KN
Size: a a a
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E
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DB
Д
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T
A
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O
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АР
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KC
select
'update '+ a.TABLE_NAME+' set '+ a.COLUMN_NAME+' = REPLACE('+a.column_name+',''"'',''замена'') where '+a.column_name+' like ''%"%'''
from
information_schema.COLUMNS a
where table_name='название таблицы'