select id,title,release_date from tbl_movies where release_date > '2014-02-20' or release_date='' order by release_date asc"id" | "title" "release_date" "49" | "Fandry" | "" "92" | "Govinda" | "" "112"| "Gadad Jambhal" | "" "114"| "Aandhali Koshimbir" | "" "118"| "Mee Aani U" | "" "125"| "Tapal" ...
似乎无法到达我的更新查询中的下一步.我能够成功查看与select no问题相关的列：SELECT sales_flat_order_grid.entity_id,sales_flat_order_grid.increment_id,sales_flat_order.coupon_code FROM sales_flat_order_grid INNER JOIN sales_flat_order ON sales_flat_order_grid.entity_id = sales_flat_order.entity_id WHERE sales_flat_order_grid.increment_id = "12345678";这显示了与正确的increment_id相关的3列. 下一步...
我在mysql中有以下两个表： 用户：+--------+-----------+ | userId | userName | +--------+-----------+ | 1 | magnus | | 2 | fabiano | | 3 | alexander | | 4 | veselin | +--------+-----------+游戏：+--------+---------+---------+ | gameId | userId1 | userId2 | +--------+---------+---------+ | 1 | 1 | 2 | | 2 | 1 | 3 | | 3 | 2 | 3 ...
我想在mysql查询的条件中将可能的值包含为null.这是我的查询：但它没有显示任何值.请建议我如何在WHERE IN条件中包含空值.SELECT * FROM TABLE1 WHERE COLUMN1 IN ('YES','NO',NULL);解决方法:尝试类似的东西：SELECT * FROM TABLE1 WHERE (COLUMN1 IN ('YES', 'NO') OR COLUMN1 IS NULL)这里的语句将被解析为COLUMN1 =’YES’或COLUMN1 =’NO’或field =’NULL’.在那里放一个NULL会得到COLUMN1 = null,这将无效.
我正在尝试创建一个聊天应用程序,我想拉出用户1和2的对话. 表：+----+---------------------+-----------+---------+ | id | message | from_user | to_user | +----+---------------------+-----------+---------+ | 1 | hello trick | 1 | 2 | | 2 | hi raf i am okay | 2 | 1 | | 3 | how is jo doing | 1 | 2 | | 4 | Hey | ...
我正在制作成员的自动滚动.我有两个mysql查询,我想显示有照片的成员并在上次登录时订购.所以在第一集我没有任何问题,因为我这样做$sql=mysql_query("SELECT * FROM accounts WHERE avatar != '' ORDER BY lastlogin DESC LIMIT 50");这工作正常.但是当用户滚动并转到页面底部时,我正在尝试加载由上面完成的相同过滤订购的下一组50个用户.所以我正在这样做$sql = mysql_query("SELECT * FROM accounts WHERE lastlogin < '$last_ms...
我有一张桌子,看起来像：+-----------+----------+ + person_id + group_id + +-----------+----------+ + 1 + 10 + + 1 + 20 + + 1 + 30 + + 2 + 10 + + 2 + 20 + + 3 + 10 + +-----------+----------+我需要一个查询,只返回组10和20和30的person_ids(只有person_id：1).我不知道如何做到这一点,因为我可以看到它需要我按person_id对行进行...
以下查询从表名返回所有new_name的结果集SELECTCASEWHEN nickname = '' THEN fullname ELSE nickname END AS new_name, FROM names我正在尝试使用以下查询通过new_name进行搜索SELECTCASEWHEN nickname = '' THEN fullname ELSE nickname END AS new_name, FROM namesWHERE new_name LIKE '%Joh%'但是我得到了错误 ‘where子句’中的未知列’new_name’ 有没有一种方法可以在不使用子查询的情况下使用此功能？解决方法...
我知道rand()的顺序不是从数据库中绘制随机值的最快方法,但我的数据库很小,而且此时;我只是想让它起作用！哈哈.这是我的代码：include('includes/dbc.php'); $top_query = "SELECT * FROM top WHERE 'occasion_id =" . $occasion . "' AND 'temperature_id = " . $temperature . "' AND 'user_id = " . $user_id . "'ORDER BY RAND() LIMIT 1"; $top_result = mysqli_query($dbc, $top_query) or die (' The top SELECT query is b...
哪些表和列应该有索引？我有一个关于flow_permanent_id和entry_id的索引,但似乎是在使用filesort？ 我该怎么做才能优化这个？mysql> explain SELECT COUNT(*) AS count_all, entry_id AS entry_id FROM `votes` WHERE `votes`.`flow_permanent_id` = '4fab490cdc1c82cfa800000a' GROUP BY entry_id; +----+-------------+-------+------+----------------------------------+----------------------------------+---------+-------...
我在MySQL中遇到一个SELECT查询有点问题,我会感谢你的一些指示.请随时指出我现有的答案(如果有的话我错过了). 查询目前如下：SELECT e.*, ie.aaa, ue.bbb, ue.ccc FROM ie LEFT JOIN e ON ie.e_id = e.e_id LEFT JOIN ue ON ie.e_id = ue.e_id WHERE ie.other_id = ? AND ue.unrelated_id = ? ORDER BY ...有三个表：即e和ue. 表ie和ue是e的关系,因此包含外键(e_id). ？表示输入参数. 问题是ue.unrelated_id =？部分.我真正想做的是...
我有以下查询：SELECT IF(TIMESTAMPDIFF(HOUR,users_codes.timestam,CURRENT_TIMESTAMP()) < 48,'no','yes') AS Expired FROM users_codes如何仅返回“否”的行？解决方法:添加WHERE条件.您不能在where子句上使用ALIAS.SELECT IF(TIMESTAMPDIFF(HOUR,users_codes.timestam,CURRENT_TIMESTAMP()) < 48,'no','yes') AS Expired FROM users_codes WHERE IF(TIMESTAMPDIFF(HOUR,users_codes.timestam,CURRENT_TIMESTAMP()) < 48,'n...
我通常会像这样加入：$query = "SELECT table1.ColA, table2.ColC "."FROM table1, table2"."WHERE table1.uid = table2.uid";但在这种情况下,WHERE部分已经用于使用PHP变量指定记录.在这种情况下如何加入表2？SELECT SQL_CALC_FOUND_ROWS ColA, ColB FROM Table1 WHERE value = $value解决方法:请试试这个$query = "SELECT table1.ColA, table2.ColC FROM table1, table2 WHERE table1.uid = table2.uid and table1.value=$value"...
我有一个数据库,我想对这样的结果进行排序：$query1 = "SELECT * FROM pictures LIMIT 0,10 "; $result = mysql_query($query1);while($row = mysql_fetch_array($result)) echo $row['picture_rating'];和$query1 = "SELECT * FROM pictures WHERE column_name = 'column_text' "; $result = mysql_query($query1);while($row = mysql_fetch_array($result)) echo $row['picture_rating'];上面的两个代码都正常工...
我有一个跟踪交易的表. 该表设置为：transactions:id, account_id, budget_id, points, type我需要返回每个budget_id的点数之和,其中type =’allocation’和point的总和,其中type =’issue’ 我知道如何在一个查询中执行每个操作,但不能同时执行. 预期结果集：budget_id allocated issued434 200000 100242 100000 5020621 45000 3940解决方法: SELECT budget_id, SUM(IF(type = 'allocatio...