Question

这是我的表

sno | city | state | stateid

1   | chennai | tamilnadu | 1

2   | dindigul | tamilnadu | 1

3   | trivandrum | Kerala | 2

4   | cochin | Kerala | 2

我想这样显示（应该在升序时选择状态ID，并且应该显示城市仍然会到达更大的州议会大厦）

塔米尔纳杜

奈

丁迪古尔

喀拉拉

特里凡得琅

知

我怎样才能实现这个目标？

我到目前为止所尝试的是

(select city from (select * from state order by statid asc))

内部查询正在运行（通过stateid asc从状态顺序中选择*），但是当我使用(select city from (innerquery)时。

投掷#1248 - Every derived table must have its own alias

我该如何解决这个问题？

Answer 1

在PHP中执行：

 <?php
 $con = mysqli_connect("localhost", "root", "", "dbname");

$rs = $con->query('select city, state from mytable');
while ($row = $rs->fetch_array(MYSQLI_ASSOC)) {
  $state_cities[$row['state']][] = $row['city'];
}

foreach ($state_cities as $state => $cities) {
  echo "<b>".$state . "</b><br>";
  foreach ($cities as $title) {
  echo $title . "<br>";
  }
}

?>

Answer 2

(select city from (select * from state order by statid asc))

它需要表名称Where和字段名称与运算符。

(select city from <TableName> WHERE <FieldName> <Operator> (select * from state order by statid asc))

Answer 3

我认为这不是一个正确的规范化表，因为state_name和id变得多余。如果有可能编辑表结构，您可以按以下方式分解表：

cities:

sno city_name state_id
1 chennai 1
2 dindigul  1
3 trivandrum 2
4 cochin 2

states:

state_id state_name
1 tamilnadu
2 kerala

然后您可以使用以下查询：

SELECT c.city_name,s.state_name FROM cities c INNER JOIN states s ON c.state_id = s.`state_id` ORDER BY c.state_id ASC;

SQL DUMP：

/*
SQLyog Ultimate v10.00 Beta1
MySQL - 5.5.24-log : Database - test
*********************************************************************
*/


/*!40101 SET NAMES utf8 */;

/*!40101 SET SQL_MODE=''*/;

/*!40014 SET @OLD_UNIQUE_CHECKS=@@UNIQUE_CHECKS, UNIQUE_CHECKS=0 */;
/*!40014 SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0 */;
/*!40101 SET @OLD_SQL_MODE=@@SQL_MODE, SQL_MODE='NO_AUTO_VALUE_ON_ZERO' */;
/*!40111 SET @OLD_SQL_NOTES=@@SQL_NOTES, SQL_NOTES=0 */;
CREATE DATABASE /*!32312 IF NOT EXISTS*/`test` /*!40100 DEFAULT CHARACTER SET latin1 */;

USE `test`;

/*Table structure for table `cities` */

DROP TABLE IF EXISTS `cities`;

CREATE TABLE `cities` (
  `sno` int(11) NOT NULL,
  `city_name` varchar(100) NOT NULL,
  `state_id` int(100) NOT NULL,
  PRIMARY KEY (`sno`,`state_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

/*Data for the table `cities` */

insert  into `cities`(`sno`,`city_name`,`state_id`) values (1,'chennai',1),(2,'dingi',1),(3,'cochin',2),(4,'efef',2);

/*Table structure for table `states` */

DROP TABLE IF EXISTS `states`;

CREATE TABLE `states` (
  `state_id` int(11) NOT NULL,
  `state_name` varchar(100) NOT NULL,
  PRIMARY KEY (`state_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

/*Data for the table `states` */

insert  into `states`(`state_id`,`state_name`) values (1,'tamilnadu'),(2,'kerala');

/*!40101 SET SQL_MODE=@OLD_SQL_MODE */;
/*!40014 SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS */;
/*!40014 SET UNIQUE_CHECKS=@OLD_UNIQUE_CHECKS */;
/*!40111 SET SQL_NOTES=@OLD_SQL_NOTES */;

SQL双精选

3 个答案: