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MySQL视图中的BLOB,而不是正确的数据

create-view blob mysql

2

cypher · 技术社区 · 14 年前

我使用 UNION ALL 从两个表中获取相同的列名,以便为这些表中的相同数据获取相同的列名(即。 tbl1.author2 AS translator tbl2.translator AS translator )以此类推,问题是当我试图从这个视图中选择一些东西时,有一半的数据显示为 BLOB

视图定义为:

SELECT e.id AS prod_id, 
e.price_vat AS price_vat, 
e.product AS title, 
e.authors AS author, 
e.isbn AS isbn, 
e.ean AS ean, 
e.page_count AS page_count, 
e.publishers AS publishers, 
e.issue_year AS issue_year,
'e' AS type
FROM ama_euromedia_products AS e
UNION ALL
SELECT
k.publishers AS publishers, 
DATE_FORMAT(k.publication_date, '%Y') AS issue_year,
k.ean AS ean, 
k.number_of_pages AS page_count, 
k.author AS author, 
k.isbn AS isbn, 
k.title_full AS title, 
k.price_amount AS price_vat, 
k.internal AS prod_id,
'k' AS type
FROM ama_kosmas_products AS k

CREATE TABLE `ama_euromedia_products` (
  `id` int(9) NOT NULL,
  `product` text COLLATE utf8_czech_ci NOT NULL,
  `isbn` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `authors` text COLLATE utf8_czech_ci,
  `publishers` text COLLATE utf8_czech_ci,
  `price` float(9,0) DEFAULT NULL,
  `price_vat` float(9,0) DEFAULT NULL,
  `vat` int(3) DEFAULT NULL,
  `availability` tinyint(1) DEFAULT NULL,
  `genres` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `ean` bigint(14) DEFAULT NULL,
  `page_count` int(7) DEFAULT NULL,
  `issue_year` int(4) DEFAULT NULL,
  `supply_date` timestamp NULL DEFAULT NULL,
  `width` int(7) DEFAULT NULL,
  `height` int(7) DEFAULT NULL,
  `weight` int(7) DEFAULT NULL,
  `binding` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `long_v` text COLLATE utf8_czech_ci,
  `short` text COLLATE utf8_czech_ci,
  `imgurl` text COLLATE utf8_czech_ci,
  `is_preliminary` tinyint(1) DEFAULT NULL,
  `stack_date` timestamp NULL DEFAULT NULL ON UPDATE CURRENT_TIMESTAMP,
  `invoice_only` tinyint(1) DEFAULT NULL,
  `new` tinyint(1) DEFAULT NULL,
  `sale` tinyint(1) DEFAULT NULL,
  `return_v` tinyint(1) DEFAULT NULL,
  PRIMARY KEY (`id`),
  KEY `indexy` (`id`,`isbn`,`genres`,`ean`,`issue_year`,`supply_date`,`stack_date`,`new`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_czech_ci;


CREATE TABLE `ama_kosmas_products` (
  `id` int(7) NOT NULL AUTO_INCREMENT,
  `internal` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `isbn` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `ean` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `contained_items` text COLLATE utf8_czech_ci,
  `title_original` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `title_full` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `subtitle` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `other_text` text COLLATE utf8_czech_ci,
  `languages` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `author` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `illustrator` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `translator` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `number_of_pages` int(7) DEFAULT NULL,
  `subject_scheme_name` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `subject_code` int(7) DEFAULT NULL,
  `subject_heading_text` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `media_file_label` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `media_file_thumbnail` varchar(255) COLLATE utf8_czech_ci DEFAULT NULL,
  `publishers` text COLLATE utf8_czech_ci,
  `publishing_status` tinyint(2) DEFAULT NULL,
  `publication_date` timestamp NULL DEFAULT NULL,
  `product_availability` tinyint(2) DEFAULT NULL,
  `on_hand` int(7) DEFAULT NULL,
  `on_order` int(7) DEFAULT NULL,
  `price_amount` int(7) DEFAULT NULL,
  `price_tax_rate_code` varchar(1) COLLATE utf8_czech_ci DEFAULT NULL,
  `price_tax_rate_percent` tinyint(2) DEFAULT NULL,
  `price_tax_amount` int(7) DEFAULT NULL,
  `price_taxable_amount` int(7) DEFAULT NULL,
  `reissue_date` timestamp NULL DEFAULT NULL,
  `invoice_only` tinyint(1) DEFAULT NULL,
  PRIMARY KEY (`id`),
  KEY `internal` (`internal`)
) ENGINE=MyISAM AUTO_INCREMENT=43341 DEFAULT CHARSET=utf8 COLLATE=utf8_czech_ci;

你知道为什么吗?

1 回复 | 直到 14 年前

1

9

Brian Hooper 14 年前

创建具有 UNION ,则必须确保相应列的数据类型相同(或至少相似到足以将一列转换为另一列)。在当前情况下,视图的第一列是 publishers price_vat ,并且除了 BLOB 这很有道理。

如果你真的需要这个作为一个观点,你可以尝试。。。

SELECT e.price_vat AS price_vat, 
       NULL        AS publishers,
       ...etc...
       'e'         AS type
    FROM ama_euromedia_products AS e;
UNION ALL
SELECT
     NULL         AS price_vat,
     k.publishers AS publishers, 
     ...etc...
     'k'          AS type
     FROM ama_kosmas_products AS k;

将单个数据类型放入每列。