如何从联接表中获取数据 [英] How to get data from a joined table
问题描述
我有两个表,其中c.id = p.category_id. 我想获取category.name,但是它给出了一个错误. 谁能告诉我如何从联接表中获取数据?
I have two table joined with c.id = p.category_id. I want to get categories.name but it gives an error. Could anyone tell me how to get data from a joined table please?
function getGalleryone(){
$data = array();
$query = 'SELECT *
FROM products AS p
JOIN categories AS c
ON c.id = p.category_id
WHERE c.name = "Galleri1"
AND p.status = "active"' ;
$Q = $this->db->query($query);
/*
$this->db->select('*');
$this->db->where('categories.name','Galleri 1');
$this->db->where('products.status', 'active');
$this->db->join('categories', 'categories.id = products.category_id');
$this->db->order_by('name','random');
$Q = $this->db->get('products');
*/
if ($Q->num_rows() > 0){
foreach ($Q->result_array() as $row){
$data = array(
"id" => $row['id'],
"name" => $row['name'],
"shortdesc" => $row['shortdesc'],
...
...
"category" => $row['categories.name']
);
}
}
$Q->free_result();
return $data;
数据库产品
CREATE TABLE IF NOT EXISTS `products` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(255) NOT NULL,
`shortdesc` varchar(255) NOT NULL,
`longdesc` text NOT NULL,
`thumbnail` varchar(255) NOT NULL,
`image` varchar(255) NOT NULL,
`class` varchar(255) DEFAULT NULL,
`grouping` varchar(16) DEFAULT NULL,
`status` enum('active','inactive') NOT NULL,
`category_id` int(11) NOT NULL,
`featured` enum('true','false') NOT NULL,
`price` float(4,2) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=20 ;
数据库类别
CREATE TABLE IF NOT EXISTS `categories` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(255) NOT NULL,
`shortdesc` varchar(255) NOT NULL,
`longdesc` text NOT NULL,
`status` enum('active','inactive') NOT NULL,
`parentid` int(11) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=15 ;
...
...
错误消息
A PHP Error was encountered
Severity: Notice
Message: Undefined index: categories.name
Filename: models/mproducts.php
Line Number: 111
谢谢.
推荐答案
categories.name
列实际上将作为结果集中的$row['name']
而不是$row['categories.name']
返回.由于products
也具有name
列,一个将替换另一个.您应该指定要返回的字段,而不是使用*
通配符选择每个字段.例如:
The categories.name
column is actually going to be returned as just $row['name']
in the result set, not $row['categories.name']
. Since products
also has a name
column, one is going to replace the other. Instead of selecting every field using the *
wildcard, you should specify which fields you want returned. For example:
SELECT p.*, c.name AS category
FROM products AS p
JOIN categories AS c
ON c.id = p.category_id
WHERE c.name = "Galleri1"
AND p.status = "active"
然后您可以将类别名称引用为$row['category']
.
Then you can reference the category name as $row['category']
.
这篇关于如何从联接表中获取数据的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!