使用相同ID＃

Question

我想從表稱爲address_contact我用的是代碼的最後ID＃3個表中選擇行的比較快的方式是：

$result = mysql_query("SELECT id FROM address_contact ORDER BY lastUpdate DESC LIMIT 1") 
or die(mysql_error()); 
$row = mysql_fetch_array($result); 
$id = .$row['id']; 

現在我想關閉連接，並打開一個新的，然後獲取所有數據的使用，我們只是得到了ID＃以下3個表：

Table 1: address_address 
Table 2: address_contact 
Table 3: address_email 

所以它會是這個樣子？？？

$result = mysql_query("SELECT address_contact.id,address_contact.lastname,address_contact.firstname,address_contact.primaryAddType,address_address.id,address_address.phone1,address_address.phone2,address_address.line2,address_email.id,address_email.email 
FROM address_address 
LEFT JOIN address_contact ON address_address.id=address_contact.id 
LEFT JOIN address_email ON address_address.id=address_email.id     
WHERE address_contact.id = ".$id) 

但是必須有一個更簡單/更快的方法來做到這一點？

Answer 1

您可以使用子查詢來查找ID一個SQL語句做到這一點。

SELECT 
    address_contact.id, 
    address_contact.lastname, 
    address_contact.firstname, 
    address_contact.primaryAddType, 
    address_address.id, 
    address_address.phone1, 
    address_address.phone2, 
    address_address.line2, 
    address_email.id, 
    address_email.email 
FROM 
    address_address 
    LEFT JOIN address_contact ON address_address.id = address_contact.id 
    LEFT JOIN address_email ON address_address.id = address_email.id     
    WHERE address_contact.id = (
    SELECT id FROM address_contact ORDER BY lastUpdate DESC LIMIT 1 
) 

Answer 2

如果這個ID是你只插入一個全新的紀錄，你應該使用msyql_last_insert_id()功能，保證您獲得最後插入這個特定的腳本/數據庫手柄一樣。你的方法受制於種族條件 - 如果某個OTHER腳本在這個腳本的後面插入了一個插入，你將得到該腳本的ID，而不是你的。

話雖這麼說，你會更好做

SELECT MAX（ID）FROM yourtable

，而不是由/限制版本的訂貨。這樣做效率更高。

的基本代碼序列是：

INSERT INTO yourtable .... 
SELECT @id := last_insert_id(); 
SELECT ... FROM yourtable WHERE id = @id; 

Answer 3

爲什麼不使用單一查詢

SELECT address_contact.id,address_contact.lastname,address_contact.firstname,address_contact.primaryAddType,address_address.id,address_address.phone1,address_address.phone2,address_address.line2,address_email.id,address_email.email 
FROM address_address 
LEFT JOIN address_contact ON address_address.id=address_contact.id 
LEFT JOIN address_email ON address_address.id=address_email.id     
ORDER BY address_contact.lastUpdate DESC 
LIMIT 1