Question

我正在使用大查询，我想创建一个用＆＃34;记录＆＃34;填充表的作业。类型列。数据将由查询填充 - 所以我如何编写一个返回＆＃34;记录＆＃34;类型列。

谢谢！

Answer 1

Pentium10提出的选项在GBQ UI或API Explorer中从未对我有用我可能会遗漏一些东西

与此同时，我发现的解决方法如下例所示

SELECT location.state, location.city FROM JS(
  (      // input table
  SELECT NEST(CONCAT(state, ',', city)) AS locations
  FROM (
    SELECT state, city FROM 
    (SELECT 'florida' AS state, 'miami' AS city),
    (SELECT 'california' AS state, 'la' AS city),
    (SELECT 'romania' AS state, 'transylvania' AS city)
    ) 
  ),
  locations,     // input columns
  "[    // output schema
    {'name': 'location', 'type': 'RECORD',
     'mode': 'REPEATED',
     'fields': [
       {'name': 'state', 'type': 'STRING'},
       {'name': 'city', 'type': 'STRING'}
     ]    
    }
  ]",
  "function(row, emit){    // function 
    for (var i = 0; i < row.locations.length; i++) {
      var c = [];
      x = row.locations[i].split(',');
      t = {state:x[0], city:x[1]}
      c.push(t);
      emit({location: c});  
    };
  }"
)

请注意：
您应该使用Allow Large Results设置目标表并取消选中Flatten Results

输出表的结果是（在JSON模式下）

[
  {
    "location": [
      {
        "state": "california",
        "city": "la"
      }
    ]
  },
  {
    "location": [
      {
        "state": "florida",
        "city": "miami"
      }
    ]
  },
  {
    "location": [
      {
        "state": "romania",
        "city": "transylvania"
      }
    ]
  }
]

添加以解决@AdiCohen的一些问题与他的真实例子   他在最近的评论中表示：

问：我的查询除了记录列之外还有其他列，但是当我运行时   查询，它们返回null。我怎么能用两者创建一个表   类型？

SELECT amount, currency, location.state, location.city FROM JS( 
  ( // input table 
    SELECT NEST(CONCAT(state, ',', city)) AS locations, 
      SUM(amount) AS amount, MAX(currency) as currency 
    FROM ( 
      SELECT state, city, amount, currency, ROW_NUMBER() OVER() as grp FROM 
        (SELECT 'florida' AS state, 'miami' AS city, 'coins' AS currency, 40 AS amount), 
        (SELECT 'california' AS state, 'la' AS city, 'coins' AS currency, 40 AS amount), 
        (SELECT 'romania' AS state, 'transylvania' AS city,'coins' AS currency, 40 AS amount) 
    ) GROUP BY grp
  ), 
  amount, currency, locations, // input columns 
  "[ // output schema 
    {'name': 'location', 'type': 'RECORD', 'mode': 'REPEATED', 
    'fields': [ 
      {'name': 'state', 'type': 'STRING'}, 
      {'name': 'city', 'type': 'STRING'} 
    ] }, 
    { 'name': 'amount', 'type': 'INTEGER'}, 
    { 'name': 'currency', 'type': 'STRING'} 
  ]", 
  "function(row, emit) { // function 
    for (var i = 0; i < row.locations.length; i++) { 
      var c = []; 
      x = row.locations[i].split(','); 
      t = {state:x[0], city:x[1]} 
      c.push(t); 
      emit({amount: row.amount, currency: row.currency, location: c}); 
    }; 
  }"
)

这里的输出是：

[
  {
    "amount": "40",
    "currency": "coins",
    "location_state": "romania",
    "location_city": "transylvania"
  },
  {
    "amount": "40",
    "currency": "coins",
    "location_state": "florida",
    "location_city": "miami"
  },
  {
    "amount": "40",
    "currency": "coins",
    "location_state": "california",
    "location_city": "la"
  }
]

Answer 2

您需要使用dot表示法将输出反映为RECORD示例查询：

select 
  'florida' as country.state, 
  'SFO' as country.city;

在此示例中，country是记录，state|city是记录中的字段。

创建一个列类型为RECORD的表

2 个答案: