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));
    }

当我尝试将数据插入数据库时会发出警告。

请帮我解决此警告。

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，以便global $connection;中不会定义它：

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

注意：您错误地拼写了$connection，不应该是$connetion。

mysqli_query（）期望参数1为mysqli，给定null

2 个答案: