Open kay1003 opened 1 week ago
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
FastJson2中格式化输出样式 JSON.toJSONString(jsonObject, JSONWriter.Feature.PrettyFormat)
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)
JSON.toJSONString(jsonObject, Boolean.TRUE)
{ "type":1001, "stream_count":0, "job_id":[0,45732], "id":[1297059520,4117498193], "is_client":false }
如何让fastjson2的PrettyFormat输出达到fastjson之前一样的效果?