在php中传递onclick Jquery变量问题

Question

我正在开发 Core php，我正在点击 div 数据即将到来，我存储了一个变量，我想通过数据库查询检查它是否匹配将出现所有记录如何做到这一点请帮助任何人一个。

HTML代码：

<div class="col-lg-3 col-md-6  test" data-idtest="Diabetes">
    <a href="" 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> 

jQuery 代码：

<script type="text/javascript">
    $(document).ready(function() { 
        $('.test').click(function (){
            var id = $(this).data('idtest') ; 
            //alert(id);
    $.ajax({
         url : "getting.php",
         type: 'POST',
         data    : {id:id },
          }).done(function(response) {
        });
      })
    });
</script>

php代码：

<?php
echo "ejejejej";
$dbHost = 'localhost';
$dbUsername = 'root';
$dbPassword = 'mysql';
$dbName = 'fre';

$id =$id;
$db = new mysqli($dbHost,$dbUsername,$dbPassword,$dbName);

      $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;

    }
?>

Answer 1

使用此代码来获得 ajax 成功的结果。

<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> 


<?php

$dbHost = 'localhost';
$dbUsername = 'root';
$dbPassword = 'mysql';
$dbName = 'fre';

$id =$_REQUEST['id'];
$db = new mysqli($dbHost,$dbUsername,$dbPassword,$dbName);

      $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;

    }
    if($projects){
       $result=array(
       "responseCode" =>1,
       "message" =>"Data found",
       "data" => $projects
       );
    }else{
       $result=array(
       "responseCode" =>0,
       "message" =>"No data found regarding this id"
       );
    }
echo json_encode($result);die();
?>
<script type="text/javascript">
    $(document).ready(function() { 
        $('.test').click(function (){
            var id = $(this).data('idtest') ; 
            //alert(id);
        $.ajax({
        type: "POST",
         url : "getting.php",
        data    : {id:id },
        cache: false,
        success: function(data){
           console.log(data);
        }
      });
      })
    });
</script>

Answer 2

您应该$_POST 来获取 PHP 中发布的参数

使用

$id =  $_POST["id"] 

另外，使用PreparedStatement 来避免SQL Injection

$db = new mysqli($dbHost,$dbUsername,$dbPassword,$dbName);
$stmt = $db->prepare("select * from tbl_users where doctor_speciality = ?");
$stmt->bind_param('s', $id));
$stmt->execute();