Question

我正在尝试使用具有更改的表来更新表。插入内容是：

-- services ------------------
insert into upd_services values
  ( 100, 'Consult- no charge',            0.00)
, ( 101, 'Routine Exam- Bird',           35.00)
, ( 102, 'Followup Exam-Bird',           32.00)
, ( 103, 'Routine Exam-Feline',          50.00)
, ( 104, 'Followup Exam-Feline',         45.00)
, ( 105, 'Routine Exam-Reptile',         25.00)
, ( 106, 'Followup Exam-Reptile',        23.00)
, ( 107, 'Rabies_V-2008 SFall-Canine',   15.00)
, ( 108, 'Rabies_V-2008 SFall-Feline',   15.00)
, ( 109, 'Rabies_Ve-2008 Spring-Canine', 25.00)
, ( 110, 'Rabies_V-2008 Spring-Feline',  25.00)
, ( 111, 'Rabies_V-Rodent',              20.00)
, ( 112, 'Rabies_V-2008 Winter-Canine',  15.00)
, ( 118, 'First Feline PCR',             20.25)
, ( 119, 'Second Feline PCR',            20.25)
, ( 120, 'Third Feline PCR',             20.25)
, ( 121, 'Flu Rhino Vacc',               26.00)
, ( 122, 'Scaly Mite',                   35.00)
, ( 123, 'Intestinal Parasite Screen',   26.00)
, ( 124, 'Tick Removal',                 15.00)
, ( 125, 'Behaviour Modification',       75.00)
, ( 126, 'Vitamin E- Concentrated',      30.00)
, ( 127, 'Sedative-Feline',              25.00)
, ( 128, 'Flea Treatment- Small Animal', 35.00)
, ( 129, 'Flea Treatment- Large Animal', 50.00)
, ( 143, 'Rabies_V-2010 SFall-Canine',   15.00)
, ( 144, 'Rabies_V-2010 SFall-Feline',   15.00)
, ( 145, 'Rabies_V-2010 Spring-Canine',  25.00)
, ( 146, 'Rabies_V-2010 Spring-Feline',  25.00)
, ( 147, 'Rabies_V-2010 Winter-Canine',  15.00)
;

insert into upd_services_changes values

  ( 128, 'Flea Treatment- Small Animal', 45.00)  
, ( 111, '',                             35.25) 
, ( 122, 'Scaly Mite Powder',             null) 
, ( 138, 'Flu Rhino Vaccine enhanced',  125.00) 
, ( 124, null,                           25.95) 
, ( 129, 'Flea Treatment- Large Animal', 65.00)  
, ( 136, 'Hazardous Materials Disposal', 10.50) 
, ( 126, 'Vitamin E- Concentrated',      45.00)
, ( 106, '',                             30.00)
, ( 105, Null,                           35.00);

我的目标是仅对现有ID进行更新，我已经完成了。我也是如果它是null，则不需要更改services表中的描述，如果它为null，则不更改price字段，但是我被卡住了。这就是我到目前为止所做的：

replace into upd_services 
select * from upd_services_changes
where .....

REPLACE INTO upd_services (srv_id, srv_desc, srv_list_price) 
values  
  ( 128, 'Flea Treatment- Small Animal', 45.00)  
, ( 111, '',                             35.25) 
, ( 122, 'Scaly Mite Powder',             null) 
, ( 124, null,                           25.95) 
, ( 129, 'Flea Treatment- Large Animal', 65.00)  
, ( 126, 'Vitamin E- Concentrated',      45.00)
, ( 106, '',                             30.00)
, ( 105, Null,                           35.00);

Answer 1

由于您的目标是 ...仅对现有ID进行更新... 尝试使用UPDATE而不是

UPDATE upd_services t JOIN upd_services_changes c
    ON t.srv_id = c.srv_id
   SET t.srv_desc = COALESCE(c.srv_desc, t.srv_desc),
       t.srv_list_price = COALESCE(c.srv_list_price, t.srv_list_price);

这是 SQLFiddle 演示

或者如果upd_services_changes还包含新记录，并且您希望将它们插入upd_services以及现有记录的更新，那么

REPLACE INTO upd_services (srv_id, srv_desc, srv_list_price)
SELECT c.srv_id, COALESCE(c.srv_desc, t.srv_desc), COALESCE(c.srv_list_price, t.srv_list_price)
  FROM upd_services_changes c LEFT JOIN upd_services t
    ON c.srv_id = t.srv_id;

这是 SQLFiddle 演示

或使用{1更合适的ON DUPLICATE KEY语法

INSERT INTO upd_services (srv_id, srv_desc, srv_list_price)
SELECT c.srv_id, COALESCE(c.srv_desc, t.srv_desc), COALESCE(c.srv_list_price, t.srv_list_price)
  FROM upd_services_changes c LEFT JOIN upd_services t
    ON c.srv_id = t.srv_id
ON DUPLICATE KEY UPDATE srv_desc = VALUES(srv_desc), 
                        srv_list_price = VALUES(srv_list_price);

这是 SQLFiddle 演示

使用单个Replace语句更新表

1 个答案: