学校
+-----+-----------+----------+
| id | the_photo | column3 |
+-----+-----------+----------+
| 11 | NULL | abc |
| 22 | NULL | asf |
| 33 | NULL | asag |
+-----+-----------+----------+
学校图像
+-----+-----------+-------+
| id | school_id | photo |
+-----+-----------+-------+
| 1 | 11 | 1 |
| 2 | 22 | 0 |
| 3 | 33 | 1 |
+-----+-----------+-------+
只有当学校图像的照片值=1时，才需要将值插入学校的_photo列：
学校
+-----+-----------+
| id | the_photo |
+-----+-----------+
| 11 | 1 |
| 22 | NULL |
| 33 | 3 |
+-----+-----------+
是否有一个简单的查询可以为所有行编写？对于一行，我知道如何插入它，但如何对多行自动惰性。
这可能就是您想要的：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;
小心 只有当您想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，然后对其进行预处理：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)
- 如果没有，那么像现在一样执行正常插入
您也可以使用内部联接来实现这一点。
Update School inner join school_images on School.id = school_images.school_id and photo=1
set the_photo = school_images.id ;
您应该使用INNER JOIN将school与school_images:连接起来
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
UPDATE school AS s, school_images AS si SET s.the_photo = si.id WHERE s.id = si.school_id AND si.photo = 1;