【问题标题】:Python Script not running from django shellPython 脚本未从 django shell 运行
【发布时间】:2015-10-15 20:10:41
【问题描述】:

我正在从 Django shell 运行 Python 脚本。在我添加 main() 函数之前,它运行良好。现在它不起作用。当我执行脚本时,什么都没有发生(没有错误或任何东西)。我做错了什么?

在 Django shell 中,我执行如下脚本:execfile('create_xml.py')

create_xml.py:

import os
import unicodedata

from django.conf import settings
from django.template.loader import render_to_string

from databank.models import Registration

def create_xml(registrations):
    for registration in registrations:
        xml = render_to_string('databank/metadata.xml', {'registration': registration})
        filename = "metadata%s.xml" % (registration.dataset.dataset_id)
        file_path = settings.XML_DATA_FILES
        with open(os.path.join(file_path, filename), 'w') as f:
            f.write(xml.encode('UTF-8'))

def main():
    while True:
        print "For which datasets would you like to create XML files?"
        dsid = input("1: All datasets\t2: Public datasets only\t3: A specific dataset- ")
        if ds == 1:
            # all datasets
            print "Creating XML files for all datasets."
            registrations = Registration.objects.all()
            create_xml(registrations)
            print "All done!"
        elif ds == 2:
            # only public datasets
            print "Creating XML files for public datasets only."
            registrations = Registration.objects.filter(dataset__status='public')
            create_xml(registrations)
            print "All done!"
        elif ds == 3:
            dsid = input("Please input a dataset id: ")
            try:
                r = Registration.objects.get(dataset__dataset_id=dsid)
                print "Creating XML file for dataset %d." % (dsid)
                registrations = [r]
                create_xml(registrations)
                print "All done!"
            except:
                print "Not a valid dataset id. please try again."
        else:
            print "Not a valid option."

if __name__ == "__main__":
    main()

【问题讨论】:

    标签: python django main django-shell


    【解决方案1】:

    这是因为__name__ == '__main__'True 仅当python 解释器将文件作为程序运行时。尝试在程序中打印__name__,然后在更改比较后打印execfile()。您将找出 __name__ 变量。检查What does if __name__ == "__main__": do?

    【讨论】:

    • aaah 我明白了,在这种情况下,__name__ 设置为 __builtin__。非常感谢@darkryder
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