PHP MySQL的選擇SUM減去SUM

Question

我想運行一個查詢，如下圖所示：

SELECT (SUM(amount) - SUM(refundAmount)) as amount FROM orders WHERE $invoiceFilter AND $websiteFilter 

它得到正確的金額列的總和，但它不是減去退款金額。

像這樣的查詢的正確語法是什麼？

Answer 1

試試這個：

SELECT SUM(amount - refundAmount) AS `amount` 
FROM orders 
WHERE 
    invoice = $invoiceFilter 
    AND website = $websiteFilter; 

Answer 2

您可能會丟失類似於參考列的內容來過濾需要處理的數據。喜歡的東西：

SELECT (SUM(amount) - SUM(refundAmount)) AS amount 
FROM orders 
WHERE 
    <a_column> = $invoiceFilter 
    AND <another_column> = $websiteFilter; 

Answer 3

您可以在FROM子句中使用子查詢。

SQL：

SELECT amount - refund 
FROM (SELECT SUM(amount) as amount 
     FROM orders 
     WHERE invoice = 'invoiceFilter' AND website = 'websiteFilter') as a, 
    (SELECT SUM(refundAmount) as refund 
     FROM orders 
     WHERE invoice = 'invoiceFilter' AND website = 'websiteFilter') as b; 

PHP：

mysql_query("SELECT amount - refund 
      FROM (SELECT SUM(amount) as amount 
        FROM orders 
        WHERE invoice = '" . mysql_real_escape_string($invoiceFilter). "' 
        AND website = '" . mysql_real_escape_string($websiteFilter). "') as a, 
        (SELECT SUM(refundAmount) as refund 
        FROM orders 
        WHERE invoice = '" . mysql_real_escape_string($invoiceFilter). "' 
        AND website = '" . mysql_real_escape_string($websiteFilter). "') as b; ");