我有数据表:
| id | status |
+-----+--------+
| 1 | 1 |
| 2 | 1 |
| 3 | 0 |
| 4 | 2 |
| 5 | 2 |
我有文件,我需要加载到此表并替换:
| id | status |
+----+--------+
| 1 | 1 |
| 2 | 0 |
| 3 | 0 |
| 4 | 0 |
| 5 | 1 |
我有一个条件:如果table = 2中的状态和file = 0中的状态,则将状态保留在table = 2中,否则将表中的状态替换为file。
查询后我需要获取新数据:
| id | status |
+-----+--------+
| 1 | 1 |
| 2 | 0 |
| 3 | 0 |
| 4 | 2 |
| 5 | 1 |
我正在尝试使用查询:
load data local
infile '".$file."'
replace
into table t1
fields terminated by ',' enclosed by '\"'
(@tid,
teacher_name,
email,
@pid,
tca_form_type,
prod_company,
prod_name,
@stts)
set status = if((select status from (select status from t1 where teacher_id=@tid and prod_id=@pid) as tmp)=2 and @stts=0,status,@stts),
teacher_id = @tid, prod_id = @pid
之后我将状态字段设为NULL 如何解决这个问题?
修改
我试过了:
set status = if((select @var:=status from (select status from t1 where teacher_id=@tid and prod_id=@pid) as tmp)=2 and @stts=0,@var,@stts),
但结果状态2更改为0.
表架构:
CREATE TABLE `table` (
`id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
`teacher_id` VARCHAR(20) NOT NULL COLLATE 'utf8_unicode_ci',
`status` INT(11) NULL DEFAULT NULL,
`prod_id` VARCHAR(10) NOT NULL COLLATE 'utf8_unicode_ci',
PRIMARY KEY (`id`),
UNIQUE INDEX `teacher_id_UNIQUE` (`teacher_id`, `prod_id`)
)
COLLATE='utf8_unicode_ci'
ENGINE=InnoDB
ROW_FORMAT=COMPACT
AUTO_INCREMENT=2053;
真实数据:
| id | teacher_id | status | prod_id |
+-----+------------+--------+---------+
| 1 | a1 | 1 | 15 |
| 2 | a1 | 1 | 16 |
| 3 | a1 | 0 | 17 |
| 4 | a2 | 2 | 16 |
| 5 | a2 | 2 | 18 |
| 6 | a3 | 0 | 15 |
| 7 | a3 | 1 | 20 |
文件数据:
| teacher_id | status | prod_id |
+------------+--------+---------+
| a1 | 0 | 15 |
| a1 | 1 | 16 |
| a1 | 0 | 17 |
| a2 | 1 | 16 |
| a2 | 0 | 18 |
| a3 | 1 | 15 |
| a3 | 1 | 20 |
我的临时解决方案:
load data local
infile '".$file."'
into table table_tmp
fields terminated by ',' enclosed by '\"'
(teacher_id,
teacher_name,
email,
prod_id,
tca_form_type,
prod_company,
prod_name,
status);
INSERT INTO table
(teacher_id, teacher_name, email, status, prod_id, tca_form_type, prod_company, prod_name)
SELECT teacher_id, teacher_name, email, `status`, prod_id, tca_form_type, prod_company, prod_name FROM table_tmp
ON DUPLICATE KEY UPDATE table.status = IF(table.status = 2 and VALUES(status) = 0, table.status, VALUES(status));
答案 0 :(得分:1)
我认为这应该足够了:
load data local
infile '".$file."'
replace
into table t1
fields terminated by ',' enclosed by '\"'
(@tid,
teacher_name,
email,
@pid,
tca_form_type,
prod_company,
prod_name,
@stts)
set status = if(status = 2 and @stts = 0, status, @stts),
teacher_id = @tid, prod_id = @pid;
如果这没有帮助,您可以尝试使用values()
function,但它会说它是INSERT ... ON DUPLICATE KEY UPDATE
声明。
在INSERT ... ON DUPLICATE KEY UPDATE语句中,您可以使用UPDATE子句中的VALUES(col_name)函数来引用语句的INSERT部分中的列值。换句话说,UPDATE子句中的VALUES(col_name)引用了将插入的col_name的值,没有发生重复键冲突。此功能在多行插入中特别有用。 VALUES()函数仅在INSERT语句的ON DUPLICATE KEY UPDATE子句中有意义,否则返回NULL。请参见第13.2.5.3节“INSERT ... ON DUPLICATE KEY UPDATE语法”。
如果这也没有帮助,请提供表格架构等等,这样我们就可以尝试自己,不必猜测。