包含在另一个文件中时,连接变量为null! [英] Connection variable null when including in another file!

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问题描述

大家好,我是一名PHP新手,但对其他编程语言有一定的经验。我试图使用不同文件中定义的连接设置将简单值输入数据库:





//DBConnect.php

Hello everyone, i am a PHP newbie but have some experience with other programming languages. I am trying to enter simple values into the database using the connection settings defined in a different file:


//DBConnect.php

$serverName="localhost";
$userName="root";
$passWord="";
$db="inventorymgmtdb";


$conn=mysqli_connect($serverName,$userName,$passWord,$db);

if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  
  exit();
  }





我已将此文件包含在我的entercategory.php文件中:





I have included this file in my entercategory.php file:

<?php
include("DBConnect.php");

function insertCategory($categoryName,$categoryDescription)
{
	global $conn;

		$insertQuery="insert into tbl_categories(CategoryName,CategoryDescription) 				values('$categoryName','$categoryDescription')";
		
		
		$retVal=mysqli_query($conn,$insertQuery);
		
		if(!$retVal)
			die('Error entering to database'.mysqli_error($conn));
		
		echo "Entered Successfully!";
			
		mysqli_close($conn);

}
?>





现在,系统抱怨$ conn为null(mysqli_query()期望参数1为mysqli,给出null ...)。



我在这里做错了什么?有人可以解释一下吗?



编辑:该函数调用如下:





Now, the system is complaining that the $conn is null ("mysqli_query() expects parameter 1 to be mysqli,null is given...").

What am i doing wrong here? Can someone please explain?

The function is called as follows:

if(isset($_POST["btnCreateCategory"]))
{
     insertCategory($_POST["txtCategoryName"],$_POST["txtCategoryDescription"]);

}





编辑:

我将DBConnect.php的代码直接放在entercategory.php中。 $ conn变量不为null并且工作得很好。但是,我无法理解为什么当代码放在单独的文件(DBConnect.php)中时,$ conn变量变为null。有人可以帮忙吗?




I have put the code of DBConnect.php directly in entercategory.php. The $conn variable is not null and working just fine. However, i am not able to understand why, when the code in placed in a separate file (DBConnect.php), the $conn variable becomes null. Can someone please assist?

推荐答案

serverName = localhost ;
serverName="localhost";


userName = root ;
userName="root";


passWord = ;
passWord="";


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