django - 如何在 {% blocktrans %} 和 {%复数 %} 标签之间保留空格而不导致 msgfmt 错误？

Question

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

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

运行后 python manage.py makemessages --all ，这是我的相关片段，例如 django.po文件为 en :

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 ，这是我收到的错误消息:

$ ./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

我知道这是因为模板文件中的换行符/空格，而且我知道如何“绕过”它——当我将模板片段更改为，例如:

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

并重新运行 makemessages ，删除 fuzzy消息中的标记，然后重新运行 compilemessages ，它编译得很好。

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，因为它极大地提高了模板文件中代码的可读性。

Answer 1

文档中提到了 blocktrans 的“trimmed”关键字，引用:

For instance, the following {% blocktrans %} tag:

{% blocktrans trimmed %}
  First sentence.
  Second paragraph.
{% endblocktrans %}

will result in the entry "First sentence. Second paragraph." in the PO file, compared to "\n First sentence.\n Second sentence.\n", if the trimmed option had not been specified.