如何在不定义内部模板的情况下使用 Grails JSON 视图?
How to use Grails JSON view without defining internal templates?
我有一个域 class:
class Business {
String name
String description
}
我有以下 JSON 个模板:
index.gson
:为对象列表
生成JSON
_business.gson
: 为业务对象
生成JSON
index.gson
import server.Business
model {
Iterable businessList
}
json {
result tmpl.business(businessList ?: [])
}
_business.gson
model {
Business business
}
json {
id business.id
name business.name
}
如何在不使用 _business.gson 模板的情况下为业务对象生成 JSON?
我想采用一种只有 index.gson 并手动渲染内部对象的方法。
import server.Business
model {
Iterable businessList
}
json {
**WHAT SHOULD I ADD HERE?**
}
json(businessList.toList()) {
**I also noticed that I can use this syntax, BUT WHAT SHOULD I ADD HERE?**
}
你可以在 json
闭包中做任何你想做的事。
json(businessList.toList()) { Business business ->
id business.id
name business.name
description business.description
}
我有一个域 class:
class Business {
String name
String description
}
我有以下 JSON 个模板:
index.gson
:为对象列表
_business.gson
: 为业务对象
index.gson
import server.Business
model {
Iterable businessList
}
json {
result tmpl.business(businessList ?: [])
}
_business.gson
model {
Business business
}
json {
id business.id
name business.name
}
如何在不使用 _business.gson 模板的情况下为业务对象生成 JSON?
我想采用一种只有 index.gson 并手动渲染内部对象的方法。
import server.Business
model {
Iterable businessList
}
json {
**WHAT SHOULD I ADD HERE?**
}
json(businessList.toList()) {
**I also noticed that I can use this syntax, BUT WHAT SHOULD I ADD HERE?**
}
你可以在 json
闭包中做任何你想做的事。
json(businessList.toList()) { Business business ->
id business.id
name business.name
description business.description
}