簡化兩個MySQL查詢到一個通過加入和/或工會

我有兩個疑問：簡化兩個MySQL查詢到一個通過加入和/或工會

(SELECT b904_Vasarlas.PrintedFigure as Image, 
    b904_Vasarlas.ProductColor as Color, 
    SUM(b904_Vasarlas.QTY) as QTY 
FROM `b904_Vasarlas` 
GROUP BY 1, 2) 
ORDER by 1, 2

具有良好的輸出：

Image;Color;QTY 
"villamos","blue","5" 
"villamos","pink","5" 
"virag","blue","5" 
"virag","pink","5" 
"virag2","blue","5" 
"virag2","pink","5" 
"vitorlas","blue","5" 
"vitorlas","pink","5" 
"vonat","blue","5" 
"vonat","pink","5" 
"zaszlo","blue","5" 
"zaszlo","pink","5" 
"zsiraf","blue","15" 
"zsiraf","pink","15"

和

(SELECT b904_Eladas.PrintedFigure as Image, 
    b904_Eladas.ProductColor as Color, 
    SUM(b904_Eladas.QTY) as QTY 
FROM `b904_Eladas` 
GROUP BY 1, 2) 
ORDER by 1, 2

具有良好的輸出：

Image;Color;QTY 
"golya","blue","20" 
"golya","pink","10" 
"hajokormany","blue","20" 
"hajokormany","pink","10" 
"macska","blue","10" 
"macska","pink","10" 
"malac","blue","10" 
"malac","pink","10" 
"villamos","blue","3" 
"villamos","pink","3" 
"virag","blue","3" 
"virag","pink","3" 
"virag2","blue","4" 
"vitorlas","blue","5" 
"vitorlas","pink","5" 
"vonat","blue","5" 
"vonat","pink","5" 
"zaszlo","blue","15" 
"zaszlo","pink","5" 
"zsiraf","blue","16" 
"zsiraf","pink","11"

如何才能獲得兩個表的數量列的差異，只有一個mysql查詢按圖像和顏色列分組？

最後我用這個較長的代碼，但我想我能解決，只有一個查詢字符串更加簡單的問題。

$Eladasok = array(); 
$ElLoop = 0; 
$GyartasQuery = '(SELECT ' . $DBConnection[1] . '_Vasarlas.PrintedFigure as RaktarKep, ' . $DBConnection[1] . '_Vasarlas.ProductColor as RaktarSzin, SUM(' . $DBConnection[1] . '_Vasarlas.QTY) as RaktarQTY FROM `' . $DBConnection[1] . '_Vasarlas` GROUP BY 1, 2) ORDER by 1, 2'; 
$EladasQuery = '(SELECT ' . $DBConnection[1] . '_Eladas.PrintedFigure as EladasKep, ' . $DBConnection[1] . '_Eladas.ProductColor as EladasSzin, SUM(' . $DBConnection[1] . '_Eladas.QTY) as EladasQTY FROM `' . $DBConnection[1] . '_Eladas` GROUP BY 1, 2) ORDER by 1, 2'; 
$GyartasResult = mysql_query($GyartasQuery); 
$EladasResult = mysql_query($EladasQuery); 

while(false !== $EladasContent = mysql_fetch_assoc($EladasResult)) {       
    $new_array[] = $EladasContent; 
    $Eladasok[$new_array[$ElLoop][EladasKep] . '-' . $new_array[$ElLoop][EladasSzin]] = $EladasContent[EladasQTY]; 
    $ElLoop++; 
} 

while(false !== $GyartasContent = mysql_fetch_assoc($GyartasResult)) { 
    if (($GyartasContent[RaktarQTY] - $Eladasok[$GyartasContent[RaktarKep] . '-' . $GyartasContent[RaktarSzin]]) <= 0) { 
     $Cellcolour = ' bgcolor="#DDDDDD"'; 
     $CellHighlight = ' bgcolor="#FFBBBB"'; 
    } else { 
     $Cellcolour = ''; 
     $CellHighlight = ''; 
    } 

    echo' 
    <tr><td align=center' . $CellHighlight . '>'.++$RoNo.'</td> 
    <td' . $Cellcolour . '>' . $GyartasContent[RaktarKep] . ' - ' . $GyartasContent[RaktarSzin] . '</td> 
    <td' . $Cellcolour . '>' . intval($GyartasContent[RaktarQTY]) . '</td> 
    <td' . $Cellcolour . '>' . intval($Eladasok[$GyartasContent[RaktarKep] . '-' . $GyartasContent[RaktarSzin]]) . '</td> 
    <td' . $Cellcolour . '>' . intval($GyartasContent[RaktarQTY] - $Eladasok[$GyartasContent[RaktarKep] . '-' . $GyartasContent[RaktarSzin]]). 
    '</td></tr>'; 
} 
echo '</tr></table></div>';

來源

2013-09-28 Perjés László

你可以加載你的數據（或至少一個樣本）到sqlfiddle.com，以便代碼可以測試？ – AgRizzo

我通過PHPMyadmin測試了代碼。工作，但總和（列）在分組後是錯誤的。我與我自己的查詢有同樣的問題。 –

以下是用我自己的代碼進行測試的鏈接，但帶有錯誤的彙總列：http://sqlfiddle.com/#!2/7b238/2 –

數據庫1中的含量：

SELECT PrintedFigure，ProductColor，QTY FROM b904_vasarlas

 
(Name, color, QTY) 
"villamos","blue","5" 
"villamos","pink","5" 
"virÃ¡g","blue","5" 
"virÃ¡g","pink","5" 
"virÃ¡g2","blue","5" 
"virÃ¡g2","pink","5" 
"vitorlÃ¡s","blue","5" 
"vitorlÃ¡s","pink","5" 
"vonat","blue","5" 
"vonat","pink","5" 
"zsirÃ¡f","blue","5" 
"zsirÃ¡f","pink","5" 
"zÃ¡szlÃ³","blue","5" 
"zÃ¡szlÃ³","pink","5" 
"zsirÃ¡f","blue","10" 
"zsirÃ¡f","pink","10"

DATABASE2中的含量：

SELECT PrintedFigure，ProductColor，QTY FROM b904_eladas

 
(Name, color, QTY) 
"vitorlÃ¡s","blue","5" 
"vitorlÃ¡s","pink","5" 
"vonat","blue","5" 
"vonat","pink","5" 
"zsirÃ¡f","blue","5" 
"zsirÃ¡f","pink","5" 
"zÃ¡szlÃ³","blue","5" 
"zÃ¡szlÃ³","pink","5" 
"zsirÃ¡f","blue","5" 
"zÃ¡szlÃ³","blue","10" 
"zsirÃ¡f","blue","6" 
"zsirÃ¡f","pink","6" 
"gÃ³lya","blue","10" 
"gÃ³lya","pink","10" 
"hajÃ³kormÃ¡ny","blue","10" 
"hajÃ³kormÃ¡ny","pink","10" 
"macska","blue","10" 
"macska","pink","10" 
"malac","blue","10" 
"malac","pink","10" 
"villamos","blue","3" 
"villamos","pink","3" 
"virÃ¡g","blue","3" 
"virÃ¡g","pink","3" 
"virÃ¡g2","blue","4" 
"gÃ³lya","blue","10" 
"hajÃ³kormÃ¡ny","blue","10"

當打印圖形和產品顏色在展位數據庫中相同時，我需要稱爲數量的列的差異。

你的結果是錯誤的：

 
Image,Color,QtyV,QtyE,Qty,QtyOtherWay 
"gÃ³lya","blue","0","0","-20","-20" 
"gÃ³lya","pink","0","0","-10","-10" 
"hajÃ³kormÃ¡ny","blue","0","0","-20","-20" 
"hajÃ³kormÃ¡ny","pink","0","0","-10","-10" 
"macska","blue","0","0","-10","-10" 
"macska","pink","0","0","-10","-10" 
"malac","blue","0","0","-10","-10" 
"malac","pink","0","0","-10","-10" 
"villamos","blue","5","5","2","2" 
"villamos","pink","5","5","2","2" 
"virÃ¡g","blue","5","5","2","2" 
"virÃ¡g","pink","5","5","2","2" 
"virÃ¡g2","blue","5","5","1","1" 
"virÃ¡g2","pink","5","5","5","5" 
"vitorlÃ¡s","blue","5","5","0","0" 
"vitorlÃ¡s","pink","5","5","0","0" 
"vonat","blue","5","5","0","0" 
"vonat","pink","5","5","0","0" 
"zÃ¡szlÃ³","blue","10","10","-5","-5" 
"zÃ¡szlÃ³","pink","5","5","0","0" 
"zsirÃ¡f","blue","45","45","13","13" 
"zsirÃ¡f","pink","30","30","8","8"

我有，當我使用加入和工會嘗試過完全相同的問題。

結果是好的使用兩個查詢和代碼while循環做磁盤陣列：

 
# Name and color  Difference 
1 villamos - blue  2 
2 villamos - pink  2 
3 virág - blue  2 
4 virág - pink  2 
5 virág2 - blue  1 
6 virág2 - pink  5 
7 vitorlás - blue  0 
8 vitorlás - pink  0 
9 vonat - blue  0 
10 vonat - pink  0 
11 zászló - blue  -10 
12 zászló - pink  0 
13 zsiráf - blue  -1 
14 zsiráf - pink  4

我問一個查詢的解決方案，因爲我想用這個循環取消：

  while(false !== $EladasContent = mysql_fetch_assoc($EladasResult)) {       
       $new_array[] = $EladasContent; 
       $Eladasok[$new_array[$ElLoop][EladasKep] . '-' . $new_array[$ElLoop][EladasSzin]] = $EladasContent[EladasQTY]; 
       $ElLoop++; 
      }

來源

2013-09-29 09:36:42

問題是當另一個數據庫包含具有相同名稱/顏色compinations的行時，彙總QTY列中的錯誤結果。 :( –

我想我終於明白了......對不起。下面是兩個解決方案（我認爲第一個是獲得更好的性能，但不積極）在任何情況下，zsiraf +粉色是15 & 11的4

SELECT MasterList.Name, MasterList.Color, E.EQty, V.VQty 
, COALESCE(V.VQty, 0) - COALESCE(E.EQty,0) AS NetQty 
FROM 
(SELECT DISTINCT Name, Color 
FROM b904_Eladas 
UNION 
SELECT DISTINCT Name, Color 
FROM b904_Vasarlas) AS MasterList 
LEFT JOIN 
(SELECT Name,Color, SUM(QTY) AS EQty 
    FROM b904_Eladas 
    GROUP BY Name,Color 
)AS E 
ON MasterList.Name = E.Name AND MasterList.Color = E.Color 
LEFT JOIN 
(SELECT Name,Color, SUM(QTY) AS VQty 
    FROM b904_Vasarlas 
    GROUP BY Name,Color 
)AS V 
ON MasterList.Name = V.Name AND MasterList.Color = V.Color 
ORDER BY 1,2

SELECT 
E.Name AS EName 
, V.Name AS VName 
, COALESCE(E.Name ,v.Name) AS Name 
, E.Color AS EColor 
, V.Color AS VColor 
, COALESCE(E.Color , V.Color) AS Color 
, COALESCE(E.Total_QTY,0) AS EQTY 
, COALESCE(V.Total_QTY, 0) AS VQTY 
, COALESCE(V.Total_QTY,0) - COALESCE(E.Total_QTY, 0) AS NETQTY 
FROM 
    (SELECT b904_Eladas.Name as `Name`, b904_Eladas.Color AS Color, SUM(b904_Eladas.QTY) AS Total_QTY 
    FROM b904_Eladas 
    GROUP BY NAME, Color) AS E 
LEFT JOIN 
    (SELECT b904_Vasarlas.Name as `Name`, b904_Vasarlas.Color AS Color, SUM(b904_Vasarlas.QTY) AS Total_QTY 
    FROM b904_Vasarlas 
    GROUP BY NAME, Color) AS V 
ON V.Name = E.Name AND V.Color = E.Color 
UNION 
SELECT 
E.Name AS EName 
, V.Name AS VName 
, COALESCE(E.Name ,v.Name) AS Name 
, E.Color AS EColor 
, V.Color AS VColor 
, COALESCE(E.Color , V.Color) AS Color 
, COALESCE(E.Total_QTY,0) AS EQTY 
, COALESCE(V.Total_QTY, 0) AS VQTY 
, COALESCE(V.Total_QTY,0) - COALESCE(E.Total_QTY, 0) AS NETQTY 
FROM 
    (SELECT b904_Eladas.Name as `Name`, b904_Eladas.Color AS Color, SUM(b904_Eladas.QTY) AS Total_QTY 
    FROM b904_Eladas 
    GROUP BY NAME, Color) AS E 
RIGHT JOIN 
    (SELECT b904_Vasarlas.Name as `Name`, b904_Vasarlas.Color AS Color, SUM(b904_Vasarlas.QTY) AS Total_QTY 
    FROM b904_Vasarlas 
    GROUP BY NAME, Color) AS V 
ON V.Name = E.Name AND V.Color = E.Color 
ORDER BY 3,6

來源

2013-10-01 16:18:46 AgRizzo

謝謝，因爲我看到的作品，我必須瞭解如何:) –

你的差異可以使用子查詢從第二個表中獲取總和。

SELECT v.PrintedFigure as Image, 
    v.ProductColor as Color, 
    SUM(v.QTY) - (
     SELECT SUM(e.QTY) 
     FROM `b904_Eladas` e 
     WHERE e.PrintedFigure = v.PrintedFigure 
      AND e.ProductColor = v.ProductColor 
    ) as QTY_diff 
FROM `b904_Vasarlas` v 
GROUP BY v.PrintedFigure, v.ProductColor 
ORDER by v.PrintedFigure, v.ProductColor

來源

2017-05-07 16:46:09

簡化兩個MySQL查詢到一個通過加入和/或工會

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