PHP Mysql選擇查詢以訂購對

Question

我正在為一家舞蹈學校制作客戶數據庫應用程序。

我需要顯示所有參加相同跳舞級別的客戶的概述。 但是我希望按夫婦而不是按客戶ID進行概覽。 為此，我要加入三個表（請看下面的查詢）：每個客戶在CRM_CONTACTS表格中都有一個唯一的ID，並且在該行中還引用了他或她的伙伴（PARTNER_ID）。

表CRM_CONTACTS

ID   CONTACTS_LNAME
1    VON KLOMPENBURG
2    Mc Donalds
3    MC Adams
4    Mr X

然后我有CRM_PRODUCTS編號PRODUCTS_NAME
1個初學者2個中級3個高級

然后在表格中我為聯系人分配產品/級別，並指出他/她的合作伙伴

ID   CONTACTS_ID    PRODUCTS_ID   PARTNER_ID
1    1              1             4
2    2              1             3
3    3              1             2
4    4              1             1

現在，我希望收到基於parter_id的一對清單訂單，因此對於初學者來說，我將獲得像這樣的清單1 VON KLOMPENBURG 2 Mr X 3 Mc Donalds 4 Mc Adams

這是我的選擇聲明

 $result = mysqli_query($coni,"SELECT CRM_PRODUCTS_PURCHASE.ID, CONTACTS_ID, 
CRM_PRODUCTS.PRODUCTS_NAME, 
CRM_PRODUCTS.PRODUCTS_PRICE,CRM_PRODUCTS_PURCHASE.PARTNER_ID, 
CRM_PRODUCTS_PURCHASE.PARTNER_NAME, 
PRODUCTS_PURCHASE_REMARKS,PRODUCTS_PURCHASE_DISCOUNT, 
PRODUCTS_PURCHASE_PAIDBYBANK,PRODUCTS_PURCHASE_PAIDBYCASH, 
CRM_CONTACTS.CONTACTS_LNAME, CRM_CONTACTS.CONTACTS_FNAME 
FROM CRM_PRODUCTS_PURCHASE
LEFT JOIN CRM_CONTACTS ON CRM_PRODUCTS_PURCHASE.CONTACTS_ID = CRM_CONTACTS.ID 
LEFT JOIN CRM_PRODUCTS ON CRM_PRODUCTS_PURCHASE.PRODUCTS_ID = CRM_PRODUCTS.ID
WHERE CRM_PRODUCTS_PURCHASE.PRODUCTS_ID = '". $PRODUCTS_ID . "'");

Answer 1

您可以執行以下操作：

$result = mysqli_query($coni,"SELECT CRM_PRODUCTS_PURCHASE.ID, CONTACTS_ID, 
CRM_PRODUCTS.PRODUCTS_NAME, 
CRM_PRODUCTS.PRODUCTS_PRICE,CRM_PRODUCTS_PURCHASE.PARTNER_ID, 
CRM_PRODUCTS_PURCHASE.PARTNER_NAME, 
PRODUCTS_PURCHASE_REMARKS,PRODUCTS_PURCHASE_DISCOUNT, 
PRODUCTS_PURCHASE_PAIDBYBANK,PRODUCTS_PURCHASE_PAIDBYCASH, 

CRM_CONTACTS.CONTACTS_LNAME, CRM_CONTACTS.CONTACTS_FNAME 
FROM CRM_PRODUCTS_PURCHASE
LEFT JOIN CRM_CONTACTS ON CRM_PRODUCTS_PURCHASE.CONTACTS_ID = CRM_CONTACTS.ID 
LEFT JOIN CRM_PRODUCTS ON CRM_PRODUCTS_PURCHASE.PRODUCTS_ID = CRM_PRODUCTS.ID
WHERE CRM_PRODUCTS_PURCHASE.PRODUCTS_ID = '". $PRODUCTS_ID . "'
ORDER BY IF (ID < PARTNER_ID, ID, PARTNER_ID)");

和您的客戶將成對訂購。