS/snowflakeingestapp/snowflakeingestapp.wsgi”，第4行，在

[Fri Jan 10 15:47:53.703176 2020][：error][pid 19755][remote 67.79.202.36:20]来自snowflakeingestapp作为应用程序导入应用程序

[Fri Jan 10 15:47:53.703181 2020][：error][pid 19755][remote 67.79.202.36:20]文件“/var/www/FLASKAPPS/snowflakeingestapp/init.py”，第6行，in

[Fri Jan 10 15:47:53.703206 2020][：错误][pid 19755][remote 67.79.202.36:20]导入雪花摄入

[Fri Jan 10 15:47:53.703210 2020][：error][pid 19755][remote 67.79.202.36:20]文件“/var/www/FLASKAPPS/snowflake Ingestapp/snowflake_ingest/init.py”，第1行，in

[Fri Jan 10 15:47:53.703225 2020][：error][pid 19755][remote 67.79.202.36:20]导入摄取\u ecomm\u json\u雪管

[Fri Jan 10 15:47:53.703230 2020][：error][pid 19755][remote 67.79.202.36:20]文件“/var/www/FLASKAPPS/snowflake Ingestapp/snowflake_ingest/ingest_ecomm_json_snowpipe.py”，第20行，in

[Fri Jan 10 15:47:53.703296 2020][：错误][pid 19755][remote 67.79.202.36:20]来自snowflake.ingest导入SimpleIngestManager

[Fri Jan 10 15:47:53.703300 2020][：error][pid 19755][remote 67.79.202.36:20]文件“/usr/local/lib/python3.7/site packages/snowflake/inset/init.py”，第1行，in

[Fri Jan 10 15:47:53.703315 2020][：error][pid 19755][remote 67.79.202.36:20]from.simple_ingest_manager导入SimpleIngestManager，StagedFile

[Fri Jan 10 15:47:53.703342 2020][：error][pid 19755][remote 67.79.202.36:20]文件“/usr/local/lib/python3.7/site packages/snowflake/inset/simple\u inset\u manager.py”，第65行

[Fri Jan 10 15:47:53.703345 2020][：error][pid 19755][remote 67.79.202.36:20]definit（self，account:Text，user:Text，pipe:Text，private_key:Text

[1月10日星期五15:47:53.703347 2020][：错误][pid 19755][远程67.79.202.36:20]

[Fri Jan 10 15:47:53.703350 2020][：错误][pid 19755][remote 67.79.202.36:20]语法错误：无效语法

Answer 1

当您从mod_wsgi执行此命令时，您使用的是Python 2而不是Python 3。您看到的错误消息是因为Snowflake的简单摄取管理器使用类型提示（通过typing模块），这在Python 2中是不可用的

下面是一个简单的片段，它在Python 2中失败并出现相同的错误，但在Python 3中成功：

from typing import Text
def foo(bar: Text):
    print(bar)
foo('huh')