How to sum mysql values in multple while PHP

Question

i like to sum all row values from the $orderfeed_query. But when i echo the $sum, i get just the sum from the last loop.

how can can i add the sum of all following loop on the $sum variable? I dont know how many loops a order have.

$checkorder = mysql_query("SELECT * FROM orders WHERE  `email` = '$email' ") or die(mysql_error());
while ($row = mysql_fetch_assoc($checkorder)) {
    $orderid = $row["orderid"];
    $check_order = $row["check_order"];

    if($check_order[0] == 1){

        $orderfeed_query = mysql_query("SELECT * FROM orderfeed WHERE `orderid` = '$orderid' AND `product` = '1'") or die(mysql_error());
        while ($row = mysql_fetch_assoc($orderfeed_query)) {
            $signaturewiz = $row["signaturewiz"];
            $flurstueckwiz = $row["flurstueckwiz"];
            $uploadwiz = $row["uploadwiz"];
            $exsignaturewiz = $row["exsignaturewiz"];
            $ibanwiz = $row["ibanwiz"];

            $sum = $signaturewiz+$flurstueckwiz+$uploadwiz+$exsignaturewiz+$ibanwiz;

            echo $sum;

        }
    }   

}
}

You should define $sum outside of the loop, you're just overriding it each time. — Mateo Barahona
– Mateo Barahona, Commented Aug 1, 2016 at 13:52

jeanj · Accepted Answer · 2016-08-01 13:56:37Z

1

This code

$sum = $signaturewiz + $flurstueckwiz + $uploadwiz + $exsignaturewiz + $ibanwiz;

overwrites sum each time. You must add to sum not to overwrite

$sum += $signaturewiz + $flurstueckwiz + $uploadwiz + $exsignaturewiz + $ibanwiz;

And declare $sum = 0; before main loop

edited Aug 1, 2016 at 13:56

jeanj

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answered Aug 1, 2016 at 13:53

nospor

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2 Comments

nospor Over a year ago

echo $sum; must be outside the loop not in the loop

nospor Over a year ago

Right now you have echo 1 and echo 2 which gives you 1 and 2 which is 12 ;)

Ben Hillier · Accepted Answer · 2016-08-01 14:04:05Z

1

Several people have pointed out that $sum should be outside the loop, and that is indeed correct. However, MySQL can do this all for you:

    $orderfeed_query = mysql_query("SELECT SUM(signaturewiz + flurstueckwiz + uploadwiz + exsignaturewiz + ibanwiz) FROM orderfeed WHERE orderid = '$orderid' AND product = '1'") or die(mysql_error());
    if ($row = mysql_fetch_row($orderfeed_query)) {
        echo $row[0];
    }

answered Aug 1, 2016 at 14:04

Ben Hillier

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4 Comments

Ben Hillier Over a year ago

@Gimo You've posted the same comment to all three answers. That kinda makes me think the actual sum of all the fields in the selection really is 12. What happens when you var_dump($row);print "<hr/>"; ' in place of the echo()` in your original example?

Gimo Over a year ago

array(16) { ["id"]=> string(1) "1" ["orderid"]=> string(10) "1468863981" ["product"]=> string(1) "1" ["check_order"]=> string(5) "11100" ["productamount"]=> string(2) "39" ["strasse"]=> string(9) "Ackerstr." ["hausnummer"]=> string(1) "1" ["plz"]=> string(5) "32051" ["ort"]=> string(7) "Herford" ["koordinaten"]=> string(0) "" ["regionalschluessel"]=> string(0) "" ["signaturewiz"]=> string(1) "1" ["flurstueckwiz"]=> string(1) "0" ["uploadwiz"]=> string(1) "0" ["exsignaturewiz"]=> string(1) "0" ["ibanwiz"]=> string(0) "" }

Gimo Over a year ago

and array(16) { ["id"]=> string(1) "7" ["orderid"]=> string(10) "1470041156" ["product"]=> string(1) "1" ["check_order"]=> string(5) "10000" ["productamount"]=> string(2) "39" ["strasse"]=> string(13) "Aachener Str." ["hausnummer"]=> string(1) "1" ["plz"]=> string(5) "10178" ["ort"]=> string(6) "Berlin" ["koordinaten"]=> string(0) "" ["regionalschluessel"]=> string(0) "" ["signaturewiz"]=> string(1) "1" ["flurstueckwiz"]=> string(1) "0" ["uploadwiz"]=> string(1) "1" ["exsignaturewiz"]=> string(1) "0" ["ibanwiz"]=> string(0) "" }

Gimo Over a year ago

when a wiz have a "1" it is activate a message and i like to sum all new messages, it works by 1 orderid but not by multiple orderids

RiggsFolly · Accepted Answer · 2016-08-01 16:53:55Z

1

Move the echo of $sum outside the loop

Use += rather than + to accumulate a total over multiple iterations

Initialize your variable before using += as if $sum has an undefined value it can mess up the count when using +=

    $sum = 0;   // init variable
    while ($row = mysql_fetch_assoc($orderfeed_query)) {
        $signaturewiz = $row["signaturewiz"];
        $flurstueckwiz = $row["flurstueckwiz"];
        $uploadwiz = $row["uploadwiz"];
        $exsignaturewiz = $row["exsignaturewiz"];
        $ibanwiz = $row["ibanwiz"];

        $sum += $signaturewiz + $flurstueckwiz + $uploadwiz + 
                $exsignaturewiz + $ibanwiz;
    }
    echo $sum;

ADDITIONAL INFO:

You see 12 and not 3 so the data from your table I assume is text and not numeric so do this to convert text numbers to integers

    $sum = 0;   // init variable
    while ($row = mysql_fetch_assoc($orderfeed_query)) {
        $signaturewiz   = (int)$row["signaturewiz"];
        $flurstueckwiz  = (int)$row["flurstueckwiz"];
        $uploadwiz      = (int)$row["uploadwiz"];
        $exsignaturewiz = (int)$row["exsignaturewiz"];
        $ibanwiz        = (int)$row["ibanwiz"];

        $sum += $signaturewiz + $flurstueckwiz + $uploadwiz + 
                $exsignaturewiz + $ibanwiz;
    }
    echo $sum;

edited Aug 1, 2016 at 16:53

answered Aug 1, 2016 at 13:55

RiggsFolly

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3 Comments

RiggsFolly Over a year ago

You do want to get the total of ALL the columns in ALL the rows in your result set, Right? Or am I confused

Gimo Over a year ago

i like to sum all values of the rows: $signaturewiz = $row["signaturewiz"]; $flurstueckwiz = $row["flurstueckwiz"]; $uploadwiz = $row["uploadwiz"]; $exsignaturewiz = $row["exsignaturewiz"]; $ibanwiz = $row["ibanwiz"];, the result of the first order is 1 the second is 2. but i see just 12 not 3.

RiggsFolly Over a year ago

See additional Info RE: Non numeric data from the database

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