Django 中模板的 if 条件这样为什么出错了?

[cheyo]

{% if 1==house.status %}
<span class="label label-success">
{% elif 2==house.status %}
<span class="label label-primary">
{% elif 3==house.status %}
<span class="label label-info">
{% elif 4==house.status %}
<span class="label label-default">
{% else %}
<span class="label label-danger">
{% endif %}
{{ house.get_status_display }}</span>

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

谢谢

[stillwater]

用ifequal?

[virusdefender]

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

[ericls]

if house.status == 1应该可以
注意必须有空格

[cheyo]

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

谢谢

[rcmerci]

模板里空格挺严格的

[mornlight]

要有空格，如果你用Pycharm的话应该会提示你

[taobeier]

不太习惯的话， 就用个IDE会稍微好点。

[tuteng]

mark

[gevin]

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]

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.