使用dataweave将对象列表转换为csv [英] Transform list of objects into csv using dataweave

问题描述

我正在尝试使用dataweave中的以下代码将对象列表转换为csv:

I am trying to transform a list of objects to csv using the following code in dataweave:

%dw 1.0
%type company = :object {class: "java.util.ArrayList"}
%input payload application/java
%output application/csv
---
{
    name: payload.name,
    address: payload.address
} as :company 

下面是执行上面的数据编织代码时得到的输出.

The below is the output that I get when I execute the above data weave code.

name,name
testName,testName2
testAddress,testAddress2

虽然我期望以下几点:(样本数据)

whilst I am expecting the following: (Sample data)

name,address
testName,testAddress
testName2,testAddress2

帮助我了解我在数据编织组件中缺少什么

Help me understand to what am I missing in the data weave component

推荐答案

一般来说，在使用DataWeave时，您使用规范表示来描述您的输出，规范表示或多或少是其他数据格式的超集.

In general terms, when using DataWeave you describe your output using a canonical representation which is more or less a super-set of other data formats.

要生成CSV输出，您需要生成一个对象数组.
这些对象均代表一个CSV行.
DataWeave中的对象是键值对的集合

To generate a CSV output you need to generate an array of objects.
Each of these objects represent a CSV row.
Objects in DataWeave are sets of key-value pairs

映射应类似于:

%dw 1.0
%output application/csv
---
payload map {
    name: $.name,
    address: $.address
}

此处的map操作为列表中的每个条目生成一个具有name和address的对象. $表示迭代(每个列表项)下的隐式变量.

The map operation here generates an object with a name and address for each entry in the list. $ represents the implicit variable under iteration (each list entry).

注意:%input payload application/java指令不是必需的，因为输入时，其内容类型(JSON，XML，CSV等)是从mule消息中获取的，如果不存在，则默认为java

Note: The %input payload application/java directive is not necessary since the content-type for your input (JSON, XML, CSV, etc) is taken from the mule message when it is set, and it defaults to java if it's not present.

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