Question

我还在学习PHP，而我正试图解决这个错误。根据{{3}}我的代码是正确的，但这是我的代码，这是我收到的错误：

$con = mysqli_connect("IP","username","passowrd","dbname");
// Check connection
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

function get_demos(){
    $result = mysqli_query($con,"SELECT * FROM demos");
    if(!$result)
    {
        die("Invalid query ".mysqli_error($con));
    }
    return $result;
}

get_demos();

警告：mysqli_query（）要求参数1为mysqli，在第10行的/home/content/83/11483383/html/php/db.php中给出null

警告：mysqli_error（）期望参数1为mysqli，null给定   在/home/content/83/11483383/html/php/db.php第13行无效的查询

我做错了什么？

感谢。

Answer 1

你应该尝试，

$con = mysqli_connect("IP","username","passowrd","dbname");
// Check connection
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

function get_demos($con){
    $result = mysqli_query($con,"SELECT * FROM demos");
    if(!$result)
    {
        die("Invalid query ".mysqli_error($con));
    }
    return $result;
}

get_demos($con);

Answer 2

您没有声明已设置变量$con。

试试这个

$con = mysqli_connect('localhost', 'root', '');
    mysqli_select_db($con, 'databse_name_here') or die ('Failed to connect to MySQL: ' . mysqli_connect_error()); 


function get_demos($con){
    $result = mysqli_query($con,"SELECT * FROM users");
    if(!$result)
    {
        die("Invalid query ".mysqli_error($con));
    }
    return $result;
}

get_demos($con);

Answer 3

您必须将连接传递给您的函数。如果你不想每次都使用单例模式来做它，那么它总是在范围内。

class DBCon {

private static $_instance = null;

 private function __construct() {
    $_instance = mysqli_connect("IP","username","passowrd","dbname");
    // Check connection
    if (mysqli_connect_errno()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
 }

 public static function get() {
   if(is_null(self::$_instance)) {
      self::$_instance = new DBCon();
   }
   return self::$_instance;
 }
}

并在您的代码中使用它：

function get_demos(){
    $result = mysqli_query(DBCon::get(),"SELECT * FROM demos");
    if(!$result)
    {
        die("Invalid query ".mysqli_error(DBCon::get()));
    }
    return $result;
}

Answer 4

在提出问题后我找到了解决方案：

$con = mysqli_connect("IP","username","passowrd","dbname");
// Check connection
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

function get_demos(){
    global $con;
    $result = mysqli_query($con,"SELECT * FROM demos");
    if(!$result)
    {
        die("Invalid query ".mysqli_error($con));
    }
    return $result;
}

get_demos();

添加全球$ con。

mysqli的PHP变量范围问题

4 个答案: