CASE in select query Posted by:jim giner Date: April 04, 2012 02:31PM I vaguely remember using a case clause in a select statement in a former occupation, but now that I am using mysql (with php) I am encounter
...student m where m.id = 1; [Err] 1064 - You have an error in your SQL syntax; 描述: delete from table 这样的句子中 3.9K30 MySQL中CASE的使用 使用演示样例: mysql> select id,name, (gender) as ‘性别’ from t_user; +—-+———+——+ | id | name | 性别 |...| 1 | |...
IF ELSE或CASE in MySQL语句? mysql、sql、if-statement 这是我的MySQL语句:tableG.id,test_group.Merged as TestGroupMerged,test_suite.Merged as TestSuiteMerged,Long_image_name as Image, test_table 我们将图像添加到另一个表( test_B ),如果test_B不包含它在test_A中应该看到...
mysql> call exp_case(1); Query OK,1row affected (0.09sec) mysql> call exp_case(2); Query OK,1row affected (0.03sec) mysql>select*fromstudentswheresidin(1,2);+---+---+---+---+---+ | sid | sname | gender | dept_id | brithday | +...
WHEN col_1 IN ('a') THEN '第二类' ELSE'其他' END === update ffmpeg_agent_node set remark = case when isnull(remark) then 'docker' else concat(remark, '123') end where name = '10.77.40.194'; mysql> select * from ffmpeg_cmd_template; +---+---+---...
WHEN pref_name in ('福冈', '佐贺', '长崎') THEN '四国' ELSE '其他' END AS district FROM poptbl 然后再加载SUM()聚合函数求解每一个分类地区的人数。原文提到先进行GROUP BY再进行SELECT的方式在一些诸如Oracle、DB2、SQL Server 等数据库里是会报错的,在MYSQL和PostgreSQL中是能顺利执行的。
在MySQL 官方文档中有如下关于SQL MODE和NOT NULL的描述: Strict mode controls how MySQL handles invalid or missing values in data-change statements such as INSERT or UPDATE. A value can be invalid for several reasons. For example, it might have the wrong data type for the column, or it might...
Description:When using built-in functions (i.e. get_lock) the digest hashing depends on the case used in the function. This makes it much harder to use a query rewrite rule, since there needs to be one-rule per case.How to repeat:The last select returns 2 rows. Ideally it should ret...
Update with the above query: mysql> update t1_hash a set a.a_code=(select case when count(1)>1 then 1 else 0 end from t1_range b where a.infom=b.infom group by b.infom); Query OK, 3 rows affected (0.00 sec) Rows matched: 3 Changed: 3 Warnings: 0 mysql> select a_code ...
I don't understand how YOU got THAT from his simple "select x=y". His statement is (to me) a simple query that returns (yes) a boolean result, but nothing else. In any case, I solved my problem, but am curious how "select x=y" could have possibly done it....