2016-09-19 4 views
-4
$sql= " 
SELECT tbl_doctors. * , 
     tbl_specialisation. * , 
     tbl_doctor_address.addr_id, 
     tbl_address . * 
FROM `tbl_doctors` INNER JOIN 
    `tbl_specialisation` 
    ON (tbl_doctors.dr_spec_id = tbl_specialisation.spec_id) INNER JOIN 
    (tbl_doctor_address JOIN 
     tbl_address 
     ON tbl_doctor_address.addr_id = tbl_address.addr_id 
    ) 
    ON tbl_doctors.dr_id = tbl_doctor_address.dr_id 
WHERE tbl_doctors.dr_id = '$id'"; 
+2

クエリを何に分割しますか? –

+2

適切なヘルプが必要な場合は、適切な質問をしてください。 – massko

+0

'where句'の 'dr_id'列にエラーが表示されます –

答えて

0

私はこの部分を改造する必要が推測:

INNER JOIN `tbl_doctor_address` ON tbl_doctors.dr_id = tbl_doctor_address.dr_id 
INNER JOIN `tbl_address` ON tbl_doctor_address.addr_id = tbl_address.addr_id 

またはアドレスがあってはならない場合は、LEFT JOINを使用することができます。このような

(tbl_doctor_address JOIN 
    tbl_address 
    ON tbl_doctor_address.addr_id = tbl_address.addr_id 
) 
ON tbl_doctors.dr_id = tbl_doctor_address.dr_id 

の中へ医師のために設定:

LEFT JOIN `tbl_doctor_address` ON tbl_doctors.dr_id = tbl_doctor_address.dr_id 
LEFT JOIN `tbl_address` ON tbl_doctor_address.addr_id = tbl_address.addr_id 

lpsかどうか?