php – 如何将行转换为列？

我有一个数据库,其中存储分组到项目中的关键字和与每个关键字相关的数据,然后我显示foreach项目的datagrids,每个关键字有一行,并且从同一个表“data”中检索了几列.我有4个表,关键字,项目,group_keywords和数据. “keywords”仅存储关键字,“投影”项目名称,“group_keywords”为分配给该项目的关键字的关键字ID,“data”是foreach关键字的所有数据所在的位置,由关键字的外键标识.id,以及用于标识数据名称的名称列.

现在要检索关键字项目的所有数据我使用此查询：

SELECT * FROM `group_keywords` 
INNER JOIN keywords on keywords.id = keyword_id 
INNER JOIN data ON data.id = keywords.id 
WHERE `group_id` = (SELECT `id` FROM `projects` WHERE `name` = 'ProjectName'

这给了我类似的东西

    id  group_id    keyword_id  id  keyword     id  name    value
    12  5   52  52  absorption food     52  data_name_x1    6
    12  5   52  52  absorption food     52  data_name_x2    8
    12  5   52  52  absorption food     52  data_name_x3    26
    12  5   52  52  absorption food     52  data_name_x4    2
...

但我想要的是得到：

id  group_id    keyword_id  id  keyword id  data_name_x1    data_name_x2    data_name_x3    data_name_x4
12  5   52  52  absorption food     52      6               8               26              2
...

所以我可以轻松地对datagrids进行分类和使用分页,否则我不知道如何去做,因为在使用大数据集时我不能将所有内容都转储到数组中,数据太多.

这是架构：

-- --------------------------------------------------------
-- Table structure for table `keywords`

CREATE TABLE IF NOT EXISTS `keywords` (
  `id` int(10) unsigned NOT NULL auto_increment,
  `keyword` varchar(255) NOT NULL,
  UNIQUE KEY `id` (`id`),
  UNIQUE KEY `keyword` (`keyword`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=115386 ;


-- --------------------------------------------------------
-- Table structure for table `data`

CREATE TABLE IF NOT EXISTS `data` (
  `id` int(10) unsigned NOT NULL,
  `name` varchar(100) NOT NULL,
  `value` varchar(15) NOT NULL,
  UNIQUE KEY `id` (`id`,`name`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;


-- --------------------------------------------------------
-- Table structure for table `projects`
--

CREATE TABLE IF NOT EXISTS `projects` (
  `id` int(10) NOT NULL auto_increment,
  `name` varchar(100) NOT NULL,
  `parent` varchar(100) default NULL,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=21 ;


-- --------------------------------------------------------
-- Table structure for table `group_keywords`

CREATE TABLE IF NOT EXISTS `group_keywords` (
  `id` int(10) NOT NULL auto_increment,
  `group_id` int(10) NOT NULL,
  `keyword_id` int(10) unsigned NOT NULL,
  PRIMARY KEY  (`id`),
  UNIQUE KEY `group_id` (`group_id`,`keyword_id`),
  KEY `keyword_id` (`keyword_id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=119503 ;


-- --------------------------------------------------------
-- Constraints for table `data`
--
ALTER TABLE `data`
  ADD CONSTRAINT `data_ibfk_1` FOREIGN KEY (`id`) REFERENCES `keywords` (`id`) ON DELETE CASCADE ON UPDATE CASCADE;

-- --------------------------------------------------------
-- Constraints for table `group_keywords`
--
ALTER TABLE `group_keywords`
  ADD CONSTRAINT `group_keywords_ibfk_1` FOREIGN KEY (`keyword_id`) REFERENCES `keywords` (`id`) ON DELETE CASCADE ON UPDATE CASCADE;

解决方法:

这个操作传统上称为“unpivot”,一些RDBMS支持它,但MysqL似乎不是其中之一.您有两个选择,在sql中执行或在PHP中执行.在MysqL中,它看起来像这样的自联接(我不知道哪个字段符合你的ID字段,所以原谅我创建我自己的例子).从性能角度来看,确保索引ID和列名称,否则这些联接将进行爬网.

shapes
ID  Name   Value
1   Color  Red
1   Shape  Circle
... for more "columns"
2   Color  Green
2   Shape  Square
... for more "columns"

SELECT
  A.ID,
  B.Value as Color,
  C.Value as Shape
  ... for more "columns"
FROM shapes A
LEFT JOIN shapes B ON B.ID = A.ID AND B.Name = 'Color'
LEFT JOIN shapes C ON C.ID = A.ID AND C.Name = 'Shape'
... for more "columns"

哪个应该净我们(除非我的head-sql-parser今晚错误运行)：

ID  Color   Shape
1   Red     Circle
2   Green   Square

对于PHP版本,您不一定要加载数组,您可以对其进行流式处理.按PK排序,然后向下设置属性.在伪代码中：

Set X to undefined
Get a Record
   Check the ID property, if it's different than X, create a new object, set X to the new ID, and yield the prevIoUs object
   Set the property of the object based on the "Name" column of our result

希望这可以帮助！

php – 如何将行转换为列？

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