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SUM values inside while loop (5 answers)

Closed 12 mins ago.

The following script shows the product and the quantity. Par example: t-shirt(1), boxershort(2).

Now I would like to show the sum of all items, in this case "3".

我怎么能够找到一个阵列——这sum?

$allItems = "";
$items = array();
$select_stmt = $db->prepare("SELECT CONCAT(product,   ( , quantity, ) ) AS ItemQty, price 
FROM database");
$select_stmt->execute();
while ($row = $select_stmt->fetch(PDO::FETCH_ASSOC))
{
    $items[] = $row["ItemQty"];
}
$allItems = implode(", ", $items);

Answer 1

you should do this:

<?php
$allItems = "";
$items = array();
$select_stmt = $db->prepare("SELECT CONCAT(product,   ( , quantity, ) ) AS ItemQty, price 
FROM database");
$select_stmt->execute();
while ($row = $select_stmt->fetch(PDO::FETCH_ASSOC))
{
    preg_match( /[0-9]+/ , $row["ItemQty"], $match);
    array_push($items, (int)$match);
}
echo array_sum($items);
$allItems = implode(", ", $items);
?>

Answer 2

Avoid selecting data in a way that fits the purpose of displaying it. What you should do instead is grab only the data:

$select_stmt = $db->prepare("SELECT product, quantity, price FROM database");

Then you can easily do calculations with raw numbers:

while ($row = $select_stmt->fetch(PDO::FETCH_ASSOC))
{
    $items[] = $row["quantity"];
}
$sum = array_sum($items);

and have the display value created in PHP:

$displayItem = sprintf( %s (%d) , $row[ product ], $row[ quantity ]);

如果你坚持,你可以直接从问询中看到这种显示,然后又把它作为另一个领域:

$select_stmt = $db->prepare("SELECT product, quantity, price, CONCAT(product,   ( , quantity, ) ) AS ItemQty FROM database");

This way you still have quantity as a number to do calculations with and ItemQty to use for displaying purposes.

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