显示某个字段的所有结果,一个显示内部连接的休息结果

时间:2013-02-27 03:15:57

标签: php mysql sql

我需要在创建区域显示商店以及订单日期。我创建了第三个表来连接已经存在的两个(商店和区域)并使用内部联接加入。首先,我得到了这样的结果:

area name: name1
shops: shop1
date: date1

area name: name1
shops: shop2
date: date1

我需要:

area name: name1
shops: shop1, shop2
date: date1

然后我找到了一些代码并将其包含在脚本中,但它仅适用于一行(下面的代码中的名称)。如果我尝试使用elseif与该代码的日期,它显示错误。代码如下。

<?php
include('db.php'); 
$result = mysql_query("SELECT * FROM area INNER JOIN area_join ON area.id = area_join.area_id INNER JOIN shops ON area_join.shops_id = shops.id ") or die("Error: " . mysql_error());; 
while($data = mysql_fetch_assoc($result)){
if($data['name'] != $groupname){
echo "<br><hr>Area name: ".$data['name']."<br />Shops: ";
$groupname = $data['name'];
}
echo "".$data['shop_name'].", ";
}
?>

2 个答案:

答案 0 :(得分:1)

您是否尝试过group_concat? (现在不能测试它,但我认为是这样的)

SELECT areaname,group_concat(shops) FROM area INNER JOIN area_join ON area.id =    area_join.area_id INNER JOIN shops ON area_join.shops_id = shops.id
Group by areaname

http://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html#function_group-concat

答案 1 :(得分:0)

使用GROUP_CONCAT

SELECT  areaName,
        GROUP_CONCAT(DISTINCT shopname) shopList
FROM    area 
        INNER JOIN area_join 
            ON area.id = area_join.area_id 
        INNER JOIN shops 
            ON area_join.shops_id = shops.id 
GROUP   BY  areaName

列名称更改为表格中找到的原始名称。