duanlan5320 2018-06-11 09:20
浏览 23

重置计算结果

I have a program that calculates a number from the database with the input of the user. I made it work fine, but my problem now is when I reset the page, the result is still there. Even if I use unset in the variable of the result, it still stays there. Here is my code:

<?php
  include 'dbconnect.php'
?>
<html>
<head>
</head>
<body>
  <br>
  <br>
  <br>
  <div align="center">
    <form method="POST">
        <?php
            unset($res);
            $fetch = "SELECT Valor FROM taxas WHERE Id = 5";
            $send = mysqli_query($con, $fetch);
            while ($row = mysqli_fetch_array($send)) {
                $value = $row['Valor'];
            }
            echo $value;
            if (isset($_POST['op'])) {
                $num1 = $_POST['n1'];
            }
            $res = $value + $num1;
        ?>
        *<input type="text" name="n1"> 
        <button name="op"> = </button>
        <?php
            echo $res;
        ?>
    </form>
</div>
</body>
</html>

if you need, code for the "dbconnect.php":

<?php
$place = "localhost";
$user = "root";
$pass = "";
$database = "teste2";
$con = mysqli_connect ($place, $user, $pass, $database);
if ($con->connect_error) {
    die("Error: " . $con->connect_error);
}
  • 写回答

2条回答 默认 最新

  • dongle7553 2018-06-11 09:28
    关注

    Try to use

    $res = '';
    $fetch = "SELECT Valor FROM taxas WHERE Id = 5";
    $send = mysqli_query($con, $fetch);
    while ($row = mysqli_fetch_array($send)) {
        $value = $row['Valor'];
    }
    echo $value;
    if(isset($_POST['op'])) {
        $num1 = $_POST['n1'];
    }
    $res = $value + $num1;
    

    maybe helps you

    评论

报告相同问题?

悬赏问题

  • ¥100 支付宝网页转账系统不识别账号
  • ¥15 基于单片机的靶位控制系统
  • ¥15 AT89C51控制8位八段数码管显示时钟。
  • ¥15 真我手机蓝牙传输进度消息被关闭了,怎么打开?(关键词-消息通知)
  • ¥15 下图接收小电路,谁知道原理
  • ¥15 装 pytorch 的时候出了好多问题,遇到这种情况怎么处理?
  • ¥20 IOS游览器某宝手机网页版自动立即购买JavaScript脚本
  • ¥15 手机接入宽带网线,如何释放宽带全部速度
  • ¥30 关于#r语言#的问题:如何对R语言中mfgarch包中构建的garch-midas模型进行样本内长期波动率预测和样本外长期波动率预测
  • ¥15 ETLCloud 处理json多层级问题