php - 如何从选项获取输入，而我必须将数据输入到同一数据库的两个表中，基于选项数据将被插入

Question

我在数据库中有两个表，我必须通过表单输入数据，但条件是 - 表单中有一个选择选项，并且作为我们必须插入数据的选项。如果选择选项 1，则数据应插入到表 x如果选择选项 2，则数据应插入到表 y

我尝试过使用 switch case 和 elseif但不能 找不到合适的条件。 html代码-

        <form method="post" class="register-form" id="register-form">
                                    <div class="form-group">
                                        <label for="name"><i class="zmdi zmdi-account material-icons-name"></i></label>
                                        <input type="text" name="aname" id="name" placeholder="Your Name"/>
                                    </div>

                                    <div class="form-group">
                                        <label for="email"><i class="zmdi zmdi-email"></i></label>
                                        <input type="email" name="email" id="email" placeholder="Your Email"/>
                                    </div>

                                    <div class="form-group">
                                        <label for="address"><i class="zmdi zmdi-home"></i></label>
                                        <input type="text" name="address" id="address" placeholder="Your Address"/>
                                    </div>
                                     <div class="form-group">
                                        <label for="user"><i class="zmdi zmdi-accounts-alt"></i></label>
                                        <select  id="person-type" name="logtype" required style="border-top: rgba(0,0,0,0.0);border-right: rgba(0,0,0,0.0);border-left: rgba(0,0,0,0.0); width: 100%;margin-left: 22px;">
                                            <option disabled selected  value="">Select your login type</option>
                                            <option value="consumer" name="consumer">Consumer</option>
                                            <option value="electrician" name="electrician">Electrician</option>
                                        </select>
                                    </div>

                                    <div class="form-group">
                                        <label for="pass"><i class="zmdi zmdi-lock"></i></label>
                                        <input type="password" name="apass" id="password" pattern=".{6,}" required placeholder="Password "  oninvalid="this.setCustomValidity('Password must be 6 digit long or more')" oninput="this.setCustomValidity('')"  />
                                    </div>
                                    <div class="form-group">
                                        <label for="re-pass"><i class="zmdi zmdi-lock-outline"></i></label>
                                        <input type="password" name="re_pass" id="confirm_password"required="@" minlength="8" placeholder="Repeat your password"/>
                                    </div>
                                    <div class="form-group">
                                        <input type="checkbox" name="agree-term" id="agree-term" class="agree-term"  required/>
                                        <label for="agree-term" class="label-agree-term"><span><span></span></span>I agree all statements in  <a href="#" class="term-service">Terms of service</a></label>
                                    </div>
                                    <div class="form-group form-button">
                                        <input type="submit" name="signup" id="signup" class="form-submit" value="Register"/>
                                    </div>
                                </form>

php 代码-

if (isset($_POST['signup'])) 
{
    if ($_POST['logtype'] == 'consumer') 
    {
        $consumer=mysqli_query($edb,"insert into consumer(name,email,address,logtype,password) values('".$_POST['aname']."','".$_POST['email']."','".$_POST['address']."','".$_POST['logtype']."',md5.
                            ('".$_POST['apass']."'))");    
                                header("Location:loginonly.php");
                }
                elseif ( $_POST['logtype'] == 'electrician' ) 
                {

                            $electrician=mysqli_query($edb,"insert into electrician(name,email,address,logtype,password) values('".$_POST['aname']."','".$_POST['email']."','".$_POST['address']."','".$_POST['logtype']."',md5.
                                ('".$_POST['apass']."'))");    
                                header("Location:loginonly.php");
                }

没有错误，但数据未插入表中

Answer 1

您应该在查询执行中添加一些错误处理，以帮助查找发生的情况。

PHP 中的基本 mysqli 错误处理类似于:

<?php
 if (!mysqli_query($edb,"YOUR QUERY HERE"))
 {
  echo("Error description: " . mysqli_error($con));
 }
?>