Php 将总数显示为0_Php_Html_Mysql

Php 将总数显示为0

php html mysql

Php 将总数显示为0,php,html,mysql,Php,Html,Mysql,我正在尝试创建一个购物车。我正试图把总数计算为输出，但它返回为0。输出应该是价格*数量，但结果是0。这些都是我已经尝试过的东西。我正在尝试创建一个购物车。我正试图把总数计算为输出，但它返回为0。输出应该是价格*数量，但结果是0。这些都是我已经尝试过的东西 <?php include 'config.php'; { if (isset($_POST["cart"])){ $field0 = $_POST["field0"]; $fie

我正在尝试创建一个购物车。我正试图把总数计算为输出，但它返回为0。输出应该是价格*数量，但结果是0。这些都是我已经尝试过的东西。我正在尝试创建一个购物车。我正试图把总数计算为输出，但它返回为0。输出应该是价格*数量，但结果是0。这些都是我已经尝试过的东西

    <?php

include 'config.php';

{
    if (isset($_POST["cart"])){

        $field0 = $_POST["field0"];
        $field1 = $_POST["field1"];
        //$field2 = $_POST["field2"];
        $field3 = $_POST["field3"];
        $field4 = $_POST["field4"];
        $field5 = $_POST["field5"];
        $or_qty = $_POST["or_qty"];
        $price = $_POST["price"];

    if (empty($field0)||empty($field1)||empty($field3)||empty($field4)||empty($field5)){
        //echo '<script>alert("Please Complete all the data")</script>';
    }else{
        $insert = mysqli_query($conn, "INSERT INTO `template` (`field0`,`field1`,`field3`,`field4`,`field5`,`or_qty`) VALUES ('$field0','$field1','$field3','$field4','$field5','$or_qty')");
        //echo '<script>alert("Order is added to cart")</script>';
    }
}
?>

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Cart</title>
</head>

<body>
    <?php
    $total = 0;
if ($field0 == 1){
    $total = $total + ($or_qty);
}
    elseif($field0 == 2){
        $total = $total + ($or_qty*$price);
    }
    elseif($field0 == 3){
        $total = $total + ($or_qty*$price);
    }
    elseif($field0 == 4){
        $total = $total + ($or_qty*$price);
    }
    echo number_format($total);
        $insert = mysqli_query($conn, "INSERT INTO `product_order` (`total`,`or_Qty`,`price`) VALUES ('$total','$or_qty','$price')");
    ?>
    <!--$total = $total + ($values["item_quantity"] * $values["item_price"]);-->

    <?php } ?>
</body>
</html>

通过给$u POST[cart]一个值并手动设置$field0、$or\u qty和$price，我可以得到$total，使之等于0以外的值。如果我删除$_POST[cart]=3；，总计始终为0

我还添加了一些逻辑，以确保只有设置了$u POST[cart]并稍微清理了代码后，才会显示总数

我认为您的问题是$u POST[cart]没有设置，因此其他变量也从未设置过

通常$\u POST值是通过表单或其他网页设置的

您对SQL注入持开放态度，应该立即解决。$price=$\u POST[price]哦，我的天哪！这可能意味着我可以以我想要的任何价格购买你们的产品。有一个列名为field0、field1、field3等的表是件奇怪的事


<?php
include 'config.php';
$_POST["cart"]=3;
    if (isset($_POST["cart"])){

        $field0 = 3;
        $field1 = $_POST["field1"];
        //$field2 = $_POST["field2"];
        $field3 = $_POST["field3"];
        $field4 = $_POST["field4"];
        $field5 = $_POST["field5"];
        $or_qty = 4;
        $price = 0.50;

    if (empty($field0)||empty($field1)||empty($field3)||empty($field4)||empty($field5)){
        //echo '<script>alert("Please Complete all the data")</script>';
    }else{
        $insert = mysqli_query($conn, "INSERT INTO `template` (`field0`,`field1`,`field3`,`field4`,`field5`,`or_qty`) VALUES ('$field0','$field1','$field3','$field4','$field5','$or_qty')");
        //echo '<script>alert("Order is added to cart")</script>';
    }
}
?>

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Cart</title>
</head>

<body>
    <?php
    if (isset($_POST["cart"])){
        $total = 0;
        if ($field0 == 1){
            $total = $total + ($or_qty);
        }
        elseif($field0 == 2){
            $total = $total + ($or_qty*$price);
        }
        elseif($field0 == 3){
            $total = $total + ($or_qty*$price);
        }
        elseif($field0 == 4){
            $total = $total + ($or_qty*$price);
        }
        echo number_format($total);
        $insert = mysqli_query($conn, "INSERT INTO `product_order` (`total`,`or_Qty`,`price`) VALUES ('$total','$or_qty','$price')");
    }
    ?>
    <!--$total = $total + ($values["item_quantity"] * $values["item_price"]);-->
</body>
</html>