mysql - 基于从 INFORMATION_SCHEMA 中选择的更新

Question

我有这个问题:

SELECT TABLE_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE column_name LIKE 'organization_id'

根据结果，我想用 'organization_id' = something; 更新所有返回的表；像这样:

UPDATE (above query results) SET `organization_id` = 'something'

Answer 1

这是动态 SQL 的工作。

试试这个:

SELECT DISTINCT CONCAT('UPDATE ',t.table_name,' SET organization_id = <ID> ;')
FROM INFORMATION_SCHEMA.COLUMNS
WHERE column_name LIKE 'organization_id'

然后复制输出并执行。