How to do this

[I-AM-OBODO]

Hi, I have a point that is assigned to an individual based on their duration in the program. I used datediff to calculate the date difference and the result I assigned to a variable.

 

Now I want to do this: if day = 0-5, point =6

what I did was

if($diff == 0 OR 1 OR 2 OR 3 OR 4 OR 5){
$point =6;
}elseif($diff == 6 OR 7 OR 8 OR 9 OR 10){
$point =4;
} else
$point =1;
}
echo $point;

[QuickOldCar]

You can tackle this the way you are

if ($diff == 0 || $diff == 1 || $diff == 2 || $diff == 3 || $diff == 4 || $diff == 5) {
    $point = 6;
} elseif ($diff == 6 || $diff == 7 || $diff == 8 || $diff == 9 || $diff == 10) {
    $point = 4;
} else {
    $point = 1;
}
echo $point;

Make a switch

switch ($diff) {
    case 0:
        $point = 6;
        break;
    case 1:
        $point = 6;
        break;
    case 2:
        $point = 6;
        break;    
    case 3:
        $point = 6;
        break;    
    case 4:
        $point = 6;
        break;    
    case 5:
        $point = 6;
        break;    
    case 6:
        $point = 4;
        break;    
    case 7:
        $point = 4;
        break;    
    case 8:
        $point = 4;
        break;    
    case 9:
        $point = 4;
        break;    
    case 10:
        $point = 4;
        break;
    default:
        $point = 1;
}
echo $point;

Do a double array checking with in_array()

$six_array = range(0, 5);
$four_array = range(6, 10);

if (in_array($diff, $six_array)) {
   
    $point = 6;
   
} elseif (in_array($diff, $four_array)) {
   
    $point = 4;
   
} else {
   
    $point = 1;
   
}
echo $point;

[Barand]

You can use "fall through" in a switch statement

switch ($diff) {
    case 0:
    case 1:
    case 2:
    case 3:
    case 4:
    case 5:
        $point = 6;
        break;    
    case 6:
    case 7:
    case 8:
    case 9:
    case 10:
        $point = 4;
        break;
    default:
        $point = 1;
}
echo $point;

[QuickOldCar]

Good point Barand.

[I-AM-OBODO]

Thanks @all

[I-AM-OBODO]

but how can I sum all the values from $point?

[Barand]

Define a total variable, say $total_points then accumulate the points values into that.

$total_points += $points;

[I-AM-OBODO]

I did so but It's saying undefined variable: total_point and yet brings the result of only one row without adding the column

[Barand]

pseudocode version:

$total_points = 0;                            // DEFINE the variable
begin loop
     $points = whatever;                      // get points value
     $total_points += $points;                // accumulate points total
endloop
echo $total_points;

[I-AM-OBODO]

still wouldn't work. could it be cos the result value is not in the database? cos diff is d result of a datediff value.

 

I.e

$stmt=$pdo->query("select datediff(due, paid) as diff where name =:name");
$stmt->execute()
 while ($row = $stmt->fetch
(PDO::FETCH_ASSOC)) {
$diff=$row['diff'] ;
} 
 switch ($diff) {
    case 0:
    case 1:
    case 2:
    case 3:
    case 4:
    case 5:
        $point = 6;
        break;
    case 6:
    case 7:
    case 8:
    case 9:
    case 10:
        $point = 4;
        break;
    default:
        $point = 1;
}
echo $point;

 $total_points = 0;
                // DEFINE the variable begin loop
     $points = $point;
                // get points value
     $total_points += $points;
                // accumulate points total endloop
echo $total_points; 

in the above code let's say I have 6 as $diff on the first row and on the second row I have $diff as 4, the $total_points ought be 10 but its giving me 6

[I-AM-OBODO]

still wouldn't work. could it be cos the result value is not in the database? cos diff is d result of a datediff value.

 

I.e

$stmt=$pdo->query("select datediff(due, paid) as diff where name =:name");
$stmt->execute()
 while ($row = $stmt->fetch
(PDO::FETCH_ASSOC)) {
$diff=$row['diff'] ;
} 
 switch ($diff) {
    case 0:
    case 1:
    case 2:
    case 3:
    case 4:
    case 5:
        $point = 6;
        break;
    case 6:
    case 7:
    case 8:
    case 9:
    case 10:
        $point = 4;
        break;
    default:
        $point = 1;
}
echo $point;

 $total_points = 0;
                // DEFINE the variable begin loop
     $points = $point;
                // get points value
     $total_points += $points;
                // accumulate points total endloop
echo $total_points; 

in the above code let's say I have 6 as $diff on the first row and on the second row I have $diff as 4, the $total_points ought be 10 but its giving me 6

[QuickOldCar]

Because you need to have everything in your while loop or make them an array and do later on.

$stmt=$pdo->query("select datediff(due, paid) as diff where name =:name");
$stmt->execute()

$total_points = 0;

while ($row = $stmt->fetch
(PDO::FETCH_ASSOC)) {
$diff=$row['diff'] ;

switch ($diff) {
    case 0:
    case 1:
    case 2:
    case 3:
    case 4:
    case 5:
        $point = 6;
        break;
    case 6:
    case 7:
    case 8:
    case 9:
    case 10:
        $point = 4;
        break;
    default:
        $point = 1;
}


$total_points += $point;

}
echo $total_points;

[Barand]

I suggest you read the pseudocode again (reply #9)

 

Setting the total to zero is done before the loop begins.

 

Inside the loop

• calculate the value for the points
• accumulate into the total

After the loop

• echo the accumulated total

[I-AM-OBODO]

It's all working now.

 

Thanks @QuickOldCar and @Barand