首页 > 解决方案 > 简单的表单输入类型提交和重置不起作用

问题描述

我一直在尝试来自互联网上不同线程的所有不同想法,但我似乎找不到任何想法为什么我的表单不会提交或重置我对数据所做的更改。该代码可以从数据库中提取数据,但是当我进行更改时,提交和重置按钮什么也不做。

<?php
session_start();
//check session first
if (!isset($_SESSION['employee_id'])){
    echo "You are not logged in!";
    exit();
}else{
    //include the header
    include ("../includes/managerheader.php");
    require_once ('../../mysqli_connect.php');
    $customer_id=$_GET['customer_id']; 
    $query = "SELECT * FROM customers WHERE customer_id=$customer_id"; 
    $result = @mysqli_query ($dbc, $query);
    $num = mysqli_num_rows($result);
    if ($num > 0) { // If it ran OK, display all the records.
        while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
?>
            <form action="customerupdate2.php" method="post">
            Customer ID: <input name="customer_id" size=50 value="<? echo $row['customer_id']; ?>"></br></br>
            First Name: <input name="first_name" size=50 value="<? echo $row['first_name']; ?>"><p>
            Last Name: <input name="last_name" size=50 value="<? echo $row['last_name']; ?>"><p>
            Email Address: <input name="email" size=50 value="<? echo $row['email']; ?>"><p>
            Password: <input name="password" size=50 value="<? echo $row['password']; ?>"><p>
            Address: <input name="address" size=50 value="<? echo $row['address']; ?>"><p>
            City: <input name="city" size=50 value="<? echo $row['city']; ?>"><p>
            State: <input name="state" size=50 value="<? echo $row['state']; ?>"><p>
            Zip: <input name="zip" size=50 value="<? echo $row['zip']; ?>"><p>
            Telephone: <input name="telephone" size=50 value="<? echo $row['telephone']; ?>"><p>
            Payment Type: <input name="payment_type" size=50 value="<? echo $row['payment_type']; ?>"><p>
            Card Number: <input name="card_number" size=50 value="<? echo $row['card_number']; ?>"><p>
            <input type="submit" value="update">
            <input type="reset" value="reset">
            <input type="hidden" name="customer_id" value="<? echo $row['customer_id']; ?>">
            </form>
<?
        } //end while statement
    } //end if statement
    mysqli_close($dbc);
    //include the footer
    include ("../includes/footer.php");
}
?>

表单的操作与此相关:

<?php
session_start();
//check the session
if (!isset($_SESSION['manager'])){
    echo "You are not logged in!";
    exit();
}else{
    //include the header
    include ("../includes/managerheader.php");
    require_once ('../../mysqli_connect.php');
    #execute UPDATE statement
    $first_name = mysqli_real_escape_string($dbc, $_POST['first_name']);
    $last_name = mysqli_real_escape_string($dbc, $_POST['last_name']);
    $email = mysqli_real_escape_string($dbc, $_POST['email']);
    $password = mysqli_real_escape_string($dbc, $_POST['password']);
    $address = mysqli_real_escape_string($dbc, $_POST['address']);
    $city = mysqli_real_escape_string($dbc, $_POST['city']);
    $state = mysqli_real_escape_string($dbc, $_POST['state']);
    $zip = mysqli_real_escape_string($dbc, $_POST['zip']);
    $telephone = mysqli_real_escape_string($dbc, $_POST['telephone']);
    $payment_type = mysqli_real_escape_string($dbc, $_POST['payment_type']);
    $card_number = mysqli_real_escape_string($dbc, $_POST['card_number']);  


    $query = "UPDATE employees SET employe_id='$employe_id',first_name='$first_name',last_name='$last_name',email='$email',
    password='$password',address='$address',city='$city',state='$state',zip='$zip',
    telephone='$telephone',payment_type='$payment_type',card_number='$card_number', WHERE id='$id'"; 
    $result = @mysqli_query ($dbc, $query); 
    if ($result){
        echo "<center><p><b>The selected customer has been updated.</b></p>"; 
        echo "<a href=managerloggedin.php>Home</a></center>"; 
    }else {
        echo "<p>The customer could not be updated due to a system error" . mysqli_error() . "</p>"; 
    }
    mysqli_close($dbc);
    //include the footer
    include ("../includes/footer.php");
}

?>

标签: phpmysqlhtml

解决方案


推荐阅读