選擇所有客戶和每個客戶的總訂單

Question

對於每個客戶，我想返回：訂單的ID，姓名，地址，城市，上限，國家和總數。表名是：顧客（10條）和訂單（20條）

我試過這個代碼，但它不工作

<?php 
$conn = mysqli_connect('127.0.0.1','...','...','...') or die("Connection failed: " . $conn->connect_error); 

$select = mysqli_query($conn,"SELECT customers.id,customers.name,customers.address,customers.city,customers.cap,customers.country COUNT(customerid) as TotalOrders FROM customers LEFT JOIN orders GROUP BY customer.id"); 

while($row = mysqli_fetch_array($select,MYSQLI_ASSOC)) 
{ 
    $tb = <<<table 
    ...{$row['totalorders']}... 
    table; 
    echo $tb; 
} 
mysqli_free_result($select); 

echo $num_record = mysqli_affected_rows($conn); 

$conn->close(); 
?> 

Answer 1

所以像...

<?php 
include('../path/to/connection/stateme.nts'); 

$query = 
" 
SELECT c.id 
    , c.name 
    , c.address 
    , c.city 
    , c.cap 
    , c.country 
     COUNT(o.customerid) TotalOrders 
    FROM customers c 
    LEFT 
    JOIN orders o 
    ON o.customerid = c.id 
GROUP 
    BY c.id; 
"; 


$result = mysqli_query($db,$query); ** where $db is something like mysqli_connect("myhost", "myusername", "mypassword"); 

while($row = mysqli_fetch_assoc($result)) 
{ 
$tb = <<<table 
...{$row['totalorders']}... 
table; 
echo $tb; 
} 

mysqli_free_result($result); 

echo $num_record = mysqli_affected_rows($db); 

$db->close(); 
?> 

（我沒有PHP的編碼器）

Answer 2

你有一個逗號缺失 -

customers.country COUNT(customerid) 
-----------------^ 

你就只剩下了一個ON條件 -

FROM customers LEFT JOIN orders //ON would be here 

有一件事你應該總是做的是創造和Ru n您對數據庫的查詢您將這些查詢放入您的代碼中。總是，總是包括錯誤檢查。