数组 – Groovy Simple JSON数组构建器

我需要在JSON中构建一个简单的JSON数组,但在循环中它会在每次迭代期间覆盖第一个值.

def jsonBuilder = new groovy.json.JsonBuilder()
contact.each {
            jsonBuilder.contact(
                    FirstName:  it.getFirstName(),LastName:  it.getLastName(),Title: it.getTitle(),)
    }

它只返回简单的JSON并覆盖每次迭代的值并保留最后一个迭代的值.在groovy中构造JSON数组的语法是什么?

解决方法

诀窍是从联系人列表中收集.假设合同列表的结构如下,请按照下面使用jsonBuilder的方式进行操作.

def contact = [ 
    [ getFirstName : { 'A' },getLastName : { 'B' },getTitle : { 'C' } ],[ getFirstName : { 'D' },getLastName : { 'E' },getTitle : { 'F' } ],[ getFirstName : { 'G' },getLastName : { 'H' },getTitle : { 'I' } ]     
]

def jsonBuilder = new groovy.json.JsonBuilder()

jsonBuilder {
    contacts contact.collect { 
        [ 
            FirstName: it.getFirstName(),LastName: it.getLastName(),Title: it.getTitle() 
        ] 
    }
}

println jsonBuilder.toPrettyString()


// Prints
{
    "contacts": [
        {
            "FirstName": "A","LastName": "B","Title": "C"
        },{
            "FirstName": "D","LastName": "E","Title": "F"
        },{
            "FirstName": "G","LastName": "H","Title": "I"
        }
    ]
}

如果您正在寻找JSONArray而不是JSONObject作为最终结构,那么使用:

jsonBuilder(
    contact.collect { 
        [ 
            FirstName: it.getFirstName(),Title: it.getTitle() 
        ]
    }
)

// OP
[
    {
        "FirstName": "A","Title": "C"
    },{
        "FirstName": "D","Title": "F"
    },{
        "FirstName": "G","Title": "I"
    }
]

它没有意义,但如果结构需要如下所示

[
    {
        "contact": {
            "FirstName": "A","Title": "C"
        }
    },{
        "contact": {
            "FirstName": "D","Title": "F"
        }
    },{
        "contact": {
            "FirstName": "G","Title": "I"
        }
    }
]

然后用

jsonBuilder(
    contact.collect { 
        [ 
            contact : [ 
                FirstName: it.getFirstName(),Title: it.getTitle() 
            ] 
        ]
    }
)

相关文章

背景:    8月29日,凌晨4点左右,某服务告警,其中一个...
https://support.smartbear.comeadyapi/docs/soapui/steps/g...
有几个选项可用于执行自定义JMeter脚本并扩展基线JMeter功能...
Scala和Java为静态语言,Groovy为动态语言Scala:函数式编程,...
出处:https://www.jianshu.com/p/ce6f8a1f66f4一、一些内部...
在运行groovy的junit方法时,报了这个错误:java.lang.Excep...