如何在不导致msgfmt错误的情况下保留{% blocktrans %}和{% plural %}标记之间的空格？

Question

我正在使用blocktrans标签渲染一些复数形式；以下是模板文件中相关的片段：

{% 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，这是我的例如django.po文件中与en相关的片段：

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，我收到了以下错误信息：

$ ./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，我知道如何“绕过”它 - 当我将模板片段更改为：

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

再次运行 makemessages ，从消息中删除 fuzzy 标记，然后重新运行 compilemessages，它会成功编译。

然而，我的问题是如何保留第一个模板语法并仍能编译消息，因为它极大地提高了模板文件中代码的可读性。

- 3cheesewheel

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你尝试过使用 '{% spaceless %}' 标签 吗？ - gdvalderrama

Answer 1

文档提到了“trimmed”关键字用于blocktrans，引用如下：

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.

- Florian

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工作并记录在（至少）Django 1.11下 https://docs.djangoproject.com/en/1.11/topics/i18n/translation/#blocktrans-template-tag - jobima

Answer 2

您可以做的最简单的事情是匹配输入字符串的格式。在您的示例中，.po 文件应如下所示：

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

这个文件可以编译通过，但是很繁琐。

据我所知，目前没有其他的解决方法。看起来django-rosetta有一个补丁可以处理这个确切的问题（参见https://github.com/mbi/django-rosetta/pull/34）。

- shezi

Answer 3

我相信使用{% spaceless %}标签应该可以解决这个问题。它的作用是删除其开始和结束之间的任何空格（和换行符）。我只测试了不使用复数形式，但它应该可以工作。

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

- gdvalderrama