小编Ric*_*mes的帖子

我正在尝试创建一个数据库,其中可能有n多个类别及其子类别.

首先,我尝试创建这样的邻接模型数据库

+-------------+----------------------+--------+
| category_id | name                 | parent |
+-------------+----------------------+--------+
|           1 | Electronics          |   NULL |
|           2 | Mobile               |      1 |
|           3 | Washing Machine      |      1 |
|           4 | Samsung              |      2 |
+-------------+----------------------+--------+

但是,我在删除节点时遇到了问题,比如如何管理已删除节点的子节点等.

然后我试图实现Joe Celko的嵌套订单集

每个图中的表结构:

Figure 1:
+----+-------------+-----+-----+
| id | name        | lft | rgt |
+----+-------------+-----+-----+
| 1  | Electronics | 1   | 2   |
+----+-------------+-----+-----+

Figure 2:
+----+-------------+-----+-----+
| id | name        | …

我正在做一个相当大的SQL所以我道歉,不能提供我的表的更大的例子.

SELECT 
customer_id,
agreement_id,
if( 'network' IN ( GROUP_CONCAT( DISTINCT services.service_code
                                 SEPARATOR ',' )  ),
                  'Yes','No') as networkservice
FROM customers
INNER JOIN agreement USING(customer_id)
INNER JOIN services USING(agreement_id)
GROUP BY customer_id

客户可以签订协议,协议可以提供很多服务.我想要弄清楚的是,"网络"是否是该协议中的服务之一.

由于GROUP_CONCAT返回逗号分隔列表,因此对我的情况来说感觉很完美.但我无法让它工作,我的想法已经不多了.

如果只有一个服务并且该服务是'network'则返回yes,但是如果有多个服务则返回No.

如果我使用(INT)service_id而不是没有区别,除非INT Im正在寻找列表中的第一个.但那仅适用于INT,如果'network'在列表中排在第一位,则返回No.

我试过了:

if( 'network' IN ( CAST(GROUP_CONCAT( DISTINCT services.service_code
                                      SEPARATOR ' ' ) AS CHAR)  ),
                   'Yes','No')

和

if( 'network' IN ( concat('\'',
                   GROUP_CONCAT(DISTINCT services.service_code
                                SEPARATOR '\', \'' ),
                   '\'') ), 'Yes','No')

如果我的解释听起来很混乱,我可以提供更多的例子.

谢谢.

有没有办法将查询应用于mysql数据库中的每个表？

就像是

SELECT count(*) FROM {ALL TABLES}
-- gives the number of count(*) in each Table

和

DELETE FROM {ALL TABLES}
-- Like DELETE FROM TABLE applied on each Table

我只是在VM上设置了debian 8.3并在本教程之后安装了xampp .一切正常,直到我尝试创建一个新表:

create table testtable
(
  id int(10) not null auto_increment,
  firstname varchar(255) collate utf8mb4_german2_ci not null,
  lastname varchar(255) collate utf8mb4_german2_ci not null,
  primary key (id),
  unique key (lastname)
)engine = innodb default charset=utf8mb4, collate=utf8mb4_german2_ci

我得到了错误:#1709 - Index column size too large. The maximum column size is 767 bytes. 然后我发现这来自于prefix limitation限制为767Byte Innodb并且我可以通过在my.cnf文件中设置innodb_large_prefix来解决这个问题.但我找不到文件,它不在下,/etc/而且没有 - 文件夹,我发现/etc/mysql/的唯一,但是,在我添加到文件并重新启动lampp之后.我仍然得到同样的错误.我做错了什么？my.cnf/opt/lampp/etc/innodb_large_prefix=1

编辑:SELECT version()返回5.6.14,所以innodb_large_prefix应该支持.

edit2:我知道我可以解决这个问题,只需将键的一部分设置为索引即可获得767Byte.但我想知道如何正确配置mysql.

我有两个独立的查询，它们具有相同的输出。现在我想了解哪一个更好？

查询1：

| id | select_type | table | type | possible_keys |    key | key_len |    ref | rows |                                              Extra |
|----|-------------|-------|------|---------------|--------|---------|--------|------|----------------------------------------------------|
|  1 |      SIMPLE |    t1 |  ALL |        (null) | (null) |  (null) | (null) |    9 |                                        Using where |
|  1 |      SIMPLE |    t2 |  ALL |        (null) | (null) |  (null) | (null) |    9 | Using where; Using join buffer (Block Nested Loop) |

查询2：

| id |        select_type | …

我想获取我周围的所有位置，但是该功能ST_Distance_Sphere不起作用。

我的查询：

select *, astext(location) as location from `locations`
where ST_Distance_Sphere(location, POINT(35.905069591297, 49.765869174153)) < 1000

错误：

SQLSTATE[42000]: Syntax error or access violation:
1305 FUNCTION app.ST_Distance_Sphere does not exist (SQL:
select *, astext(location) as location from `locations`
where ST_Distance_Sphere(location, POINT(35.905069591297, 49.765869174153)) < 1000)

我的本地服务器上有一个 MySQL 表。该表包括用户标记的地点的纬度和经度。我正在尝试获取距离所提供的纬度和经度 1 公里范围内标记其位置的所有 id。但我的结果却出乎我的意料。

表：map_locations

id  user_id lat     lng     place_name
1   1   28.584688   77.31593    Sec 2, Noida
2   2   28.596026   77.314494   Sec 7, Noida
3   5   28.579876   77.356131   Sec 35, Noida
4   1   28.516831   77.487405   Surajpur, Greater Noida
5   1   28.631451   77.216667   Connaught Place, New Delhi
6   2   19.098003   72.83407    Juhu Airport, Mumbai

这是 PHP 脚本

$lat = '28.596026';
$long = '77.314494';
$query = "SELECT  id,
                  (6371 * acos( cos( radians($lat) ) * cos( radians('lat') ) *
                        cos( …

CREATE TABLE `files` (
  `did` int(10) unsigned NOT NULL DEFAULT '0',
  `filename` varbinary(200) NOT NULL,
  `ext` varbinary(5) DEFAULT NULL,
  `fsize` double DEFAULT NULL,
  `filetime` datetime DEFAULT NULL,
  PRIMARY KEY (`did`,`filename`),
  KEY `fe` (`filetime`,`ext`),          -- This?
  KEY `ef` (`ext`,`filetime`)           -- or This?
) ENGINE=InnoDB DEFAULT CHARSET=utf8 ;

表中有一百万行.文件时间大多不同.数量有限ext.因此,filetime具有高基数并且ext具有低得多的基数.

该查询涉及ext和filetime:

WHERE ext = '...'
  AND filetime BETWEEN ... AND ...

这两个指标中的哪一个更好？为什么？

我有一个返回响应速度的查询.当timediff计算结果超过24小时时,它可以正常工作.我收到一个错误"java.sql.SQLException:非法小时值'31'为java.sal.Time类型的值"

"31"的数量随时间戳之间的小时差异而变化.

我不知道从哪里开始.我尝试timestampdiff()但尝试运行查询时出错.

任何意见或建议将不胜感激.

SELECT a.*
FROM
(SELECT
databunker.purchases.id                                           AS 'pur_id',
 databunker.purchases.business                                    AS 'pur_business',
 databunker.purchases.time                                        AS 'pur_time',
 databunker.customers.id                                          AS 'customer_id',
 databunker.customers.phone_number                                AS 'customer_phone_#',
 databunker.customers.is_primary                                  AS 'customer_is_primary',
 map.object_salesforce_id,
 databunker.five9_calls.campaign_name,
 databunker.five9_calls.start_timestamp AS 'start_of_call',
 TIMEDIFF(databunker.five9_calls.start_timestamp, databunker.purchases.time) AS 'speed_to_response'

FROM databunker.purchases
 LEFT OUTER JOIN databunker.customers ON (databunker.purchases.customer_id = databunker.customers.id)
 LEFT OUTER JOIN (SELECT
                    databunker.mappings.object_salesforce_id,
                    databunker.mappings.object_id AS 'map_customer_id'
                  FROM databunker.mappings
                  WHERE databunker.mappings.object_class = 'customer') AS map
   ON (databunker.purchases.customer_id = map.map_customer_id)
 LEFT OUTER JOIN databunker.five9_calls ON (map.object_salesforce_id = 
databunker.five9_calls.salesforce_id)
WHERE databunker.purchases.business = 'uma'
     AND databunker.purchases.outcome_type = …

我有 SQL 查询（参见示例）。但我找不到如何在查询生成器中编写它的方法。你有什么想法吗？这怎么可能？

WITH main AS (
    SELECT id FROM table1
)
SELECT * FROM table2
WHERE 
  table2.id IN (SELECT * FROM main)

我想获得如下格式：

$latestPosts = DB::table('posts')
                   ->select('user_id', DB::raw('MAX(created_at) as last_post_created_at'))
                   ->where('is_published', true)
                   ->groupBy('user_id');

$users = DB::table('users')
        ->joinSub($latestPosts, 'latest_posts', function ($join) {
            $join->on('users.id', '=', 'latest_posts.user_id');
        })->get();

但对于WITH