转换数据库结果以匹配accounts-api.raml文件中指定的Account模式

Question

我正在尝试使用dataweave可视映射器将名称（从d​​b返回）映射为名字和姓氏，因为我需要拆分字符串（name），我正在使用dataweave的splitBy（）函数splitBy（String，String） ：数组，如您所见，它返回字符串数组，我想对名字键示例输入打个招呼：firstName：“ hello world” splitBy（“”）示例输出：firstname：[“ hello”，“ world”] ，

mycode：

%dw 2.0
output application/json

---
payload map ( payload01 , indexOfPayload01 ) -> {
    id: payload01.accountID default "",
    firstName: "payload01.name" splitBy(" "), 
    address: ((payload01.street default "") ++ (payload01.city default "")) ++ (payload01.state default ""),
    postal: payload01.postal default "",
    country: payload01.country default "",
    creationDate: payload01.creationDate as String default "",
    accountType: payload01.accountType default "",
    miles: payload01.miles default 0
}

更新：我通过在dataweave中实现了这一点，方法是制作一个用户定义的函数，该函数调用splitBy（）函数。在下面的代码中，我解析名字和姓氏从db返回的名称

%dw 2.0
output application/json
fun split(data) = data splitBy (" ")
---
payload map ( payload01 , indexOfPayload01 ) -> {
    id: payload01.accountID default "",
    address: ((payload01.street default "") ++ (payload01.city default "")) ++ (payload01.state default ""),
    postal: payload01.postal default "",
    country: payload01.country default "",
    creationDate: payload01.creationDate as String default "",
    accountType: payload01.accountType default "",
    firstName: split(payload01.name)[0],
    lastName: split(payload01.name)[1],
    miles: payload01.miles default 0
}

Answer 1

听起来您只需要选择结果数组中的第一项即可：

firstName : (payload01.name splitBy " ")[0]