Question

我的PDO查询引发错误

42000 1064 SQL语法错误

$sql = "SELECT * FROM {$this->config->__get('table_medicine')} WHERE patient_id = ? AND medicine LIKE %?%";
    $query = $this->dbh->prepare($sql);
    $data = array($patient_id, $medicine);
    $response = $query->execute($data) or die(implode(" ", $query->errorInfo()));

有人能看出我做错了什么吗？

Answer 1

%需要位于LIKE的字符串参数内。在SQL中使用CONCAT()：

$sql = "SELECT * FROM {$this->config->__get('table_medicine')} 
        WHERE patient_id = ? AND medicine LIKE CONCAT('%', ?, '%')";

或在PHP中进行连接：

$data = array($patient_id, '%'.$medicine.'%');

php pdo sql query错误：1064 with LIKE

1 个答案: