django - 如何在{％blocktrans％}和{％duplex％}标记之间保持空白而又不引起msgfmt错误？

我正在渲染一些pluralization using the blocktrans tag;这是模板文件中的相关代码段:

{% blocktrans count choice_count=choice_count %}
  You have {{ choice_count }} choice:
{% plural %}
  You have {{ choice_count }} choices:
{% endblocktrans %}

运行python manage.py makemessages --all后，这是我的相关代码段，例如django.po的en文件:

msgid ""
"\n"
"  You have %(choice_count)s choice:\n"
msgid_plural ""
"\n"
"  You have %(choice_count)s choices:\n"
msgstr[0] "You have one choices:"
msgstr[1] "You have %(choice_count)s choice(s):"

但是，当我运行python manage.py compilemessages时，这是我得到的错误消息:

$ ./manage.py compilemessages
processing file django.po in /home/yiqing/repos/training/site/training/locale/en/LC_MESSAGES
/home/yiqing/repos/training/site/training/locale/en/LC_MESSAGES/django.po:60: `msgid' and `msgstr[0]' entries do not both begin with '\n'
msgfmt: found 4 fatal errors

我知道这是由于模板文件中的换行符/空格所致，而且我知道如何“解决”它-当我将模板代码段更改为例如时:

{% blocktrans count choice_count=choice_count %}You have {{ choice_count }} choice:{% plural %}You have {{ choice_count }} choices:{% endblocktrans %}

然后重新运行makemessages，从消息中删除fuzzy标记，然后重新运行compilemessages，这样编译就可以了。

但是，我的问题是如何保留第一个模板语法并仍然能够编译消息，因为它大大提高了模板文件中代码的可读性。

最佳答案

该文档提到blocktrans的“trimmed”关键字，并引用: