直到數據耗盡時從不同行中刪除值？

Question

我有一個表像這樣

+----+-------+------+ 
| id | apple | pear | 
+----+-------+------+ 
| 1 | 3 | 10 | 
+----+-------+------+ 
| 2 | 5 | 8 | 
+----+-------+------+ 
| 3 | 4 | 7 | 
+----+-------+------+ 

我不得不刪除9蘋果和梨11。 結果這樣

+----+-------+------+ 
| id | apple | pear | 
+----+-------+------+ 
| 1 | 0 | 0 | 
+----+-------+------+ 
| 2 | 0 | 7 | 
+----+-------+------+ 
| 3 | 1 | 7 | 
+----+-------+------+ 

有沒有辦法做到這一點，而不必做了很多疑問？

我做了這樣的事情。 我想知道是否有更好的方法。

$eat["apple"]=9; 
    $eat["pear"]=11; 
    $id=0; //I will increment this id during the cycle 

    while ($eat["apple"] > 0 && $eat["pear"] > 0) { 
     $get_fruit=mysql_query("SELECT id,apple,pear FROM fruits WHERE 
      id>'".$id."' 
    ORDER BY id LIMIT 1") or die (mysql_error()); 

     if((mysql_num_rows($get_fruit))==0){ 
      die; 
     } else { 

      $fruit = mysql_fetch_assoc($get_fruit); 
      $id = $fruit["id"]; 

      foreach ($eat as $key => $value) { 

      if ($fruit[$key]>$eat[$key]) { 
       $fruit[$key]-=$eat[$key]; 
       $eat[$key]=0; 
      } else { 
       $eat[$key]-=$fruit[$key]; 
       $fruit[$key]=0; 
      } 
     } 

    $update=mysql_query("UPDATE fruits SET 
      apple='".$fruit["apple"]."', 
      pear='".$fruit["pear"]."' 
      WHERE id='".$id."'LIMIT 1") or die (mysql_error()); 
} 

}

Answer 1

CREATE TABLE t 
    (`id` int, `apple` int, `pear` int) 
; 

INSERT INTO t 
    (`id`, `apple`, `pear`) 
VALUES 
    (1, 3, 10), 
    (2, 5, 8), 
    (3, 4, 7) 
; 

UPDATE t 
INNER JOIN (
    SELECT 
    id, apple, pear, 
    GREATEST(0, apple - GREATEST(0, @atr)) AS new_apples, 
    @atr := @atr - apple, 
    GREATEST(0, pear - GREATEST(0, @ptr)) AS new_pears, 
    @ptr := @ptr - pear 
    FROM t 
    , (SELECT @atr := 9, @ptr := 11) variable_initialization_subquery /*@atr = ApplesToRemove...*/ 
    ORDER BY id 
) sq ON t.id = sq.id 
SET t.apple = sq.new_apples, 
t.pear = sq.new_pears; 

SELECT * FROM t; 

| ID | APPLE | PEAR | 
|----|-------|------| 
| 1 |  0 | 0 | 
| 2 |  0 | 7 | 
| 3 |  3 | 7 | 

• DEMO

Answer 2

這裏的蘋果......

SET @apple = 9; 
UPDATE fruits x 
    JOIN 
    (SELECT id 
      , apple 
      , GREATEST(apple-@apple,0)n,@apple:=@apple-apple 
     FROM fruits 
     ORDER 
      BY id 
    ) y 
    ON y.id = x.id 
    SET x.apple = y.n;