duanqian3953 2016-07-28 13:25
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PHP Javascript数组

Hi I'm a novice PHP developer to say the least and I'm pretty stuck.

I'm trying to get the below code working I know I need a foreach loop or similar but I'm way out of my depth, I'm really not sure how to get this working, I know to most of you this is basic stuff but I'm lost.

Basically this is a receipt when a tab is settled. It searches the MySql Database for the relevant items and should print them on the receipt.

I know the MySql Query is not linked to the output at the bottom but I don't know how to do it.

<html lang="en">
    <head>
        <title>Receipt</title>
    </head>

    <body>
        <?php

            $invoicenum = $_POST['invoicenum'];
            $name = $_POST['name'];

            $netrev=$invoicenum - 1;

        ?>
        Items:
        <?
            $itemQuery = mysql_query("SELECT * FROM sales WHERE invoicenum = '$netrev' AND tabname = '$name'");
    $result = array();

    while($row = mysql_fetch_array($itemQuery))
    {
        $result[] = $row['itemname'];
    }

    echo json_encode($result);

            $amounts = json_decode($result['amounts']);
            $items = json_decode($result['items']);
            $prices = json_decode($result['prices']);
        ?>
            <br>
            <?
                for ($i = 0; $i < count($items); $i++)
                {
                    echo $amounts[$i] . "x " . $items[$i] . " - " . $prices[$i] . "<br>";
                }
            ?>

    </body>
</html>

I've removed code that isn't relevant to this array, if you can help I'd be forever grateful.

</div>
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2条回答 默认 最新

  • dpr77335 2016-07-28 13:35
    关注

    no need of encode and decode.

    $itemQuery = mysql_query("SELECT * FROM sales WHERE invoicenum = '$netrev' AND tabname = '$name'");
    
    while($row = mysql_fetch_array($itemQuery))
    {
        echo $row['amounts'] . "x " . $row['items'] . " - " . $row['prices'] . "<br>";
    }
     ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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