Question

我需要使用oop和mysqli将值插入数据库

我的代码是

<?php 
class dbcon
{
    function __construct()
    {
        $link = mysqli_connect("localhost","root","","oop_testing") or die(mysqli_error());
    }

    function insert_table($fname,$city)
    {
        $res=mysqli_query($link,"INSERT INTO tbl_user(name,city) values ('$fname','$city')");
        return $res;
    }
}

?>

<html>
    <head>
        <title>OOP CRUD</title>
    </head>
    <body>
        <form action="" method="POST" name="form1">
            <table>
                <tr>
                    <td>USERNAME:</td>
                    <td><input type="text" name="username" /></td>
                </tr>
                <tr>
                    <td>CITY:</td>
                    <td><input type="text" name="city" /></td>
                </tr>
                <tr>
                <td colspan="2" align="center"> <input type="submit" name="submit" value="Insert" /> </td>
                </tr>
            </table>    
        </form>
        <?php 
        $con = new dbcon();

        if(isset($_POST['submit']))
        {
            $fname=$_POST['username'];
            $city=$_POST['city'];
            $con->insert_table($fname, $city);
        }
        ?>
    </body>
</html>

但它返回

注意：未定义的变量：第11行的C：\ xampp \ htdocs \ CRUD_OOP \ index.php中的链接

警告：mysqli_query（）期望参数1为mysqli，在第11行的C：\ xampp \ htdocs \ CRUD_OOP \ index.php中给出空值

有什么帮助吗？预先感谢

Answer 1

class dbcon
{
    private $link; //private by design
    function __construct()
    {
        $this->link = mysqli_connect("localhost","root","","oop_testing") or 
        die(mysqli_error());
    }

    function insert_table($fname,$city)
    {
         $res=mysqli_query($this->link,"INSERT INTO tbl_user(name,city) values ('$fname','$city')");
    return $res;
    }
}

您的$ link仅在__construct函数中可用。在对象内部使用变量

在PHP和mysqli中使用oop概念插入数据库

1 个答案: