Question

大家好，先谢谢。如果X is 50, Y is 0 or 1, Z is some integer

示例：

X is 40, Y is 1, Z is 2000

我想在MySQL中保存价值

0 to 39 = -2000 and 40 to infinite = 2000

稍后X is 60, Y is 1 and Z is 3000

值应该保存在MySQL中

0 to 39 = -5000
40 to 59 = -1000
60 to infinite = 5000

我想在PHP中实现这一点。

Answer 1

我希望这会有所帮助。我试图涵盖给定X值的所有可能性。

/*Your Table structure should be like this 
Here min and max are Bigint as we cannot use infinite so using the maximum value of bigint 

id  min     max                     value
1   0       39                      -2000
2   40      9223372036854775807     2000
*/
$x= from user;
$y=from User;
$z=from User;
$negZ=$z*-1; // just to make negative value of $z
$newid=0; // we will use these for update2 as well. 
//first we will check wether the give x range is exist  or not IF exist then we dont need to add any more new row

$sql=$conn->prepare("SELECT id FROM table FROM table WHERE min=:x");
$sql->execute(array(':x'=>$x));
while($col=$sql->fetch())
{
    $existRangeid=$col['id'];
}
if($existRangeid==0)//If these range doesn't exist then we need to add new row
{
    //Now we will select the column which we gonna split. 
    $flag=false;
    $sql=$conn->prepare("SELECT id, min,max,value FROM table WHERE min<:x AND max>:x");
    $sql->execute(array(':x'=>$x));
    while($row=$sql->fetch())
    {
        $max=$row['max'];
        $min=$row['min'];
        $id=$row['id'];
        $value=$row['value'];
        $newVal=$value + $negZ;
        $newMax=$x-1;
        $flag=true;
    }
    if($flag==FALSE) // IF the given $x is same as max value of any row in DB then we need to follow //following logic. 
    // as an example if $x is 39 then it will run following code
    {
        $sql=$conn->prepare("SELECT id, min,max,value FROM table WHERE max=:x"); // we will find the row with max value is same as given x
        $sql->execute(array(':x'=>$x));
        while($row=$sql->fetch())
        {
            $max=$row['max'];
            $min=$row['min'];
            $id=$row['id'];
            $value=$row['value'];
            $newVal=$value + $negZ;
            $newMax=$x-1;

        }
        //Now we will insert new range, it will have min and max's value same.
        //e.g. Min = 39; Max= 39; 
        $insertsql=$conn->prepare('INSERT into table(min,max,value) VALUES(:min,:max,:value)');
        $insertsql->execute(array(
            ':min'=>$x,
            ':max'=>$x,
            ':value'=>$newVal
            ));
        $newid=$conn->lastInsertId(); // we will use this in Update 2
        // Now we will update the range's max value. e.g. if x=39. first row with id 1 will now have 38 as a max value
        $update1=$conn->prepare("UPDATE table SET max=:newmax WHERE id=:id");
        $update1->execute(array(':newmax'=>$newMax, ':id'=>$id));
    }
    else
    {
        //Insert the new row with given range
        $insertsql=$conn->prepare('INSERT into table(min,max,value) VALUES(:min,:max,:value)');
        $insertsql->execute(array(
            ':min'=>$min,
            ':max'=>$newMax,
            ':value'=>$newVal
            ));
        $newid=$conn->lastInsertId(); // we will use this in Update 2

        //here we will update the minmun value of the row which we split.
        $update1=$conn->prepare("UPDATE table SET min=:min WHERE id=:id");
        $update1->execute(array(':min'=>$x, ':id'=>$id));
    }
}
//if the give range exist into the system then condition id!=:newid will be id!=0 so it will update all the rows which satisfy max<:x condition.
// IF the range doesnt exist then we wont update newly inserted row because it already contain correct value check insertsql 
$update2=$conn->prepare("UPDATE table SET value=(value+:negZ) WHERE max<:x AND id!=:newid");
$update2->execute(array(
    ':negZ'=>$negZ,
    ':x'=>$x,
    ':newid'=>$newid
    ));

// This will update all the ranges above given x by adding give z to the existing value
$update3=$conn->prepare("UPDATE table SET value=(value+:z) WHERE min>=:x");
$update3->execute(array(
    ':z'=>$z,
    ':x'=>$x
));

如果您对此逻辑有任何疑问，请与我们联系。

将数值保存到数值范围，并根据新值更新以前保存的范围

1 个答案: