I am trying to access an index position in list dynamically.
我正在尝试动态访问列表中的索引位置。
{{ allAppsList.{{app.id}}.link }}
{ { { app.id } } { allAppsList。链接} }
It is not working.
它不工作。
{{ allAppsList.0.link }}
{ { allAppsList.0。链接} }
worked.
工作。
It should be some grammar issue. I google online, but didn't find anything helpful.
这应该是一些语法问题。我在网上烧烤，但是没有发现任何有用的东西。
BTW: This is what I want to do if using in Java.
顺便说一句:如果在Java中使用，这就是我想做的。
int index = app.id;
int指数= app.id;
String link = appList[index].getLink();
(指数)字符串链接= appList .getLink();
2 个解决方案
#1
2
You can't do such a thing in django template, But it's not impossible also.
在django模板中不能做这样的事情，但也不是不可能的。
If you are really into this, you can create a custom template filter. The implementation would be easy, Just pass the list and index to your custom filter and do exactly what you would do with Java.
如果您真的对此感兴趣，您可以创建自定义模板过滤器。实现起来很简单，只需将列表和索引传递到自定义过滤器中，然后就可以像使用Java那样做了。
Start to writing custom template filter
开始编写自定义模板过滤器。
#2
0
You can't lookup lists like that in Django's template language - it's a design decision. Method calls, dictionary keys, attribute calls and list indexes are all handled by the . (dot) notation. Have a look at the docs for more about how Django decides on what to do when it encounters a dot
在Django的模板语言中，不能像这样查找列表——这是一个设计决策。方法调用、字典键、属性调用和列表索引都由。(点)符号。如果想了解更多关于Django如何在遇到点时决定要做什么，请查看文档
#1
2
You can't do such a thing in django template, But it's not impossible also.
在django模板中不能做这样的事情，但也不是不可能的。
If you are really into this, you can create a custom template filter. The implementation would be easy, Just pass the list and index to your custom filter and do exactly what you would do with Java.
如果您真的对此感兴趣，您可以创建自定义模板过滤器。实现起来很简单，只需将列表和索引传递到自定义过滤器中，然后就可以像使用Java那样做了。
Start to writing custom template filter
开始编写自定义模板过滤器。
#2
0
You can't lookup lists like that in Django's template language - it's a design decision. Method calls, dictionary keys, attribute calls and list indexes are all handled by the . (dot) notation. Have a look at the docs for more about how Django decides on what to do when it encounters a dot
在Django的模板语言中，不能像这样查找列表——这是一个设计决策。方法调用、字典键、属性调用和列表索引都由。(点)符号。如果想了解更多关于Django如何在遇到点时决定要做什么，请查看文档