Django 中模板的 if 条件这样为什么出错了?

{% if 1==house.status %}

{% elif 2==house.status %}

{% elif 3==house.status %}

{% elif 4==house.status %}

{% else %}

{% endif %}
{{ house.get_status_display }}

错误信息：
TemplateSyntaxError at /project/19607/
Could not parse the remainder: '==house.status' from '1==house.status'

谢谢

Label

elif

class=

10 replies • 2014-10-31 17:24:27 +08:00

stillwater

Oct 30, 2014

用ifequal?

virusdefender

Oct 30, 2014

{% ifequal a "111" %}{% endifequal %}判断相等

ericls

Oct 30, 2014 via Android

if house.status == 1应该可以
注意必须有空格

cheyo

Oct 30, 2014

@ericls 果然加一个空格就行了
@stillwater
@virusdefender 用ifequal也行

谢谢

rcmerci

Oct 30, 2014

模板里空格挺严格的

mornlight

Oct 30, 2014

要有空格，如果你用Pycharm的话应该会提示你

taobeier

Oct 31, 2014

不太习惯的话，就用个IDE会稍微好点。

tuteng

Oct 31, 2014

mark

gevin

Oct 31, 2014

如果house.status的类型是str，你的代码那个就成了：
{% if 1=='1' %}

这种情况下就会报错

需要保持 '==' 两边值得类型一致

nooper

Oct 31, 2014

Try to remove the complex in the if cases.Just judge by integer , what if the status change , that's should not be hard code in the template. and you'd better add the logic into the views. or templatetags.