如何將兩個查詢與不同的WHERE/LIKE條件結合起來？

Question

我的兩個查詢：

$compareTotals1 = mysqli_query($con," 
    SELECT *, (SUM(rent_amount)+SUM(late_fees)-SUM(discount_amount)) AS total, 
       SUM(late_fees) AS latefees, 
       SUM(discount_amount) AS discounts 
    FROM  transaction 
    WHERE  paid_on LIKE '%2014%' 
"); 
$compareTotals2 = mysqli_query($con," 
    SELECT *, (SUM(rent_amount)+SUM(late_fees)-SUM(discount_amount)) AS total, 
       SUM(late_fees) AS latefees, 
       SUM(discount_amount) AS discounts 
    FROM  transaction 
    WHERE  paid_on LIKE '%2015%' 
"); 

我如何輸出結果：

if ($row = mysqli_fetch_array($compareTotals1)) { 
    echo CURRENCY.number_format($row['total'],2); 
    echo CURRENCY.number_format($row['latefees'],2); 
    echo CURRENCY.number_format($row['discounts'],2); 
} else { 
    echo "No Records."; 
} 

if ($row = mysqli_fetch_array($compareTotals2)) { 
    echo CURRENCY.number_format($row['total'],2); 
    echo CURRENCY.number_format($row['latefees'],2); 
    echo CURRENCY.number_format($row['discounts'],2); 
} else { 
    echo "No Records."; 
} 

的paid_on LIKE '% %'通過下拉框和一些JavaScript動態生成的。這是變化的唯一部分。

我怎樣才能凝結成一個查詢這個，所以我只需要使用一個mysqli_fetch_array？

Answer 1

假設你想多行工會很可能在清潔

$compareTotals1 = mysqli_query($con," 
    SELECT *, (SUM(rent_amount)+SUM(late_fees)-SUM(discount_amount)) AS total, 
       SUM(late_fees) AS latefees, 
       SUM(discount_amount) AS discounts, 
       '2014' as Yr 
    FROM  transaction 
    WHERE  paid_on LIKE '%2014%' 
    UNION ALL 
    SELECT *, (SUM(rent_amount)+SUM(late_fees)-SUM(discount_amount)) AS total, 
       SUM(late_fees) AS latefees, 
       SUM(discount_amount) AS discounts, 
       '2015' as Yr 
    FROM  transaction 
    WHERE  paid_on LIKE '%2015%' 
"); 

您可以到這裏的合併成一個OR並確保通過拼出一組或你不會知道這是2015年或2014年，除非*包括這些細節。

$compareTotals1 = mysqli_query($con," 
    SELECT *, (SUM(rent_amount)+SUM(late_fees)-SUM(discount_amount)) AS total, 
       SUM(late_fees) AS latefees, 
       SUM(discount_amount) AS discounts, 
       case when paid_on like '%2014%' then '2014' 
        when paid_on like '%2015%' then '2015' end as yr 
    FROM  transaction 
    WHERE  paid_on LIKE '%2014%' 
     OR  paid_on LIKE '%2015%' 
    --GROUP BY all fields from select relevant to group by... without structure and sample data from table can't figure out. 
    -- This might work though I'd be concerned all the * columns could be returning improper results. 
    GROUP BY case when paid_on like '%2014%' then '2014' when paid_on like '%2015%' then '2015' end 
"); 

也許...... group by case when paid_on like '%2014%' then '2014' when paid_on like '%2015%' then '2015' end但是這是非常具體的。

，我們也許能夠按paid_on，但它似乎它不只是一年...所以你可能會得到每年多行......如此反覆，而不對結構的樣本數據想不出該怎麼辦。

或者，也許你想要一個交叉連接更多的列...不...行

$compareTotals1 = mysqli_query($con," 
    Select * from (
    SELECT *, (SUM(rent_amount)+SUM(late_fees)-SUM(discount_amount)) AS total, 
       SUM(late_fees) AS latefees, 
       SUM(discount_amount) AS discounts 
    FROM  transaction 
    WHERE  paid_on LIKE '%2014%') CROSS JOIN 

    (SELECT *, (SUM(rent_amount)+SUM(late_fees)-SUM(discount_amount)) AS total, 
       SUM(late_fees) AS latefees, 
       SUM(discount_amount) AS discounts 
    FROM  transaction 
    WHERE  paid_on LIKE '%2015%') B 
"); 

Answer 2

爲什麼不在你的WHERE中使用OR？

$compareTotals1 = mysqli_query($con," 
    SELECT *, (SUM(rent_amount)+SUM(late_fees)-SUM(discount_amount)) AS total, 
       SUM(late_fees) AS latefees, 
       SUM(discount_amount) AS discounts 
    FROM  transaction 
    WHERE  paid_on LIKE '%2014%' 
     OR  paid_on LIKE '%2015%' 
    GROUP BY YEAR(paid_on) -- is `paid_on` a date? 
");