我有2个表设计和商店,我需要设计商店ID是'某事'的设计,特别是设计应允许的商店ID。
示例 - 商店ID是'401274'然后我想要设计应该属于商店ID ='401274'并且在商店ID(401274)上,设计ID应该在商店的design_id_allowed列中可用,我已经写了一个子查询但它在这里不起作用是语法错误
我的查询是,
SELECT `designs`.`id`, (SELECT `stores`.`id` as 'store_id' FROM `stores` WHERE `stores`.`id`='401274' And `stores`.`design_id_allowed` like '%#'.`designs`.`id`.'#%') FROM `designs` WHERE `designs`.`store_id` = '401274'
错误是,
1064 - 您的SQL语法出错;检查与MySQL服务器版本对应的手册,以获得正确的语法
在'SELECT
stores
附近。id
为'store_id'FRNstores
WHERE 第1行stores
。id
='401274'
答案 0 :(得分:2)
尝试将您的查询更改为
SELECT
`designs`.`id`,
(SELECT `stores`.`id` as 'store_id' FROM `stores` WHERE `stores`.`id`='401274' And `stores`.`design_id_allowed` like '%2629%')
FROM `designs`
WHERE `designs`.`store_id` = '401274'
PHP
$db = new PDO($dsn, $user, $password);
$sql = "
SELECT
`designs`.`id`,
(SELECT `stores`.`id` as 'store_id' FROM `stores` WHERE `stores`.`id`='".$db->quote($var1)."' And `stores`.`design_id_allowed` like '%".$db->quote($var2)."%')
FROM `designs`
WHERE `designs`.`store_id` = '401274'
";
$data = $db->query($sql)->fetchAll();
答案 1 :(得分:1)
列名由<body>
<div id="myContainer"></div>
<script src="Main.js"></script>
<script>
var app = Elm.Main.init( {
node: document.getElementById("myContainer"),
flags: window.options
} );
</script>
</body>
,
答案 2 :(得分:1)
在查询中使用正则表达式,如
WHERE `designs`.`store_id` REGEXP '^#[0-9]+,#[0-9]+$'
答案 3 :(得分:1)
试试这个好友 -
SELECT
`designs`.`id`,
(SELECT `stores`.`id` as 'store_id' FROM `stores` WHERE `stores`.`id`='401274' And `stores`.`design_id_allowed` like '%2629%')
FROM `designs`
WHERE `designs`.`store_id` = '401274'