我在mysql中有一个名为“ tables”的表,该表包含所有表名和表ID的名称。
我需要对此表进行INNER JOIN并替换值{table}
和{column_id}
SELECT *
FROM attributes_categories AS ac
INNER JOIN attributes_codes AS aco ON aco.id_attribute_code =
ac.attributes_codes_id_attribute_code
INNER JOIN {$table} AS b
LEFT JOIN translations AS t ON t.id = b.{$column_id}
我正在尝试类似的操作,但是,我不能替换“ INNER JOIN {$ table}”,因为我不能在表名的位置插入结果:
SELECT COLUMN_NAME
FROM attributes_categories AS ac
INNER JOIN attributes_codes AS aco ON aco.id_attribute_code =
ac.attributes_codes_id_attribute_code
INNER JOIN tables AS tab ON tab.id_table = ac.tables_id_table
INNER JOIN INFORMATION_SCHEMA.TABLES ON TABLE_NAME = tab.table
INNER JOIN tab.table AS b
LEFT JOIN translations AS t ON t.id = b.column_id
WHERE TABLE_TYPE = 'BASE TABLE' AND TABLE_SCHEMA='knskjhpumw1'
结果是:
Mensagens do MySQL :
#1146 - Tabela 'tab.table' does not exist
谢谢!
在存储过程中,您可以使用CONCAT()
(等)构造查询并“执行”它。这样,您可以“轻松地”插入动态表名(等)。
另一种方法是在客户端代码中构造SQL,例如PHP或Java。