Question

学

+-----+-----------+----------+
| id  | the_photo | column3  |
+-----+-----------+----------+
| 11  |     NULL  |     abc  |
| 22  |     NULL  |     asf  |
| 33  |     NULL  |     asag |
+-----+-----------+----------+

school_images

+-----+-----------+-------+    
| id  | school_id | photo |
+-----+-----------+-------+
| 1   | 11        |  1    |
| 2   | 22        |  0    |
| 3   | 33        |  1    |
+-----+-----------+-------+

...

只有当照片值= 1时才需要将值插入学校的the_photo列，例如：school_images：

学校

+-----+-----------+
| id  | the_photo |
+-----+-----------+
| 11  |     1     |
| 22  |     NULL  |
| 33  |     3     |
+-----+-----------+

是否可以编写一个简单的查询来为所有行执行此操作？对于一行我知道如何插入它，但我怎么能自动惰性多行。

Answer 1

UPDATE school AS s, school_images AS si SET s.the_photo = si.id WHERE s.id = si.school_id AND si.photo = 1;

Answer 2

这可能是您正在寻找的内容：SELECT INTO。取自链接：

INSERT INTO tbl_temp2 (fld_id)
    SELECT tbl_temp1.fld_order_id
    FROM tbl_temp1 WHERE tbl_temp1.fld_order_id > 100;

Answer 3

请注意!! 只有在您想要在新插入时执行上述请求的操作时，此答案才有效。否则，您将需要一个不同的策略

你应该有这样的东西：

从school_images中选择所有值，然后预测它们：

foreach($result as $res) { if($res['photo'] == 1) { // see step 2 } else { // see step 3 } }
如果你在school_photos中的值为1，那么：

INSERT INTO school (the_photo, colum3) VALUES ($res['id'], $some_value)

如果你没有，那么就像你现在一样进行正常插入。

Answer 4

您也可以使用内部联接来实现这一目标。

 Update School inner join school_images on School.id = school_images.school_id and photo=1
set the_photo = school_images.id ;

Answer 5

您应该使用INNER JOIN将school加入school_images：

UPDATE
  school INNER JOIN school_images
  ON school.id = school_images.school_id
     AND school_images.photo=1
SET
  school.the_photo=school_images.id

在另一个表mysql的一列中插入值

5 个答案: