在php中列出页面重定向问题

Question

我正在使用 php，我已经从数据库中获取了所有数据，并且我必须显示列出页面，我必须在列表页面中显示所有动态日期并且我正在从数据库中获取所有数据但是如何重定向和显示列表页面，我如何尝试这个我没有得到请检查我的代码并帮助我。

这是我的 HTML 代码：

<div class="col-lg-3 col-md-6  test" data-idtest="Diabetes">
                    <a href="javascript:void(0);" class="box_cat_home">
                        <i class="icon-info-4"></i>
                        <img src="assets/img/icon_cat_3.svg" width="60" height="60" alt="">
                        <h3>Diabetes</h3>
                        <ul class="clearfix">
                        <li><strong>124</strong>Doctors</li>
                        <!-- <li><strong>60</strong>Clinics</li> -->
                        </ul>
                    </a>
                </div>

我的 Jquery 代码：

<script type="text/javascript">
 $(document).ready(function() { 
        $('.test').click(function (){
            var id = $(this).data('idtest') ; 
            alert(id);
        $.ajax({
        type: "POST",
         url : "speciallist.php",
        data    : {id:id },
        cache: false,
        success: function(data){
           console.log(data);
        }
      });
      })
    });
</script>

这是我的 PHP 代码：

<?php
$dbHost = 'localhost';
$dbUsername = 'root';
$dbPassword = 'mysql';
$dbName = 'fre';
$db = new mysqli($dbHost,$dbUsername,$dbPassword,$dbName);

$id =$_REQUEST['id'];

      $men ="select * from tbl_users where doctor_speciality='$id'";
      $men_result=$db->query($men);
      $projects=array();
      while($row=mysqli_fetch_assoc($men_result)){ 
      $projects[] = $row;
  }
?>

动态列表页面：speciallist.php

<div class="col-lg-12 list-items">
                    <?php foreach($projects as $row){
                    $userId = $row['userId'];?>
                <div class="strip_list wow fadeIn">
                    <a href="#0" class="wish_bt"></a>
                    <figure>
                        <!-- <a href="detail.php"><img src="http://via.placeholder.com/565x565.jpg" alt=""></a> -->
                        <a href="detail.php?userId=<?php echo $userId;?>"><img src="<?php echo "admin/uploads"."/".$row['doctor_image']; ?>"</a>
                    </figure>
                    <a href="detail.php?userId=<?php echo $userId;?>"><h3><?php echo $row['name']; ?></h3></a>
                    <p><!-- Doctor Education : --> <?php echo $row['doctor_education']; ?>,<?php echo $row['doctor_speciality']; ?></p>
                    <p> 9 years experience</p>
                    <p>Gynecologist/Obstetrician</p>
                    <!-- <p> Doctor Speciality : <?php echo $row['doctor_speciality']; ?></p> -->
                    <!-- <p><?php echo $row['doctor_profile']; ?></p> -->
                    <p><!--  Mobile : --> <?php echo $row['mobile']; ?></p>
                    <!-- <p><?php echo $row['doctor_profile']; ?></p> -->
                    <!-- <p>Dr. Mr is a Senior Endodontist, specializes in Smile Designing, Single Visit Root Canal Treatments, Implants, Cosmetic Dentistry.</p> -->
                    <div class="sideleft">
                        <p>Clinic Name:<?php echo $row['clinic_name']; ?></p>
                        <p>Clinic Address:<?php echo $row['clinic_address']; ?></p>
                    </div>
                    <ul>
                        <li>MON-SAT 9:00 AM - 9:30 PM</li>
                        <li><a href="detail.php?userId=<?php echo $userId;?>">Book now</a></li>
                    </ul>
                </div>
                <?php }?>
                </div>

Answer 1

将“PHP代码”添加到expert.php的顶部并将您的javascript代码更改为：

<script type="text/javascript">
  $(document).ready(function() { 
    $('.test').click(function (){
        var id = $(this).data('idtest') ; 
        alert(id);
        document.location = "speciallist.php?id=" + id;
    });
  });
</script>