为什么这个离开加入工作？

Question

我遇到了这段代码的麻烦。错误接近LIKE。这是代码......

try {
    $st = $db->query('SELECT', "SELECT a.`fast_cash` as `fast`, a.`urgent_cash` as `urgent`, b.`store_name` as `store` FROM `Locations` a LEFT JOIN `loan_balances` b ON b.`store_name` LIKE '%' + a.`nick` + '%'", 'array'); 
} catch (Exception $e) {
    echo $e->getMessage();
}

Answer 1

您正在寻找的on子句是：

ON b.`store_name` LIKE concat('%', a.`nick`, '%')

此查询不需要where子句。