Php 我的sql查询是对的还是错的？它向我显示类似警告的错误：mysqli_num_rows（）期望参数1是mysqli_result，布尔值

我的问题是

$statement = "SELECT * FROM profile_details
       WHERE    YEAR(CURDATE())-YEAR(dob) BETWEEN '$search_age'
                                              AND '$search_age1'
            AND gender LIKE '$search_gender'
        OR main_caste LIKE '$search_relegion'
        OR education_type LIKE '$search_qualification'
        OR occupation LIKE '$search_occupation'";

当我回显mysqli_错误（$con）时它正在显示消息

You have an error in your SQL syntax; check the manual that corresponds
         to your MariaDB server version for the right syntax
         to use near
     'SELECT * FROM profile_details
      WHERE YEAR(CURDATE())-YEAR(dob) BETWEEN '20' AND '' at line 1

下面给出了我的连接文件

<?php

/*Default time zone ,to be able to send mail */
date_default_timezone_set('Asia/Kolkata');

//connect database
$con = mysqli_connect ("localhost","root","root","testdb") or die ("Unable to connect. Check your connection parameters.");         //host, username, password, database name

        //database connect error
        if (mysqli_connect_errno())
        {   
        echo "Failed to connect to MySql: " . mysqli_connect_errno();
        }

?>

搜索结果php文件为。。我正在对搜索结果使用分页。。。在这一页上，我得到了一个错误：您的SQL语法有一个错误；查看与您的MariaDB服务器版本相对应的手册，以了解在“从概要文件中选择*详细信息”附近使用的正确语法，其中

    <?php 
                date_default_timezone_set('Asia/Kolkata');
                require_once("includes/config.php");
                require_once("includes/user_pagination.php");
                global $con;
                //if(isset($_POST['home_quick_search'])){

                //get serch data from index.php
                    $search_gender = $_GET['s_gender']; 
                    $search_relegion = $_GET['s_religion']; 
                    $search_occupation = $_GET['s_occupation']; 
                    $search_age = $_GET['s_age']; 
                    $search_age1 = $_GET['s_age1']; 
                    $search_qualification = $_GET['s_qualification']; 

                $page = (int)(!isset($_GET["page"]) ? 1 : $_GET["page"]);
                if ($page <= 0) $page = 1;

                $per_page = 10; // Set how many records do you want to display per page.

                $startpoint = ($page * $per_page) - $per_page;

                $statement = "`profile_details` WHERE `DATEDIFF(YEAR,dob,CURDATE())` AS dob BETWEEN '$search_age' AND '$search_age1' AND `gender` LIKE '$search_gender' OR `main_caste` LIKE '$search_relegion' OR `education_type` LIKE '$search_qualification' OR `occupation` LIKE '$search_occupation' ORDER BY `user_id` ASC"; 

//$statement = "SELECT * FROM profile_details WHERE YEAR(CURDATE())-YEAR(dob) BETWEEN '$search_age' AND '$search_age1' AND gender LIKE '$search_gender' AND main_caste LIKE '$search_relegion' AND education_type LIKE '$search_qualification' AND occupation LIKE '$search_occupation'";



                $results = mysqli_query($con,"SELECT * FROM {$statement} LIMIT {$startpoint} , {$per_page}") or die (mysqli_error($con));

                    if (mysqli_num_rows($results) != 0) {

                    // displaying records.
                    $i = 0; 
                        while ($row_user = mysqli_fetch_array($results)) {

                        $u_id = $row_user['user_id'];
                        $u_pid = $row_user['profile_id'];
                        $u_gender = $row_user['gender'];
                        $u_fname = $row_user['first_name'];
                        $u_relegion = $row_user['main_caste'];
                        $u_city = $row_user['city'];
                        $u_image = $row_user['photo'];
                        $u_dob = $row_user['dob'];

                        //age calculation
                        $dateOfBirth = $u_dob;
                        $today = date("Y-m-d");
                        $diff = date_diff(date_create($dateOfBirth), date_create($today));
                        //echo 'Age is '.$diff->format('%y');
                        $age = $diff->format('%y');
                        //age calculation ends

                        $u_status_all = $row_user['user_status'];
                        $u_status_of = $row_user['user_status']=='OFFLINE';
                        $u_status_on = $row_user['user_status']=='ONLINE';


                     echo'
                          <div class="col-sm-6 paid_people-left">
                            <ul class="profile_item">
                              <a href="view_profile.php?userdetail_id=$u_id">

                               <li class="profile_item-img">';

                              if (!empty($u_image)){
                                  echo '<img src="users-photo/resized_'.$u_image.'" class="img-responsive" alt="'.$u_fname.'"/>';
                              }
                              else{
                                  echo '<img src="images/s2.jpg" class="img-responsive"  alt=""/>';
                              }



                            echo ' </li>   
                                <li class="profile_item-desc">
                                  <h4>Profile ID: '.$u_pid.'</h4>
                                  <p>Age: '.$age.' Yrs, '.$u_gender.'</p>
                                  <h5>View Full Profile</h5>
                                  <p>';

    if (!isset($u_status_on) && $u_status_of){
                            echo "Nope";
                        }
                        elseif (isset ($u_status_of) && $u_status_on){
                            echo "<img src='images/online.png' /> <span class='label label-success'>"; 
                        }
                        else {  
                        if (isset ($u_status_of)){
                            echo "<img src='images/offline.png' /> <span class='label label-default'>"; 
                            }   
                        }

                        echo "
                            $u_status_all</span><p>
                            </li>
                            <div class='clearfix'> </div>
                            </a>
                            </ul>
                        </div>";                       

                    }

                    } else {
                                            echo "Sorry ! No Profiles are found.";
                                    }

            //  }   

            ?>

               <div class="pagination-div">
                <ul class="pagination">
                    <?php echo pagination($statement,$per_page,$page,$url='?'); ?>
                </ul>
           </div> 

几个注意事项：

和
运算符的优先级高于
或
运算符。添加paren可以让未来的读者更清楚地了解您的意图。您是否打算仅在性别匹配时检查年龄，否则忽略年龄条件

年龄的计算似乎有点奇怪，因为它完全忽略了月份和日期。标准模式是将裸日期列与返回日期值的表达式进行比较

 WHERE ( t.dob >  DATE(NOW()) - INTERVAL 25 YEAR AND
         t.dob <= DATE(NOW()) - INTERVAL 19 YEAR
       )

返回

b           e           
----------  ----------
1992-02-10  1998-02-10  

为了调试SQL问题，echo或var_在提交SQL文本以供执行之前转储该文本的值

然后检查SQL


问题中显示的代码模式，包括SQL文本中的值，是我们在易受SQL注入攻击的代码中经常看到的模式。我们无法判断此代码是否易受攻击，因为我们无法判断这些值是否已正确转义

SELECT ...
    FROM ( SELECT ... ) AS x  -- You need parens around derived query

a AND b OR c
is treated as
( a AND b ) OR c

请在布尔表达式中添加paren，让我们相信您做得对。
好吧，执行echo$语句并打印结果。它就在那里！语法错误。检查您的SQL查询，它显然是错误的。@Dimi谢谢。。。我正在尝试将出生日期转换为年龄格式…如何转换…TIMESTAMPDIFF（YEAR，dob，CURDATE（））为age@Dimi谢谢我在上面试过，但它显示出同样的错误。。。。我的
dob
类型是
date
。。。。。。。。
SELECT ...
    FROM ( SELECT ... ) AS x  -- You need parens around derived query

a AND b OR c
is treated as
( a AND b ) OR c