关闭PHP中包含的文件

Question

我正在尝试通过将pro_id传递给方法来更改MySQL查询中的pro_id，它正在打印表格。

注意：第2行上的C：\\ WAMP64 \\ WWW \\ DASHBOARD \\ CONFIG2.PHP中已经定义了恒定服务器
注意：常量用户名已经在C：\\ WAMP64 \\ WWW \\ DASHBOARD \\ CONFIG2.PHP在线3中定义
注意：在第4行的C：\\ WAMP64 \\ WWW \\ DASHBOARD \\ CONFIG2.PHP中已经定义了恒定密码
注意：常量数据库已经在第5行的C：\\ WAMP64 \\ WWW \\ DASHBOARD \\ CONFIG2.PHP中定义

这些错误。

我的代码是这样的，

<?php
// Defining function
function project($a){

            echo '<DOCTYPE html>
            <head>
                <link rel="stylesheet" type="text/css" href="http://localhost/Dashboard/css/table_css.css">

            </head>
            <section>
            <body>';

         include("C:\wamp64\www\DashBoard\Config2.php");

          $sql = "select total_tc,passed,failed,blocked, (total_tc-passed-failed-blocked) as notrun from (select * from (select version_id as version_id from testcases) as t0, (select count(tc_id) as total_tc from testcases where testcases.pro_id=$a) as t,(select count(tc_result) as passed from executions join testcases on testcases.id=executions.parent_id where tc_result='p' and testcases.pro_id=$a  )as t2,
                        (select count(tc_result) as failed from executions join testcases on testcases.id=executions.parent_id where tc_result='f' and testcases.pro_id=$a)as t3,(select count(tc_result) as blocked from executions join testcases on testcases.id=executions.parent_id where tc_result='b' and testcases.pro_id=$a)as t4) as final where version_id= 1 limit 1;";

          $query = mysqli_query($db, $sql);
          echo'<div class="tbl-header">
                <table cellpadding="0" cellspacing="0" border="0">
                    <thead>
                        <tr>
                         <th>Total Testcase</th>
                         <th>Passed</th>
                         <th>Failed</th>
                         <th>Blocked</th>
                         <th>Not run</th>
                      </tr>
                    </thead>
                </table>
          </div>
       <div class="tbl-content">
         <table cellpadding="0" cellspacing="0" border="0">';

             while ($row = mysqli_fetch_array($query))
               {
               echo '<tr>

                <td>'.$row['total_tc'].'</td>
                <td>'.$row['passed'].'</td>
                <td>'.$row['failed'].'</td>
                <td>'.$row['blocked'].'</td>
                <td>'.$row['notrun'].'</td>

              </tr>';
             }
             echo'
        </tbody>
      </table>
    </div>
  </section>
</html>';
}
// Calling function
for ($x = 1; $x <8; $x++) {
    project($x);
} 
?>

并且Config2.php文件包含

<?php
   define('SERVER', '192.168.0.7:3306');
   define('USERNAME', 'Deepak');
   define('PASSWORD', 'test@123');
   define('DATABASE', 'dashtest');
   $db = mysqli_connect(SERVER,USERNAME,PASSWORD,DATABASE);
?>

谁能帮我解决这个问题？

Answer 1

您将在函数中包含带有常量的文件：

     include("C:\wamp64\www\DashBoard\Config2.php");

因此，第二次调用函数时，您将收到这些警告。

将文件包含在函数外部（并在需要时将数据库连接作为参数传递）或使用include_once 。

关闭PHP中包含的文件

问题描述

1 个解决方案

解决方案1
3 2017-11-14 13:55:42

关闭PHP中包含的文件

问题描述

1 个解决方案

解决方案1 3 2017-11-14 13:55:42

解决方案1
3 2017-11-14 13:55:42