Php mysql\u数组中出现错误

我在运行我的

视图\u reservation.php

时遇到了这个问题。它没有在数据库中查看我的以下记录。 下面是

view\u reservation.php

错误是第89行。是“WHILE”所在的位置：

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\FINAL_THESIS\admin\view_reservation.php on line 89

守则：

<?php 

include('dbcon.php');

$id = $_GET['reservation_id'];



$customer_query=mysql_query("SELECT reservation.res_type                                            ,reservation.type                                           ,reservation.fname                                          ,reservation.mname                                          ,reservation.lname                                          ,reservation.age                                            ,reservation.address                                            ,reservation.wfname                                         ,reservation.wmname                                         ,reservation.wlname                                         ,reservation.sched_time                                         ,reservation.sched_month                                            ,reservation.sched_date                                         ,reservation.sched_year                                         ,reservation.cfname                                         ,reservation.pd_name                                            ,reservation.bd_month                                           ,reservation.bd_date                                            ,reservation.bd_year                                            ,reservation.wbd_month                                          ,reservation.wbd_date                                           ,reservation.wbd_year                                           ,reservation.w_age                                          ,reservation.w_address                                          ,reservation.status                                         ,customer.firstname                                         ,customer.middlename                                            ,customer.lastname                                          ,customer.email                                         ,customer.age                                           ,customer.gender                                            ,customer.barangay                                          ,customer.com_address                                           ,reservation.res_id FROM customer Inner Join reservation 
ON customer.cus_id = reservation.cus_id where res_id = '$id'");

    while($cust_rows=mysql_fetch_array($customer_query)) {
        if($cust_rows['gender'] == 'Male') {    
            $gender ='Mr.';
        } else {
            $gender ='Ms.';
        }

        $id=$cust_rows['cus_id']; 
        $fn=$cust_rows['firstname']." ".$cust_rows['middlename']." ".$cust_rows['lastname'];
        $email=$cust_rows['email'];
        $age=$cust_rows['age'];
        $brgy=$cust_rows['barangay'];
        $comp=$cust_rows['com_address'];

我将您的代码复制到，它打印了一个关于您的第一行的错误。我删除了不必要的空格，现在语法正常了。请尝试以下代码：

<?php 

include('dbcon.php');

$id = (int)$_GET['reservation_id'];



$customer_query=mysql_query("SELECT reservation.res_type, reservation.type, reservation.fname, reservation.mname, reservation.lname, reservation.age, reservation.address, reservation.wfname, reservation.wmname, reservation.wlname, reservation.sched_time, reservation.sched_month, reservation.sched_date, reservation.sched_year, reservation.cfname, reservation.pd_name, reservation.bd_month, reservation.bd_date, reservation.bd_year, reservation.wbd_month, reservation.wbd_date, reservation.wbd_year, reservation.w_age, reservation.w_address,reservation.status, customer.firstname, customer.middlename, customer.lastname, customer.email, customer.age, customer.gender, customer.barangay, customer.com_address, reservation.res_id FROM customer INNER JOIN reservation
ON customer.cus_id = reservation.cus_id WHERE res_id ='$id'");

    while($cust_rows=mysql_fetch_array($customer_query))
    {
        if($cust_rows['gender'] == 'Male')
        {   
            $gender ='Mr.';
            }
            else
            {
            $gender ='Ms.';
            }

    $id=$cust_rows['cus_id']; 
    $fn=$cust_rows['firstname']." ".$cust_rows['middlename']." ".$cust_rows['lastname'];
    $email=$cust_rows['email'];
    $age=$cust_rows['age'];
    $brgy=$cust_rows['barangay'];
    $comp=$cust_rows['com_address'];

?>



也可以考虑从<代码> MySqL**<代码>函数移动到<代码> MySqLix*< /Cord>或./P>这是不安全的应用程序。另外，您正在使用
mysql.*
函数。不推荐。请不要贬低我的帖子。我是PHP初学者，所以我还在学习。谢谢。如果有任何关于如何解决或更改安全编码的建议。你能告诉我吗？你好，丹尼尔，还有其他编码或代码可以解决这个问题吗-这是不安全的-用户可以破坏您的数据库。您可以使用
$id=（int）$\u GET['reservation\u id']取而代之。您还应该使用
mysqli.*
函数或PDO。Mysqli编写的语句很棒。很明显，由于查询失败，您将得到一个错误。请尝试将此查询复制到例如phpMyAdmin中，并查看您得到的错误@vlzvl我同意。否决一个有问题的行动有什么意义？这个问题是否不清楚或没有表现出任何努力？在回答中指出他使用的是
mysqli
或
PDO
，而不是否决他。我只是添加了它并应用了它，然后我将使用mysqli
。谢谢，没问题。祝你好运