我有一个
MySQL查询,直接在我的本地
MySQL数据库上执行时工作正常,但在通过
PHP执行时显示不同的结果.
SELECT a.id,a.title,a.public,a.sysstamp,a.password,t.thumbURL,t.count FROM 0_lychee_albums AS a LEFT JOIN (SELECT id,album,thumbURL,@num := IF(@group = album,@num + 1,0) AS count,@group := album AS dummy from 0_lychee_photos WHERE album != 0 ORDER BY album,star DESC) AS t ON a.id = t.album WHERE count <= 2 OR count IS NULL;
或作为一个班轮:
SELECT a.id,t.count FROM 0_lychee_albums AS a LEFT JOIN (SELECT id,@group := album AS dummy FROM 0_lychee_photos WHERE album != 0 ORDER BY album,star DESC) AS t ON a.id = t.album WHERE count <= 2 OR count IS NULL;
结果:
| id | title | public | sysstamp | password | thumbURL | count | | 71 | [Import] 01 | 0 | 1415091268 | NULL | cad008943372d984a9b74378874128f8.jpeg | 0 | | 72 | [Import] 9n401238 | 0 | 1415091268 | NULL | 7b832b56f182ad3403521589e2815f67.jpeg | 0 | | 72 | [Import] 9n401238 | 0 | 1415091268 | NULL | f058f379ce519f1d8a2ff8c0f5003631.jpeg | 1 | | 72 | [Import] 9n401238 | 0 | 1415091268 | NULL | a4d59377bed059e3f60cccf01a69c299.jpeg | 2 | | 73 | Untitled | 0 | 1415114200 | NULL | NULL | NULL |
PHP结果:
| id | title | public | sysstamp | password | thumbURL | count | | 71 | [Import] 01 | 0 | 1415091268 | NULL | cad008943372d984a9b74378874128f8.jpeg | 0 | | 72 | [Import] 9n401238 | 0 | 1415091268 | NULL | 7b832b56f182ad3403521589e2815f67.jpeg | 0 | | 72 | [Import] 9n401238 | 0 | 1415091268 | NULL | f058f379ce519f1d8a2ff8c0f5003631.jpeg | 0 | | 72 | [Import] 9n401238 | 0 | 1415091268 | NULL | a4d59377bed059e3f60cccf01a69c299.jpeg | 0 | | 72 | [Import] 9n401238 | 0 | 1415092318 | NULL | 7b832b56f182ad3403521589e2815f67.jpeg | 0 | | 72 | [Import] 9n401238 | 0 | 1415092369 | NULL | cad008943372d984a9b74378874128f8.jpeg | 0 | | 72 | [Import] 9n401238 | 0 | 1415092369 | NULL | 84030a64a1f546e223e6a46cbf12910f.jpeg | 0 | | 73 | Untitled | 0 | 1415114200 | NULL | NULL | NULL |
a)计数没有像它应该增加的那样
b)因为a)它显示的行数多于应有的数量(应限制为每个ID 3个)
我多次检查它,两个查询完全相同. PHP中没有用户输入或任何差异.
我已经检查了similar questions,但他们没有帮助.以下查询在MysqL和PHP上显示相同的结果:
SHOW VARIABLES LIKE 'character_set%'; SHOW VARIABLES LIKE 'collation%';
是否有人意识到存在这种差异的问题?
编辑更多信息:
$database = new MysqLi($host,$user,$password,$database); $query = "SELECT a.id,star DESC) AS t ON a.id = t.album WHERE count <= 2 OR count IS NULL"; $albums = $database->query($query); while ($album = $albums->fetch_assoc()) { print_r($album); }
$database->set_charset('utf8'); $database->query('SET NAMES utf8;');
对.不保证select子句中表达式的计算顺序.因此,变量赋值可以按不同的顺序发生,具体取决于调用查询的方式.
您可以通过将所有变量赋值放入单个表达式来解决此问题.尝试将此子查询用于t:
(SELECT id,(@num := IF(@group = album,if(@group := album,0) ) ) as count FROM 0_lychee_photos CROSS JOIN (SELECT @num := 0,@group := NULL) vars WHERE album <> 0 ORDER BY album,star DESC ) t
documentation中的具体说明是:
As a general rule,other than in SET statements,you should never
assign a value to a user variable and read the value within the same
statement. For example,to increment a variable,this is okay:06001
For other statements,such as SELECT,you might get the results you
expect,but this is not guaranteed. In the following statement,you
might think that MysqL will evaluate @a first and then do an
assignment second:06002
However,the order of evaluation for expressions involving user variables is undefined.