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这是一个创建于 4053 天前的主题，其中的信息可能已经有所发展或是发生改变。
{% 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
<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
 | 1 stillwater 2014-10-30 20:34:40 +08:00 用ifequal? |
 | 2 virusdefender 2014-10-30 20:35:24 +08:00  1 {% ifequal a "111" %}{% endifequal %}判断相等 |
 | 3 ericls 2014-10-30 21:43:11 +08:00 via Android 1 if house.status == 1应该可以 注意必须有空格 |
 | 4 cheyo OP |
 | 5 rcmerci 2014-10-30 22:03:23 +08:00 模板里空格挺严格的 |
 | 6 mornlight 2014-10-30 22:50:29 +08:00 要有空格，如果你用Pycharm的话应该会提示你 |
 | 7 taobeier 2014-10-31 00:00:18 +08:00 不太习惯的话， 就用个IDE会稍微好点。 |
 | 8 tuteng 2014-10-31 07:18:59 +08:00 mark |
 | 9 gevin 2014-10-31 13:43:52 +08:00 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 |
 | 10 nooper 2014-10-31 17:24:27 +08:00 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. |
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