mysqli_query（）期望参数1为mysqli，null给出

Question

/*connection file*/ 
    function connect(){ 
      $server = "localhost"; 
      $user = "xxxx"; 
      $password = "xxxx"; 
      $db = "xxxx"; 
      $connetion = mysqli_connect($server,$user,$password,$db); 
    } 

这是我的连接文件。我使用MVC连接到数据库。

/*function declaration and insert query*/ 
    function insert($table,$value){ 
     $fld = ""; 
     $val = ""; 
     $i = 0; 
     foreach ($value as $k => $v) { 
      if($i == 0){ 
       $fld .= $k; 
       $val .= "'" . $v ."'"; 
      } 
      else{ 
       $fld .= "," . $k; 
       $val .= ",'" .$v . "'"; 
      } 
      $i++; 
     } 
     global $conn; 
     return mysqli_query($conn,"INSERT INTO $table($fld) VALUES($val)") or die(mysqli_error($conn)); 
    } 

当我尝试向数据库中插入数据时发出警告。

请帮我解决这个警告。

Answer 1

您需要从函数connect返回连接对象。

function connect(){ 
     $server = "localhost"; 
     $user = "xxxx"; 
     $password = "xxxx"; 
     $db = "xxxx"; 
     $connection = mysqli_connect($server,$user,$password,$db); 
     return $connection; // return connection object 
} 

Answer 2

您需要返回$connection，这样就不会在global $connection;是不确定的：

function connect(){ 
    $server = "localhost"; 
    $user = "xxxx"; 
    $password = "xxxx"; 
    $db = "xxxx"; 
    $connection = mysqli_connect($server,$user,$password,$db); 
    return $connection; // return $connection 
} 

注意：你拼写错误$connection，不应该是$connetion。