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[QUESTION]如何让fastjson2的PrettyFormat输出达到fastjson之前一样的效果? #2972

@kay1003

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@kay1003

如何让fastjson2的PrettyFormat输出达到fastjson之前一样的效果?

FastJson2中格式化输出样式
JSON.toJSONString(jsonObject, JSONWriter.Feature.PrettyFormat)

{
	"type":1001,
	"job_id":[
		0,
		45732
	],
	"id":[
		1297059520,
		4117498193
	],
	"is_client":false,
	"stream_count":0
}

FastJson中格式化输出样式
JSON.toJSONString(jsonObject, Boolean.TRUE)

{
	"type":1001,
	"stream_count":0,
	"job_id":[0,45732],
	"id":[1297059520,4117498193],
	"is_client":false
}
  • 我该如何在fastjson2中做到fastjson之前的输出样式,不想让Json中数组的val展开换行

因为项目有解析前后对比换行后样式错乱

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