Question

function isValidUser($dbc,$email)
    {

        $sql=sprintf("select id from user where email = '%s' and password ='%s' ", $_POST['email'],$_POST['password'] ); 
        $qresult = mysqli_query($dbc,$sql);//line 18
        $test = mysqli_fetch_array($qresult);//line 19
        echo $test[0];
        if(mysqli_num_rows($qresult)>0)//line21
        {
            echo "User Exists";
            return true;
        }
        else
        {
            echo "Not a User";
            return false;
        }
    }

我想也许我有一些范围问题（根据类似问题的其他stackoverflow响应）。但我似乎不明白。我是php的新手。可以有人告诉我如何纠正这个。？

 mysqli_query() expects parameter 1 to be mysqli, object given in... line 18
 mysqli_fetch_array()  expects parameter 1 to be mysqli, null given in line 19
 mysqli_num_rows()  expects parameter 1 to be mysqli, null given in line 21

Answer 1

试试这个：

function isValidUser($dbc,$email)
    {
        $email = $_POST['email'];
        $host = "";
        $username = "";
        $password_mysql = "";
        $database ="";

        $password  = $_POST['password'];
        $link = mysqli_connect($host, $username, $password_mysql, $database); /* hard code the connection variables in to test if something is wrong here. */
        $sql="select id from user where email = '".$email."' and password ='".$password."'"; 
        $qresult = mysqli_query($link,$sql);//line 18
        $test = mysqli_fetch_array($qresult, MYSQLI_BOTH);//line 19
                while($test = mysqli_fetch_array($qresult, MYSQLI_BOTH)){    
                echo  $test['id']."<br>";
                };
        if(mysqli_num_rows($qresult)>0)//line21
        {
            echo "User Exists";
            return true;
        }
        else
        {
            echo "Not a User";
            return false;
        }
    }

mysqli_query（）期望参数1是mysqli，对象是给定的

1 个答案: