Question

我试图根据位置从数据库中显示此表信息，但不管是什么电子邮件ID，它只显示来自lacion牛津的数据。任何人都可以让我知道我是新来的代码有什么问题，并试图获得一些expirience手提前致谢这是我的代码片段

<?php
 if (isset($_POST['submitbutton']))
        {
            $email = $_POST['email'];
            $password = $_POST['password'];
            $sql = "SELECT email,password from users WHERE email = '".$email."' AND password ='".$password."'";
            $result = $conn->query($sql);
                     if ($email = "oxford@connectone.com")
                     {
                         echo 'The email is'. $email;
                         $query = "Select * from stock WHERE location = 'oxford'";
                         $result1 = $conn->query($query);
                          if ($result1->num_rows > 0){
                              ?>
                           <table class="reports">>
                              <thead>
                              <tr> 
                              <th> Item name </th>
                              <th> Price </th>
                              <th> Quantity </th>
                              <th> Quantity Damaged </th>
                              </tr> </thead> <tbody>
                              <?php
                            while( $row = $result1->fetch_assoc() ){
                            echo
                                 "<tr>
                                     <td>{$row['item_name']}</td>
                                     <td>{$row['price']}</td>
                                     <td>{$row['quantity']}</td>
                                     <td>{$row['quantity_damaged']}</td>
                                     </tr>\n";
                          }  ?> </tbody>
                </table>  
                    <?php  
                          }

                     }
                     elseif ($email = "bridgport@connectone.com")
                     {
                         echo 'The email is'. $email;
                         $query1 = "Select * from stock WHERE location = 'bridgeport'";
                         $resut2 = $conn->query($query1);
                         if ($result2->num_rows > 0){
                               ?>
                           <table class="reports">
                              <thead>
                              <tr> 
                              <th> Item name </th>
                              <th> Price </th>
                              <th> Quantity </th>
                              <th> Quantity Damaged </th>
                              </tr> </thead> <tbody>
                              <?php
                            while( $row = $result2->fetch_assoc() ){
                            echo
                                 "<tr>
                                     <td>{$row['item_name']}</td>
                                     <td>{$row['price']}</td>
                                     <td>{$row['quantity']}</td>
                                     <td>{$row['quantity_damaged']}</td>
                                     </tr>\n";
                          }  ?>

Answer 1

您在if语句中分配值：

if ($email = "oxford@connectone.com")

分配值时，将返回赋值（"oxford@connectone.com"）。这个字符串是＆＃34; truthy＆＃34; （评估为true），以便第一个if语句运行，并忽略所有else。

您应该使用comparison operator - ==或===。

if ($email === "oxford@connectone.com") { //... }
elseif ($email === "bridgport@connectone.com") { //... }

php html数据库表冗余

1 个答案: