6

我试图使用 USQL 从 JSON 文件中提取数据。查询成功运行而不产生任何输出数据或导致“顶点失败快速错误”。

JSON 文件如下所示：

{
  "results": [
    {
      "name": "Sales/Account",
      "id": "7367e3f2-e1a5-11e5-80e8-0933ecd4cd8c",
      "deviceName": "HP",
      "deviceModel": "g6-pavilion",
      "clientip": "0.41.4.1"
    },
    {
      "name": "Sales/Account",
      "id": "c01efba0-e0d5-11e5-ae20-af6dc1f2c036",
      "deviceName": "acer",
      "deviceModel": "veriton",
      "clientip": "10.10.14.36"
    }
  ]
}

我的 U-SQL 脚本是

REFERENCE ASSEMBLY [Newtonsoft.Json];
REFERENCE ASSEMBLY [Microsoft.Analytics.Samples.Formats];

DECLARE @in string="adl://xyz.azuredatalakestore.net/todelete.json";

DECLARE @out string="adl://xyz.azuredatalakestore.net/todelete.tsv";

@trail2=EXTRACT results string FROM @in USING new Microsoft.Analytics.Samples.Formats.Json.JsonExtractor();

@jsonify=SELECT Microsoft.Analytics.Samples.Formats.Json.JsonFunctions.JsonTuple(results,"name","id","deviceName","deviceModel","clientip") AS rec FROM @trail2;

@logSchema=SELECT rec["name"] AS sysName,
              rec["id"] AS sysId,
              rec["deviceName"] AS domainDeviceName,
              rec["deviceModel"] AS domainDeviceModel,
              rec["clientip"] AS domainClientIp 
       FROM @jsonify;

OUTPUT @logSchema TO @out USING Outputters.Tsv();
4

2 回答 2

9

实际上，JSONExtractor支持以 JSONPath 表示的 rowpath 参数，使您能够识别要映射到行的 JSON 对象或 JSON 数组项。因此，您可以使用 JSON 文档中的单个语句提取数据：

@logSchema = 
    EXTRACT name string, id string, deviceName string, deviceModel string, clientip string
    FROM @input
   USING new Microsoft.Analytics.Samples.Formats.Json.JsonExtractor("results[*]");
于 2016-06-30T00:43:04.280 回答
2

萨拉特，

问题是您的@trail2 输出是一个json 数组“[{...},{...}]”，据我所知，JsonFunction 无法解析。所以我将它输出到一个文件并用输入器重新读取它，它可以解析数组。

REFERENCE ASSEMBLY [Newtonsoft.Json];
REFERENCE ASSEMBLY [Microsoft.Analytics.Samples.Formats];

DECLARE @in string="adl://xyz.azuredatalakestore.net/todelete.json";
DECLARE @out string="adl://xyz.azuredatalakestore.net/todelete.tsv";
DECLARE @mid string="adl://xyz.azuredatalakestore.net/intermediate.txt";


@trail2=EXTRACT results string FROM @in USING new Microsoft.Analytics.Samples.Formats.Json.JsonExtractor();

OUTPUT @trail2 TO @mid USING Outputters.Text(quoting:false);

@jsonify=EXTRACT name string,
                id string, 
                deviceName string ,
                deviceModel string,
                clientip string
FROM @mid USING new Microsoft.Analytics.Samples.Formats.Json.JsonExtractor();

@logSchema=SELECT name AS sysName,
              id AS sysId,
              deviceName AS domainDeviceName,
              deviceModel AS domainDeviceModel,
              clientip AS domainClientIp 
       FROM @jsonify;

OUTPUT @logSchema TO @out USING Outputters.Tsv();
于 2016-04-22T20:34:55.067 回答