如何使用JsonBuilder构造一个带有变量名和值的键的json? [英] How to construct json using JsonBuilder with key having the name of a variable and value having its value?

问题描述

如何使用具有相同名称的键和值使用JsonBuilder构造json?

How to construct json using JsonBuilder with key and value having same name?

import groovy.json.JsonBuilder

def userId = 12 // some user id obtained from else where.

def json = new JsonBuilder()
def root = json {
    userId userId
}
print json.toString()

哪个会产生错误

groovy.lang.MissingMethodException:没有方法的签名: java.lang.Integer.call()适用于参数类型: (java.lang.Integer)值:[12]可能的解决方案:wait()，any()， abs()，wait(long)，wait(long，int)和(java.lang.Number)

groovy.lang.MissingMethodException: No signature of method: java.lang.Integer.call() is applicable for argument types: (java.lang.Integer) values: [12] Possible solutions: wait(), any(), abs(), wait(long), wait(long, int), and(java.lang.Number)

引用密钥无效.任何想法如何使这项工作.

Quoting the key does has no effect. Any idea how to make this work.

我希望JSON像{ userId: 12 }.另外，为什么不能将密钥写为字符串?

I want the JSON to be like { userId: 12 }. Also, why does writing the key as string not work?

long userId = 12   
def json = new JsonBuilder()
def root = json {
    "userId" userId
}

提供的示例仅是一个片段.情况是我有很多控制器动作，这些动作已经具有各种变量.现在，我要添加一个部分，在其中尝试创建变量具有的各种值的JSON字符串.因此，更改现有的变量名称不是很实际，如果我可以用相同的名称构造JSON字符串，则将更加一致.为我想要的所有变量编写访问器方法也不是一种优雅的方法.我目前所做的是对userId使用不同的命名方案，例如user_id，但同样，它与我遵循的其他约定不一致.因此，我正在寻找一种优雅的方法以及JsonBuilder如此行为的原因.

The example provided is just a snippet. The situation is that I have a lot of controller actions, which has various variables already. Now I am adding a part where I am trying to create a JSON string with various values the variables hold. So it's not very practical to change existing variable names and if I could construct the JSON string with the same name, it would be more consistent. Writing accessor methods for all the variables I wanted is also not an elegant method. What I did at present is to use different naming scheme like user_id for userId but again, it's not consistent with rest of the conventions I follow. So I am looking for an elegant approach and the reason why JsonBuilder behaves in this manner.

如果使用JavaScript，

In case of JavaScript,

var a = 1
JSON.stringify({a: a})    // gives "{"a":1}"

这是预期的结果.

推荐答案

• 为变量userId声明访问器，如果您需要JSON使其类似于{userId:12}
• Declare accessors for the variable userId, if you need the JSON to look like {userId:12}

为

import groovy.json.JsonBuilder

def getUserId(){
    def userId = 12 // some user id obtained from else where.
}

def json = new JsonBuilder()
def root = json{
    userId userId
}
print json.toString()

• 如果您需要JSON类似于{12:12}，这是最简单的情况:
• If you need the JSON to look like {12:12} which is the simplest case:

然后

import groovy.json.JsonBuilder

def userId = 12 // some user id obtained from else where.

def json = new JsonBuilder()
def root = json{
    "$userId" userId
}
print json.toString()

• 仅出于常规脚本的考虑，您可以将def从userId中删除以获得第一个行为. :)
• Just for the sake of the groovy script you can remove def from userId to get the first behavior. :)

为

import groovy.json.JsonBuilder

userId = 12

def json = new JsonBuilder()
def root = json{
    userId userId
}
print json.toString()

更新

在构建JSON时，局部变量也可用作映射键(默认情况下为String).

Local variables can also be used as map keys (which is String by default) while building JSON.

import groovy.json.JsonBuilder

def userId = 12 
def age = 20 //For example
def email = "[e-mail]"

def json = new JsonBuilder()
def root = json userId: userId, age: age, email: email

print json.toString() //{"userId":12,"age":20,"email":"[e-mail]"}

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