SQL查詢不返回預期值

Question

我正在爲我的客戶端和頁面之一（用PHP和HTML編寫）的數據庫工作，我需要在其客戶端上顯示所有信息，但是當我運行下面的查詢（假設選擇'family'表中的所有行），它返回總共0行。當它應該返回所有行的表中的

SELECT 
families.id AS fam_id, 
families.last_name AS fam_surname, 
families.address_1 AS fam_address_1, 
families.address_2 AS fam_address_2, 
families.city_id AS fam_city, 
families.phone AS fam_phone, 
families.mobile AS fam_mobile, 
families.email AS fam_email, 
families.f_d_worker_1 AS fam_fdw_1, 
families.f_d_worker_2 AS fam_fdw_2, 
families.status_id AS fam_status_id, 
families.trans_date AS fam_trans_date, 
families.entry_date AS fam_entry_date, 
families.exit_date AS fam_exit_date, 
families.eligible_date AS fam_eligible_date, 
families.active_date AS fam_active_date, 
families.lga_loc_id AS fam_lga_id, 
families.facs_loc_id AS fam_facs_id, 
families.ind_status_id AS fam_indig_id, 
families.referral_id AS fam_ref_id, 
families.active_status AS fam_act_status, 
families.comm_org_id AS fam_com_org, 
city.id AS city_id, 
city.name AS city_name, 
city.state_id AS city_state, 
city.post_code AS post_code, 
states.id AS state_id, 
states.long_name AS state_name, 
states.abbrev AS state_abbrev, 
client_status.id AS client_stat_id, 
client_status.name AS client_stat_name, 
community_org.id AS com_org_id, 
community_org.name AS com_org_name, 
facs_location.id AS facs_id, 
facs_location.name AS facs_name, 
lga_location.id AS lga_id, 
lga_location.name as lga_name, 
indig_status.id AS indig_id, 
indig_status.name AS indig_name, 
referrals.id AS ref_id, 
referrals.name AS ref_name, 
f_d_workers.id AS fdw_id, 
f_d_workers.first_name AS fdw_first_name, 
f_d_workers.last_name AS fdw_last_name, 
client_status.id AS client_id, 
client_status.name AS client_name 
FROM 
`families`, 
`city`, 
`client_status`, 
`community_org`, 
`facs_location`, 
`f_d_workers`, 
`indig_status`, 
`lga_location`, 
`referrals`, 
`states` 
WHERE 
families.city_id = city.id AND 
families.f_d_worker_1 = f_d_workers.id AND 
families.f_d_worker_2 = f_d_workers.id AND 
families.status_id = client_status.id AND 
families.lga_loc_id = lga_location.id AND 
families.facs_loc_id = facs_location.id AND 
families.ind_status_id = indig_status.id AND 
families.referral_id = referrals.id AND 
families.comm_org_id = community_org.id 

Answer 1

沒有看到你的架構或數據，我會猜測，一個或多個查詢中的聯接不工作的方式，你認爲這是去上班。例如，查詢意味着每個家庭記錄對於city_id，f_d_worker_1，f_d_worker_2，status_id，lga_loc_id，facs_loc_id，ind_status_id，referral_id和comm_org_id都將具有非空值。如果家庭記錄具有這些字段中的每個字段的值，則看起來在一個或多個對應表（城市，f_d_workers，client_status，lga_locations，facs_locations，indig_status，引用，community_org）中沒有匹配的id字段。

當我認爲應該工作的複雜查詢無法正常工作時，我通常會使用的第一步是將內部聯接轉換爲外部聯接，然後只查找缺失的記錄。

另一方面，如果您的一個或多個聯接元素是可選的（即可能沒有f_d_worker_2 id），那麼您應該在查詢本身中使用外部聯接。我希望這有幫助。

修訂

SELECT 
families.id AS fam_id, 
families.last_name AS fam_surname, 
families.address_1 AS fam_address_1, 
families.address_2 AS fam_address_2, 
families.city_id AS fam_city, 
families.phone AS fam_phone, 
families.mobile AS fam_mobile, 
families.email AS fam_email, 
families.f_d_worker_1 AS fam_fdw_1, 
families.f_d_worker_2 AS fam_fdw_2, 
families.status_id AS fam_status_id, 
families.trans_date AS fam_trans_date, 
families.entry_date AS fam_entry_date, 
families.exit_date AS fam_exit_date, 
families.eligible_date AS fam_eligible_date, 
families.active_date AS fam_active_date, 
families.lga_loc_id AS fam_lga_id, 
families.facs_loc_id AS fam_facs_id, 
families.ind_status_id AS fam_indig_id, 
families.referral_id AS fam_ref_id, 
families.active_status AS fam_act_status, 
families.comm_org_id AS fam_com_org, 
city.id AS city_id, 
city.name AS city_name, 
city.state_id AS city_state, 
city.post_code AS post_code, 
states.id AS state_id, 
states.long_name AS state_name, 
states.abbrev AS state_abbrev, 
client_status.id AS client_stat_id, 
client_status.name AS client_stat_name, 
community_org.id AS com_org_id, 
community_org.name AS com_org_name, 
facs_location.id AS facs_id, 
facs_location.name AS facs_name, 
lga_location.id AS lga_id, 
lga_location.name as lga_name, 
indig_status.id AS indig_id, 
indig_status.name AS indig_name, 
referrals.id AS ref_id, 
referrals.name AS ref_name, 
workers1.id AS fdw1_id, 
workers1.first_name AS fdw1_first_name, 
workers1.last_name AS fdw1_last_name, 
workers2.id AS fdw2_id, 
workers2.first_name AS fdw2_first_name, 
workers2.last_name AS fdw2_last_name, 
client_status.id AS client_id, 
client_status.name AS client_name 
FROM 
`families` 
LEFT OUTER JOIN `city` ON families.city_id = city.id 
LEFT OUTER JOIN `client_status` ON families.status_id = client_status.id 
LEFT OUTER JOIN `community_org` ON families.comm_org_id = community_org.id 
LEFT OUTER JOIN `facs_location` ON families.facs_loc_id = facs_location.id 
LEFT OUTER JOIN `f_d_workers` AS workers1 ON families.f_d_worker_1 = workers1.id 
LEFT OUTER JOIN `f_d_workers` AS workers2 ON families.f_d_worker_2 = workers2.id 
LEFT OUTER JOIN `indig_status` ON families.ind_status_id = indig_status.id 
LEFT OUTER JOIN `lga_location` ON families.lga_loc_id = lga_location.id 
LEFT OUTER JOIN `referrals` ON families.referral_id = referrals.id 
LEFT OUTER JOIN `states` ON city.state_id = states.id 

我想這是你想要知道的。