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如何通过 SQL BigQuery 从长字符串中提取

[英]How to EXTRACT from long string by SQL BigQuery

通过使用SQL的BigQuery(#standardSQL),我想提取0.1e1和0.55e6到另一列转换数据类型float - > Int64的。

   \nsub_total:\n- !ruby/object:BigDecimal 18:0.1e1\n- !ruby/object:BigDecimal 18:0.55e6\ninvoice_number:\n- \n- '

我的预期:

字符串 | 1 | 550000

使用regexp_extract_all()提取字符串。 然后转换为浮点数,然后转换为整数以求和:

select t.*,
       (select sum(cast(cast(val as float64) as int64))
        from unnest(regexp_extract_all(str, '[0-9][.][0-9]*e[0-9]+')) val
       )
from (select '   \nsub_total:\n- !ruby/object:BigDecimal 18:0.1e1\n- !ruby/object:BigDecimal 18:0.55e6\ninvoice_number:\n- \n- ' as str
     ) t;

编辑:

如果要将这些提取到数组中,只需使用:

select t.*,
       (select array_agg(cast(cast(val as float64) as int64))
        from unnest(regexp_extract_all(str, '[0-9][.][0-9]*e[0-9]+')) val
       ) as int_array
from (select '   \nsub_total:\n- !ruby/object:BigDecimal 18:0.1e1\n- !ruby/object:BigDecimal 18:0.55e6\ninvoice_number:\n- \n- ' as str
     ) t

如果您想将它们放在单独的列中,只需使用数组操作。

编辑:

如果您想要单独列中的值:

select t.*, ar[safe_ordinal(1)] as col1, 
       ar[safe_ordinal(2)] as col2
from (select t.*,
             array_agg(cast(cast(val as float64) as int64)) as ar
      from (select '   \nsub_total:\n- !ruby/object:BigDecimal 18:0.1e1\n- !ruby/object:BigDecimal 18:0.55e6\ninvoice_number:\n- \n- ' as str
           ) t
     ) t

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