标签: join

在标题中是否可以将多个数组连接在一起而无需复制并仅使用指针？我花费了大量的计算时间将较小的数组复制到较大的数组中.注意我不能使用向量,因为umfpack(一些矩阵求解库)不允许我或我不知道如何.

举个例子:

int n = 5;

// dynamically allocate array with use of pointer

int *a = new int[n];


// define array pointed by *a as [1 2 3 4 5]

for(int i=0;i<n;i++) {
    a[i]=i+1;
}


// pointer to array of pointers ???  --> this does not work

int *large_a = new int[4];
for(int i=0;i<4;i++) {
    large_a[i] = a;
}

注意:我已经知道了一个简单的解决方案,只是迭代地将它们复制到一个新的大型数组中,但很高兴知道是否不需要复制在整个程序期间存储的重复块.我正处于学习曲线中.

感谢大家阅读

我在ubuntu 12.10和MySQL Workbench上使用MySQL 5.1.41.

我有2个产品表,t1和t2.t1是实时数据,t2是准备更新到t1的导入数据,用于更新所有新产品价格.所以我跑:

SELECT * FROM t1
JOIN t2 ON t1.id = t2.id
WHERE t1.price != t2.price;

这将返回1201记录,其中价格不同并需要更新.所以我跑:

UPDATE t1 JOIN t2 ON t1.id = t2.id
SET t1.price = t2.price
WHERE t1.price != t2.price;

这完成没有错误并报告1143行受影响,行匹配:1143更改:1143警告:0

所以这里已有的东西不对.选择查询中有1201条记录不同,但只有1143条使用相同的连接和条件进行了更改？

运行初始选择查询我希望看到仍有不同价格的58条记录.但是当它运行时,我得到了与我最初相同的1201.就像没有提交更新一样.

有任何想法吗？

我不知道为什么它不起作用.

"SELECT id GROUP_CONCAT(skill SEPARATOR ', ') FROM resume_skills GROUP BY id";

我有一个表resume_skills,然后有两个字段id和技能.它返回一个像这样的错误:

ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'GROUP_CONCAT(skill SEPARATOR ', ') FROM resume_skills GROUP BY id' at line 1

我还没搞清楚.救命？

CentOS 4或5,MySQL 5.0.77:

mysql> UPDATE
   ->  `userTable`
   -> INNER JOIN `anotherTable`
   ->   ON `userTable.userid`=`anotherTable.userid`
   -> SET `userTable.someField`="Jimmy Page"
   -> WHERE `userTable.someField`="Jim Morrison"
   ->   AND `anotherTable.date` < NOW();
ERROR 1054 (42S22): Unknown column 'userTable.someField' in 'field list'
mysql>
mysql> SELECT count(someField) FROM userTable;
+---------------+
| count(someField) |
+---------------+
|          5076 |
+---------------+
1 row in set (0.00 sec)

mysql>

我确实通过终端中的鼠标复制/粘贴将someField和userTable输入SELECT语句,所以我知道这不是拼写错误.这可能是什么问题？

谢谢!

我可能错过了一些非常明显的东西.我试图在Cake中将多个表连接在一起,但我只从第一个表中获取字段.请考虑以下代码..

$joins = array();
$joins[] =  array(
            'table' => 'products', 
            'alias' => 'Product', 
            'type' => 'LEFT',
            'conditions' => array(
                'Device.id = Product.device_id'
            )
    );

$joins[] =  array(
            'table' => 'pricepoints', 
            'alias' => 'Pricepoints', 
            'type' => 'LEFT',
            'conditions' => array(
                'Pricepoints.product_id = Product.id'
            )
    ); 
$all_products = $this->Device->find('all', array("joins" => $joins); 

这将返回以下SQL

SELECT `Device`.`id`, `Device`.`manufacturer_id`, `Device`.`name`, `Device`.`type_id`, `Manufacturer`.`id`, `Manufacturer`.`name` FROM `devices` AS `Device` LEFT JOIN products AS `Product` ON (`Device`.`id` = `Product`.`device_id`) LEFT JOIN pricepoints AS `Pricepoints` ON (`Pricepoints`.`product_id` = `Product`.`id`) …

我去桌子了。

TABLE 1: [..fields..] [CATEGORIE] [..fields..]

TABLE 2: [..fields..] [ID] [CATEGORIE] [..fields..]

我想连接有点特殊，并尝试这样：

SELECT [..other fields..], CATEGORIE, (SELECT ID FROM TABLE2 WHERE TABLE2.CATEGORIE = TABLE1.CATEGORIE) FROM TABLE1;

我想在主查询的列中有子查询的ID

这样（[]代表列）

[resultfield1] [resultfield2] [resultfield3] [ID1，ID2，ID3，ID4，...]

有没有办法负担得起？

非常感谢您的帮助，

提前致谢

当我们在Oracle中的两个表之间创建连接时,在一个或两个表上有一些额外的过滤条件,oracle将首先连接表,然后过滤或首先过滤条件然后加入.

或者简单来说,这两个中哪一个是更好的查询

假设我们有2个员工和部门表,我希望员工所有员工+部门详细信息,员工薪水大于50000

查询1:从员工e,部门d中选择e.name,d.name,其中e.dept_id = d.id,e.salary> 50000;

查询2:选择e.name,d.name from(select*from employee where salary> 50000)e,department d where e.dept_id = d.id;

我是SQL的新手.我想在SQL中加入三个表.以下是我的查询,请检查并纠正我错误的地方 -

表:

• 卡:ID,代码,名称,CC
• PGM:ID,代码
• PGMeCode:ID,Code,CC

查询:

Select *
FROM CARD
INNER JOIN PGMeCode PGMeCode.Code = CARD.Code AND PGMeCode.CC = CARD.CC
INNER JOIN PGM PGM.Code = Card.Code
WHERE Card.ID = 'SomeThing'

我不知道我做错了什么.请建议我!!

提前致谢.

嗨,我需要能够链接相关的应用程序,并试图找出保存,更新和删除的最佳实践表结构.

我有下表:

APPLICATION{ApplicationId, Name, Description}

我需要能够说应用程序1链接到2和3.因此,如果打开应用程序2,您会看到它链接到应用程序1和3.然后应用程序3链接到1和2.

链表的最佳表结构是什么？

编辑

我的主要查询是我需要每个连接的记录,即对于应用程序1,2和3,我需要6条记录吗？1-> 2,1-> 3,2-> 1,2-> 3,3-> 1,3-> 2 ?? 如果不是,返回给定ID的所有链接应用程序的最佳查询是什么？

谢谢,你能帮助我吗,我正在迁移一个应用程序并转到mysql 5.5非常慢的查询返回非常感谢你的帮助.

SELECT DISTINCT b.cc46_cc55_field_id_id AS row_id, t0.cc46_value AS f_264, t1.cc46_value AS f_265, t2.cc46_value AS f_266
FROM tc46_fields_values b 
LEFT OUTER JOIN tc46_fields_values t0 ON b.cc46_cc55_field_id_id = t0.cc46_cc55_field_id_id 
  AND t0.cc46_cc26_field_id = 264 
LEFT OUTER JOIN tc46_fields_values t1 ON b.cc46_cc55_field_id_id = t1.cc46_cc55_field_id_id 
  AND t1.cc46_cc26_field_id = 265 
LEFT OUTER JOIN tc46_fields_values t2 ON b.cc46_cc55_field_id_id = t2.cc46_cc55_field_id_id 
  AND t2.cc46_cc26_field_id = 266
WHERE b.cc46_cc55_field_id_id IN (5924,5925,5926,5927,5928,5929,5930,5931,5932,5933,5934,5935,5936,5937,5938,5939,5940,5941,5942,5943,5944,5945,5946,5947,5948,5949,5950,5951,5952,5953,5954,5955,5956,5957,5958,5959,5960,5961,5962,5963,5964,5965,5966,5967,5968,5969,5970,5971,5972,5973,5974,5975,5976,5977,5978,5979,5980,5981,5982,5983,5984,5985,5986,5987,5988,5989,5990,5991,5992,5993,5994,5995,5996,5997,5998,5999,6000,6001,6002,6003,6004,6005,6006,6007,6008,6009,6010,6011,6012,6013,6014,6015,6016,6017,6018,6019,6020,6021,6022,6023,6024,6025,6026,6027,6028,6029,6030,6031,6032,6033,6034,6035,6036,6037,6038,6039,6040,6041,6042,6043,6044,6045,6046,6047,6048,6049,6050,6051,6052,6053,6054,6055,6056,6057,6058,6059,6060,6061,6062,6063,6064,6065,6066,6067,6068,6069,6070,6071,6072,6073,6074,6075,6076,6077,6078,6079,6080,6081,6082,6083,6084,6085,6086,6087,6088,6089,6090,6091,6092,6093,6094,6095,6096,6097,6098,6099,6100,6101,6102,6103,6104,6105,6106,6107,6108,6109,6110,6111,6112,6113,6114,6115,6116,6117,6118,6119,6120,6121,6122,6123,6124,6125,6126,6127,6128,6129,6130,6131,6132,6133,6134,6135,6136,6137,6138,6139,6140,6141,6142,6143,6144,6145,6146,6147,6148,6149,6150,6151,6152,6153,6154,6155,6156,6157,6158,6159,6160,6161,6162,6163,6164,6165,6166)