Jinja2检查字典列表中是否存在值 [英] Jinja2 check if value exists in list of dictionaries
问题描述
我正在尝试检查包含字典的列表中是否存在值.我使用烧瓶1.0.2.请参见下面的示例:
I am trying to check if a value exists inside a list with dictionaries. I use flask 1.0.2. See example below:
person_list_dict = [
{
"name": "John Doe",
"email": "johndoe@mydomain.com",
"rol": "admin"
},
{
"name": "John Smith",
"email": "johnsmith@mydomain.com",
"rol": "user"
}
]
我找到了解决此问题的两种方法，您能告诉我哪个更好吗?:
I found two ways to solve this problem, can you tell me which is better?:
<pre>{% if "admin" in person_list_dict|map(attribute="rol") %}YES{% else %}NOPE{% endif %}</pre>
# return YES (john doe) and NOPE (john smith)
第二个选项:烧瓶模板过滤器
烧瓶代码:
Second option: Flask template filter
Flask code:
@app.template_filter("is_in_list_dict")
def is_any(search="", list_dict=None, dict_key=""):
if any(search in element[dict_key] for element in list_dict):
return True
return False
模板代码:
<pre>{% if "admin"|is_in_list_dict(person_list_dict, "rol") %} YES {% else %} NOPE {% endif %}</pre>
# return YES (john doe) and NOPE (john smith)
谢谢:-).
推荐答案
如果可能的话，在将其添加到Jinja中之前，我将把此逻辑移到脚本的python部分.因为，如 Jinja文档:毫无疑问，您应该尝试从模板中删除尽可能多的逻辑."
If possible, I would move this logic to the python part of the script before rendering it in Jinja. Because, as stated in the Jinja documentation: "Without a doubt you should try to remove as much logic from templates as possible."
any([person['role'] == 'admin' for person in person_dict_list])乍一看要容易得多.
如果这不是一个选择，我可能会使用第一个内置函数，因为我认为在您自己的解决方案中，不太可能在极端情况下出错，并且代码减少了大约6倍.
If that's not an option, I would probably use the first, build in function, because I think it's less prone to errors in edge cases as your own solution, and is about 6x less code.
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