我有一些价值观连字符分隔的一排:连字符分隔的值和LEFT_JOIN
表:live_customers
行:地区
id | areas
1 | 10-20-30
2 | 40-50-60
...
使用此...
LEFT JOIN $table5 AS table5 ON live.areas REGEXP CONCAT('(^|-) ?',table5.id,' ?($|-)')
我的结果看起来像:
(tab id:1) area: 10
(tab id:1) area: 20
...
(tab id:2) area: 40
...
但我相信:
(tab id:1) area: 10,20,30
(tab id:2) area: 40,50,60
我怎么能解决?
编辑:
完整的查询看起来像:
SELECT live.*,
live.id AS lid,
table1.id, table1.value AS tn_val,
table2.id, table2.value AS tp_val,
table3.id, table3.value AS ht_val,
table5.id, table5.value AS ar_val
FROM $dblist AS live
LEFT JOIN $table1 AS table1 ON live.town = table1.id
LEFT JOIN $table2 AS table2 ON live.htype = table2.id
LEFT JOIN $table3 AS table3 ON live.ht = table3.id
LEFT JOIN $table5 AS table5 ON live.areas REGEXP CONCAT('(^|-) ?',table5.id,' ?($|-)')
ORDER BY live.id ASC
PHP呼应:
...
if ($post['areas']){ // Debugging areas stuff
echo '<strong>'.$_areas.': (ar_val)</strong> '.$post['ar_val'].'<p>';
echo '<strong>'.$_areas.': (areas)</strong> '.$post['areas'].'<p>';
}
...
EDIT2:
这是相当我很难解释英语我的问题,但我想尽我所能:)
表中的“live_customers”我确实有这样的:
id | areas
1 | 10-20-30
2 | 40-50-60
...
表中的“区域”
(这是一个完全不同的表):
id | value
38 | Zone1
39 | Zone2
40 | Zone3
...
在你看到的只是SQL查询表变量,因为我previousvly宣布他们在页面的顶部:
$table5 = 'areas';
$dblist = 'live_customers';
等。
解决方案
感谢任何人的答案和让我知道“GROUP_CONCAT”。
这里是我的解决方案:
SELECT live.*,
live.id AS lid,
table1.id, table1.value AS tn_val,
table2.id, table2.value AS tp_val,
table3.id, table3.value AS ht_val,
table5.id, GROUP_CONCAT(table5.value) AS ar_val
FROM $dblist AS live
LEFT JOIN $table1 AS table1 ON live.town = table1.id
LEFT JOIN $table2 AS table2 ON live.htype = table2.id
LEFT JOIN $table3 AS table3 ON live.ht = table3.id
LEFT JOIN $table5 AS table5 ON FIND_IN_SET(table5.id, REPLACE(live.areas, '-', ','))
GROUP BY live.id
结果是我所期待的^^
最有可能使用'GROUP_CONCAT()'。请发布其余的查询,我们可以帮助您加入。 – 2013-03-11 14:11:32
感谢您的回答,只是对其进行了编辑。 – Imero 2013-03-11 14:17:47
我没有看到有类似'areas'的任何列。你想列出哪一列?这将有助于查看实际的表和列,而不是PHP变量。 – 2013-03-11 14:25:07