【问题标题】:How to replace string from a snowflake table如何从雪花表中替换字符串
【发布时间】:2021-05-21 15:11:08
【问题描述】:

我有一个雪花表作为 TEST_TABLE

SOURCE_RECORD_ID                                 COL_A  COL_B COL_C
paragon_altdata:model:2021-04-29_003108:1          A     A     A
paragon_altdata:model:2021-04-29_003108:2          A     A     A
paragon_altdata:model:2021-04-29_003108:3          A     A     A
paragon_altdata:model:2021-04-29_003108:4          A     A     A
paragon_altdata:model:2021-04-29_003108:5          A     A     A
paragon_altdata:model:2021-04-29_003108:6          A     A     A
paragon_altdata:model:2021-04-29_003108:7          A     A     A
paragon_altdata:model:2021-04-29_003108:8          A     A     A
paragon_altdata:model:2021-04-29_003108:9          A     A     A

在此表中,我想将第一列中的部分字符串 paragon_altdata 替换为 paragonintel_altdata

我的表格应该如下所示

SOURCE_RECORD_ID                                 COL_A  COL_B COL_C
paragonintel_altdata:model:2021-04-29_003108:1         A     A     A
paragonintel_altdata:model:2021-04-29_003108:2         A     A     A
paragonintel_altdata:model:2021-04-29_003108:3         A     A     A
paragonintel_altdata:model:2021-04-29_003108:4         A     A     A
paragonintel_altdata:model:2021-04-29_003108:5         A     A     A
paragonintel_altdata:model:2021-04-29_003108:6         A     A     A
paragonintel_altdata:model:2021-04-29_003108:7         A     A     A
paragonintel_altdata:model:2021-04-29_003108:8         A     A     A
paragonintel_altdata:model:2021-04-29_003108:9         A     A     A

请任何人帮助我解决这个问题

【问题讨论】:

    标签: snowflake-cloud-data-platform


    【解决方案1】:

    你可以只使用replace,然后创建一个新表做一个create table as select

    create NEW_TABLE as 
    select 
         replace(SOURCE_RECORD_ID, 'paragon_', 'paragonintel_') as SOURCE_RECORD_ID
        ,COL_A
        ,COL_B
        ,COL_C
    from TEST_TABLE
    ;
    

    然后您可以交换它们并删除旧的。

    【讨论】:

    • 嗨@gregpavlik 有什么方法可以避免在查询中写入 COL_A、COL_B 等。因为我有很多专栏
    • @PavanK 使用 select * 代替
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